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Diffstat (limited to 'Fundamental_Of_Physics_by_D_Haliday')
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diff --git a/Fundamental_Of_Physics_by_D_Haliday/1-Measurement.ipynb b/Fundamental_Of_Physics_by_D_Haliday/1-Measurement.ipynb new file mode 100644 index 0000000..fa53edd --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/1-Measurement.ipynb @@ -0,0 +1,150 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Measurement" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"velocityP = 23 //rides per h\n", +"c1 = 4 //from ride to stadia\n", +"c2 = 6 //from stadia to plethra\n", +"c3 = 30.8 //from plethra to meter\n", +"c4 = 10^-3 //from meter to kilometer\n", +"c5 = 60 * 60 //from h to sec\n", +"\n", +"//Sample Problem 1-1\n", +"printf('**Sample Problem 1-1**\n')\n", +"velocityC = velocityP * c1 * c2 * c3 * c4 / c5\n", +"printf('The speed is %e km/s', velocityC)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"conv1 = 170.474 //conversion from crans to liters\n", +"conv2 = 48.26 //from covido to cm\n", +"V1 = 1255 //in crans\n", +"\n", +"//Sample Problem 1-2\n", +"printf('**Sample Problem 1-2**\n')\n", +"VC = V1 * conv1 * 10^3 / (conv2^3)\n", +"printf('The required declaration is %e cubic covidos', VC)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"//the crossection to be approximately squre\n", +"Radius = 2 //in meter\n", +"side = 4 * 10^-3 //converted from mm to meter\n", +"\n", +"//Sample Problem 1-3\n", +"printf('**Sample Problem 1-3**\n')\n", +"//making the volume equal\n", +"Length = 4/3 * %pi * Radius^3 / side^2\n", +"L_km = Length/10^3\n", +"order = round(log(L_km)/log(10)) //will give us order of magnitude\n", +"printf('The order of length of string is 10^%d km', order)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"height = 1.70 //in meter\n", +"elapsed_time = 11.1 //in sec\n", +"\n", +"//Sample Problem 1-4\n", +"printf('**Sample Problem 1-4**\n')\n", +"//the angle between the two radius is\n", +"theta = elapsed_time / (24 * 3600) * %pi*2 //in radians\n", +"//we also have d^2 = 2 * r *h\n", +"//as d is very small hence can be considered as a arc\n", +"//d = r * theta\n", +"//=> r * theta^2 = 2 * h\n", +"radius = 2 * height /theta^2\n", +"printf('The radius of earth is %e m', radius)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/10-Collisions.ipynb b/Fundamental_Of_Physics_by_D_Haliday/10-Collisions.ipynb new file mode 100644 index 0000000..acb6d68 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/10-Collisions.ipynb @@ -0,0 +1,216 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: Collisions" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"m = 140 * 10^-3 //in kg\n", +"Vi = -39 //in m/s\n", +"Vf = 39 //in m/s\n", +"\n", +"//Sample Problem 10-1a\n", +"printf('**Sample Problem 10-1a**\n')\n", +"//J = Pf - Pi\n", +"J = m *(Vf - Vi)\n", +"printf('The magnitude of impulse acted on the ball from bat is equal to %fN-s\n', J)\n", +"\n", +"//Sample Problem 10-1b\n", +"printf('\n**Sample Problem 10-1b**\n')\n", +"t = 1.20* 10^-3 //in sec\n", +"Favg = J/t\n", +"printf('The average force during the collision is %fN\n', Favg)\n", +"\n", +"//Sample Problem 10-1c\n", +"printf('\n**Sample Problem 10-1c**\n')\n", +"Vf = 45* [cos(dtor(30)), sin(dtor(30))]\n", +"Vi = [-39, 0]\n", +"J = m* (Vf - Vi)\n", +"printf('The magnitude of new inpulse is %fN-s\n', norm(J))\n", +"printf('The new impulse makes an angle of %f degress with the horizontal', rtod(atan(J(2)/ J(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 5.4 //in kg\n", +"m = 9.5* 10^-3 //in kg\n", +"g = 9.8 //in m/s^2\n", +"h = 6.3* 10^-2 //in meter\n", +"\n", +"//Sample Problem 10-2\n", +"printf('**Sample Problem 10-2**\n')\n", +"//Mechanical energy conservation-\n", +"//0.5*(M+m)*Va^2 = (M+m)*g*h\n", +"Va = sqrt(g*h/0.5)\n", +"//Momentum conservation for the collision\n", +"Vb = (M+m)*Va/m\n", +"printf('The velocity of the bullet before collision is %fm/s', Vb)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m1 = 0.70 //in kg\n", +"m = [0.14, 3.2] //in kg\n", +"k = [4.1* 10^4, 2.6* 10^6] //in N/m\n", +"d = [16* 10^-3, 1.1* 10^-3] //in meter\n", +"\n", +"//Sample Problem 10-3a\n", +"printf('**Sample Problem 10-3a**\n')\n", +"name = ['board', 'block']\n", +"U = zeros(2,1)\n", +"for count = 1:2\n", +" U(count) = 0.5* k(count)* d(count)^2\n", +" printf('The energy stored in %s is %fJ\n', name(count), U(count))\n", +"end\n", +"\n", +"//Sample Problem 10-3b\n", +"printf('\n**Sample Problem 10-3b**\n')\n", +"for count = 1:2\n", +" //Energy conservation\n", +" Vf = sqrt(U(count)/(0.5*(m1+m(count))))\n", +" //Momentum conservation\n", +" Vi = (m1 + m(count))*Vf/m1\n", +" printf('The minimum velocity required to break the %s is %fm/s\n', name(count), Vi)\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('collision.sci', -1)\n", +"\n", +"//Given that\n", +"m1 = 30*10^-3 //in kg\n", +"h1 = 8*10^-2 //in m\n", +"m2 = 75*10^-3 //in kg\n", +"g = 9.8 //in m/s^2\n", +"e = 1\n", +"\n", +"//Sample Problem 10-4\n", +"printf('**Sample Problem 10-4**\n')\n", +"//velocity just before collision\n", +"Vi = zeros(1,2)\n", +"Vi(1) = sqrt(2*g*h1)\n", +"Vi(2) = 0\n", +"Vf= fsolve([0,0], collision)\n", +"printf('The velocity of m1 after collision is %fm/s', abs(Vf(1)))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Ma = 83 //in kg\n", +"Va = [6.2, 0] //in km/h\n", +"Mb = 55 //in kg\n", +"Vb = [0, 7.8] //in km/h\n", +"\n", +"//Sample Problem 10-5a\n", +"printf('**Sample Problem 10-5a**\n')\n", +"Vf = (Ma*Va + Mb*Vb)/(Ma+Mb)\n", +"printf('The common velocity after the collision is %fm/s\n', norm(Vf))\n", +"\n", +"//Sample Problem 10-5b\n", +"printf('\n**Sample Problem 10-5**\n')\n", +"printf('The velocity of center of mass is not altered due to collision')" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/11-Rotation.ipynb b/Fundamental_Of_Physics_by_D_Haliday/11-Rotation.ipynb new file mode 100644 index 0000000..46a1c52 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/11-Rotation.ipynb @@ -0,0 +1,341 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11: Rotation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.10: Sample_Problem_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 1 //(say)\n", +"R = 0.15 //in meter\n", +"L = 2.0 * R\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 11-10\n", +"printf('**Sample Problem 11-10**\n')\n", +"I = 0.5*m*R^2 + m*L^2/12 + m*(L/2+R)^2\n", +"deltaU = m* g* (L + 2*R)\n", +"//deltaK = 0.5*I*w^2\n", +"//therefore-\n", +"w = sqrt(deltaU/(0.5*I))\n", +"printf('The angular speed is equal to %frad/s', w)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"t = poly(0, 't')\n", +"A = -1.00-0.600*t+0.250*t^2\n", +"\n", +"//Sample Problem 11-1a\n", +"printf('**Sample Problem 11-1a**\n')\n", +"Ts = [-3:0.5:6]\n", +"As = horner(A, Ts)\n", +"xset('window', 1)\n", +"xtitle( 'angular variable for the disk v/s time', 'time(sec)', 'Y-axis')\n", +"plot(Ts, As, 'm-o')\n", +"\n", +"//Sample Problem 11-1b\n", +"printf('\n**Sample Problem 11-1b**\n')\n", +"To = roots(derivat(A))\n", +"printf('At t=%fsec, theta approaches its minimum value equal to %f\n', To, horner(A, To))\n", +"\n", +"//Sample Problem 11-1c\n", +"printf('\n**Sample Problem 11-1c**\n')\n", +"Os = horner(derivat(A), Ts)\n", +"plot(Ts, Os, 'r-+')\n", +"legend('theta(rad)', 'omega(rad/s)')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"alpha = 0.335 //in rad/s^2\n", +"Wo = -4.6 //in rad/s\n", +"Ao = 0 //in rad\n", +"Af = 5* 2*%pi //in rad\n", +"\n", +"//Sample Problem 11-2a\n", +"printf('**Sample Problem 11-2a**\n')\n", +"//Using newton's second equation of motion\n", +"t = poly(0, 't')\n", +"p = Ao + Wo*t + 0.5*alpha*t^2 - Af\n", +"to = roots(p)\n", +"printf('At time equal to %fsec, the reference line will be at given position\n', to(2))\n", +"\n", +"//Sample Problem 11-2c\n", +"printf('\n**Sample Problem 11-2c**\n')\n", +"p = Wo + alpha*t\n", +"ts = roots(p)\n", +"printf('At time equal to %fsec, the disk momentarily stops', ts)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"W1 = 3.40 //in rad/s\n", +"W2 = 2.00 //in rad/s\n", +"rev_taken = 20\n", +"\n", +"//Sample Problem 11-3a\n", +"printf('**Sample Problem 11-3a**\n')\n", +"angle_traversed = 2*%pi*rev_taken\n", +"//Using newton's third equation of motion\n", +"//Wf^2 = Wi^2 + 2*alpha*theta\n", +"alpha = (W2^2 - W1^2)/(2*angle_traversed)\n", +"printf('The angular acceleration during the stop is %frads^2\n', alpha)\n", +"\n", +"//Sample Problem 11-3b\n", +"printf('\n**Sample Problem 11-3b**\n')\n", +"//Using newton's first equation of motion\n", +"time_taken = (W2 - W1)/alpha\n", +"printf('The time taken in decreasing the speed is %fsec', time_taken)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r = 15 //in meter\n", +"g = 9.8 //in m.s^2\n", +"a = 11 * g //in m.s^2\n", +"\n", +"//Sample Problem 11-4a\n", +"printf('**Sample Problem 11-4a**\n')\n", +"w = sqrt(a/r)\n", +"printf('The angular speed should be %frad/s\n', w)\n", +"\n", +"//Sample Problem 11-4b\n", +"printf('\n**Sample Problem 11-4b**\n')\n", +"t = 120 //in sec\n", +"alpha = w/t\n", +"at = alpha*r\n", +"printf('The tangential acceleration will be %fm/s^2', at)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 272 //in kg\n", +"R = 38*10^-2 //in meter\n", +"w = 14000* 2*%pi/60 //in rad/s\n", +"\n", +"///Sample Problem 11-6\n", +"printf('**Sample Problem 11-6**\n')\n", +"I = 0.5* M* R^2\n", +"E = 0.5* I* w^2\n", +"printf('The energy released during the explosion is %eJ', E)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 2.5 //in kg\n", +"R = 0.20 //i meter\n", +"m = 1.2 //in kg\n", +"g = 9.8 //in m/s^2\n", +"I = 0.5*M*R^2\n", +"\n", +"//Sample Problem 11-7\n", +"printf('**Sample Problem 11-7**\n')\n", +"//mg - T = ma\n", +"//T*R = I*a/R\n", +"//T = I*a/R^2\n", +"//on adding =>\n", +"a = m*g/(m+I/R^2)\n", +"T = m*(g-a)\n", +"alpha = a/R\n", +"printf('The acceleration of the block is %fm/s^2\n', a)\n", +"printf('The angular acceleration of the pulley is %frad/s^2\n', alpha)\n", +"printf('The tension in the string is %fN', T)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 80 //in kg\n", +"d1 = 0.30 //in meter\n", +"alpha = 6 //in rad/s^2\n", +"I = 15 //in kg.m^2\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 11-8a\n", +"printf('**Sample Problem 11-8a**\n')\n", +"F = I*alpha/d1\n", +"printf('The magnitude of F is %fN\n', F)\n", +"\n", +"//Sample Problem 11-8b\n", +"printf('\n**Sample Problem 11-8b**\n')\n", +"d2 = 0.12 //in meter\n", +"//F*d1 - M*g*d2 = I*alpha\n", +"F = I*alpha + M*g*d2\n", +"F = F/d1\n", +"printf('The magnitude of F in second case is %fN', F)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Example11_7.sce', -1)\n", +"clc\n", +"\n", +"//Given that\n", +"t = 2.5 //in sec\n", +"\n", +"//Sample Problem 11-9\n", +"printf('\n**Sample Problem 11-9**\n')\n", +"w = 0 + alpha*t\n", +"RE = 0.5* I* w^2\n", +"printf('The rotational kinetic energy of the disk will be %fJ', RE)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/12-Rolling_Torque_and_Angular_Momentum.ipynb b/Fundamental_Of_Physics_by_D_Haliday/12-Rolling_Torque_and_Angular_Momentum.ipynb new file mode 100644 index 0000000..3d01c78 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/12-Rolling_Torque_and_Angular_Momentum.ipynb @@ -0,0 +1,313 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12: Rolling Torque and Angular Momentum" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 1.4 //in kg\n", +"R = 8.5*10^-2 //in meter\n", +"v = 15*10^-2 //in meter\n", +"\n", +"//Sample Problem 12-1\n", +"printf('**Sample Problem 12-1**\n')\n", +"I = 0.5*M*R^2\n", +"w = v/R\n", +"K = 0.5*M*v^2 + 0.5*I*w^2\n", +"printf('The total kinetic energy of the disk is %fJ', K)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"M = 6 //in kg\n", +"A = 30 //in degrees\n", +"h = 1.20 //in meter\n", +"g = 9.8 //in m/s^2\n", +"//Radius of the disk is equal to R\n", +"\n", +"//Sample Problem 12-2a\n", +"printf('**Sample Problem 12-2a**\n')\n", +"//using mechanical energy conservation\n", +"//M*g*h = 0.5*I*w^2 + 0.5*M*v^2\n", +"//w = v/R\n", +"//I = (2/5)M*R^2\n", +"v = sqrt(M*g*h/(0.5*M*(2/5) + 0.5*M))\n", +"printf('The speed of the disk at the bottom is %fm/s\n', v)\n", +"\n", +"//Sample Problem 12-2b\n", +"printf('\n**Sample Problem 12-2b**\n')\n", +"//at bottom\n", +"a = -g*sin(dtor(A))/(1+(2/5)*M/M)\n", +"fs = -(2/5)*M*a\n", +"printf('The frictinal force on the disk is equal to %fN', fs)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"exec('cross_product.sci', -1)\n", +"\n", +"//Given that\n", +"A = dtor(30) //in rad\n", +"r = 3*[cos(A), 0, sin(A)]\n", +"F1 = 2*[1, 0, 0]\n", +"F2 = 2*[0, 0, 1]\n", +"F3 = 2*[0, 1, 0]\n", +"\n", +"//Sample Problem 12-3\n", +"printf('**Sample Problem 12-3**\n')\n", +"T1 = crossproduct(r, F1)\n", +"T2 = crossproduct(r, F2)\n", +"T3 = crossproduct(r, F3)\n", +"printf('The torque due to F1 is equal to %fN.m\n', norm(T1))\n", +"printf('The torque due to F2 is equal to %fN.m\n', norm(T2))\n", +"printf('The torque due to F3 is equal to %fN.m', norm(T3))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"p1 = 5 //in kg.m/s\n", +"d1 = 2.0 //in meter\n", +"p2 = 2 //in kg.m/s\n", +"d2 = 4.0 //in meter\n", +"\n", +"//Sample Problem 12-4\n", +"printf('**Sample Problem 12-4**\n')\n", +"L = p1*d1 - p2*d2 //the direction of L2 is opposite to L1\n", +"printf('The net angular momentum of the two particle system is %dkg.m/s^2', L)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 38 //in meter\n", +"Mc = 1.1 *10^4 //in kg\n", +"Mw = 6.0* 10^5 //in kg\n", +"Mp = 70 //in kg\n", +"Nc = 36 //number of cars\n", +"Np = 60*Nc //total number of person\n", +"Wf = 2*%pi/(2*60) //in rad/s\n", +"\n", +"//Sample Problem 12-6a\n", +"printf('**Sample Problem 12-6a**\n')\n", +"Mpc = Mc*Nc + Mp*Np\n", +"Ipc = Mpc*R^2\n", +"Iw = (Mw/2)*R^2 //considering the mometum due to only the outer hoop\n", +"It = Ipc+Iw\n", +"L = It*Wf\n", +"printf('The angular momentum is %ekg.m/s^2\n', L)\n", +"\n", +"//Sample Problem 12-6b\n", +"printf('\n**Sample Problem 12-6b**\n')\n", +"//Using the definiiton of the torque\n", +"deltat = 5.0 //in sec\n", +"Tavg = L/deltat\n", +"printf('The magnitude of average torque is %eN.m', Tavg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Iwh = 1.2 //in kg.m/s^2\n", +"Wwh = 3.9*2*%pi //in rad/s\n", +"Ib = 6.8 //in kg.m/s^2\n", +"\n", +"//Sample Problem 12-7\n", +"printf('**Sample Problem 12-7**\n')\n", +"//angular momentum conservation along the axis of rotation of the system\n", +"//Ib*Wb - Iwh*Wwh = Iwh*Wwh\n", +"Wb = 2*Iwh*Wwh/Ib\n", +"printf('The angular speed of the composite system is %frev/s', Wb/(2*%pi))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"t = 1.87 //in sec\n", +"I1 = 19.9 //in kg.m^2\n", +"I2 = 3.93 //in kg.m^2\n", +"//From the figure\n", +"A1 = 0.5*2*%pi //in rad\n", +"A2 = 3.5*2*%pi //in rad\n", +"\n", +"//Sample Problem 12-8\n", +"printf('**Sample Problem 12-8**\n')\n", +"//w1 = I2*w2/I1\n", +"//w1*t1 = A1\n", +"//w2*t2 = A2\n", +"//t = t1 + t2\n", +"//t = A1/w1 + A2/w2\n", +"//t = A1/(I2*w2/I1) + A2/w2\n", +"w2 = (1/t) * (A1*I1/I2 + A2)\n", +"printf('The angular speed during the tuck should be %frev/s', w2/(2*%pi))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"M = 1 //(say)\n", +"d = 0.50 //in meter\n", +"Wi = -2 //in rad/s\n", +"m = 1/3*M\n", +"Vi = 12 //in m/s\n", +"A = dtor(60) //in rad\n", +"\n", +"//Sample Problem 12-9\n", +"printf('**Sample Problem 12-9**\n')\n", +"//initial angular momentum\n", +"Ii = M*d^2/3 * 4\n", +"Lti = Ii*Wi //of the turnstile\n", +"Lbi = m* Vi * d * cos(A)\n", +"//final angular momentum\n", +"//Lf = Itotal * Wf\n", +"Itotal = Ii + m*d^2\n", +"//Using angular momentum conservation\n", +"Wf = (Lti + Lbi)/Itotal\n", +"printf('The final angular velocity of the ball-turnstile is %frad/s', Wf)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/13-Equilibrium_and_Elasticity.ipynb b/Fundamental_Of_Physics_by_D_Haliday/13-Equilibrium_and_Elasticity.ipynb new file mode 100644 index 0000000..6735a21 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/13-Equilibrium_and_Elasticity.ipynb @@ -0,0 +1,251 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: Equilibrium and Elasticity" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 1 //(say)\n", +"m = 1.8 //in kg\n", +"M = 2.7 //in kg\n", +"d = L/4\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 13-1\n", +"printf('**Sample Problem 13-1**\n')\n", +"//From the figure\n", +"//Balancing torque from A\n", +"//N2*L = M*g*L/4 + m*g*L/2\n", +"N2 = M*g/4 + m*g/2\n", +"//similarly\n", +"N1 = M*g*(3/4) + m*g/2\n", +"printf('The Normal from point A is %fN\n', N1)\n", +"printf('The Normal from point B is %fN', N2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 12 // in meter\n", +"m = 45 //in kg\n", +"h = 9.3 //in meter\n", +"M = 72 //in kg\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 13-2\n", +"printf('**Sample Problem 13-2**\n')\n", +"//From the figure\n", +"N2 = (M+m)*g\n", +"A = asin(h/L)\n", +"//Balancing torque from the bottom point of the ladder\n", +"N1 = (m*g*L/3*cos(A)+M*g*L/2*cos(A))/(L*sin(A))\n", +"f = N1\n", +"printf('The normal force from the wall is equal to %fN\n', N1)\n", +"printf('The normal force from the pavement is equal to %fN\n', N2)\n", +"printf('The frictional force from the pavement is equal to %fN', f)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 430 //in kg\n", +"a = 1.9 //in meter\n", +"b = 2.5 //in meter\n", +"m = 85 //in kg\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 13-3a\n", +"printf('**Sample Problem 13-3a**\n')\n", +"Tr = M*g\n", +"//Balancing torque from the hinge\n", +"Tc = (m*g*b/2 + Tr*b)/a\n", +"printf('The magntude of force on the beam from the cable is %fN\n', Tc)\n", +"\n", +"//Sample Problem 13-3b\n", +"printf('\n**Sample Problem 13-3b**\n')\n", +"Fu = Tr + m*g\n", +"Fd = Tc\n", +"printf('The magnitude of force on the beam from the hinge is %fN', sqrt(Fu^2+Fd^2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 55 //in kg\n", +"w = 1 //in meter\n", +"d = 0.20 //in meter\n", +"u1 = 1.1\n", +"u2 = 0.70\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 13-4a\n", +"printf('**Sample Problem 13-4a**\n')\n", +"//From the figure\n", +"//fr + fl = m*g\n", +"//Nr = Nl\n", +"//u1*Nr + u2*Nl = m*g\n", +"Nr = m*g/(u1+u2)\n", +"Nl = Nr\n", +"fr = u2*Nr\n", +"fl = u1*Nl\n", +"printf('The minimum horiontal push on the wall should be %fN\n', Nr)\n", +"\n", +"//Sample Problem 13-4b\n", +"printf('\n**Sample Problem 13-4b**\n')\n", +"//balancing torque from the shoulder side wall\n", +"l = (fl*w - m*g*d)/Nl\n", +"printf('The vertical distance between her shoulder and her feet should be %fm', l)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 9.5*10^-3 //in meter\n", +"L = 81*10^-2 //in meter\n", +"F = 62*10^3 //in N\n", +"Y = 2*10^11 //in N/m^2\n", +"\n", +"//Sample Problem 13-5\n", +"printf('**Sample Problem 13-5**\n')\n", +"//From the definition of stress, strain & hook's law\n", +"sigma = F/(%pi*R^2)\n", +"strain = sigma/Y\n", +"deltaL = strain*L\n", +"printf('The stress in the rod is %en/m^2\n', sigma)\n", +"printf('The strain in the rod is %e\n', strain)\n", +"printf('The elongation in the rod is %em', deltaL)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 290 //in kg\n", +"A = 1*10^-4 //in m^2\n", +"Y = 1.3*10^10 //in N/m^2\n", +"l = 0.50*10^-3 //in meter\n", +"g = 9.8 //in m/s^2\n", +"L = 1.00 //in meter\n", +"\n", +"//Sample Problem 13-6\n", +"printf('**Sample Problem 13-6**\n')\n", +"//We know F = A*Y*(delataL/L)\n", +"l1 = poly(0,'l1')\n", +"p = M*g - 3*(A*Y*(l1/L)) - (A*Y*((l1+l)/L))\n", +"l1 = roots(p)\n", +"F3 = A*Y*(l1/L)\n", +"F1 = (A*Y*((l1+l)/L))\n", +"printf('The force on the leveled legs is %fN\n', F3)\n", +"printf('The force on the unleveled legs is %fN', F1)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/14-Gravitation.ipynb b/Fundamental_Of_Physics_by_D_Haliday/14-Gravitation.ipynb new file mode 100644 index 0000000..f19df24 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/14-Gravitation.ipynb @@ -0,0 +1,323 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14: Gravitation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec ('Gravitation.sci',-1)\n", +"\n", +"//Given that\n", +"m1 = 6 //kg\n", +"m2 = 4 //kg\n", +"m3 = 4 //kg\n", +"a = 2 * (10^-2)\n", +"\n", +"//Sample Problem 14-1\n", +"printf('**Sample Problem 14-1**\n')\n", +"//F1 = F12 + F13\n", +"F12 = [0,-GForce(m1,m2,a)]\n", +"F13 = [GForce(m1,m3,2*a),0]\n", +"F1 = F12 + F13\n", +"printf('The magnitude of net force is approximately equal to %e N', norm(F1))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Gravitation.sci',-1)\n", +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"//masses in kg\n", +"m1 = 8\n", +"m2 = 2\n", +"m3 = 2\n", +"m4 = 2\n", +"m5 = 2\n", +"a = 2*(10^-2); //in meter\n", +"Theta = dtor(30) //in radians\n", +"\n", +"//Sample Problem 14-2\n", +"printf('**Sample Problem 14-2**\n')\n", +"//The net force will be equal to the vector eum of all the forces acting on the particle due to the rest of the particles i.e F1 = F12 + F13 + F14 + F15\n", +"F12 = [GForce(m1,m2,(2*a))*sin(Theta), GForce(m1,m2,(2*a))*cos(Theta)]\n", +"F13 = [GForce(m1,m3,a)*sin(Theta), -GForce(m1,m3,a)*cos(Theta)]\n", +"F14 = [-GForce(m1,m4,(2*a))*sin(Theta), -GForce(m1,m4,(2*a))*cos(Theta)]\n", +"F15 = [-GForce(m1,m5,a)*sin(Theta),-GForce(m1,m5,a)*cos(Theta)]\n", +"F1 = F12 + F13 + F14 + F15\n", +"printf('The net force on particle 1 is approimately equal to %e N', norm(F1))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3_a: Sample_Problem_3a.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec ('Gravitation.sci',-1)\n", +"\n", +"//Given that\n", +"r = 6.77 * 10^6 //in meter\n", +"dr = 1.7 //in meter\n", +"\n", +"//Sample Problem 3a\n", +"printf('**Sample Problem 3a**\n')\n", +"dg = -2 * G * Me * dr /(r^3)\n", +"printf('The difference in acceleration is approximately equal to %e m/sec*sec', dg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3_b: Sample_Problem_3b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec ('Gravitation.sci',-1)\n", +"\n", +"//variavles with their values\n", +"Mh = 1.99 * 10^31 //in kg\n", +"R = 6.77 * 10^6 //in meter\n", +"DR = 1.7 //in meter\n", +"\n", +"//Sample Problem 3b\n", +"printf('**Sample Problem 3b**\n')\n", +"//the difference in gravitational acceleration is given by\n", +"DG = -2 * G * Mh * DR /(R^3)\n", +"printf('The difference in acceleration is approximately equal to %em/s^2', DG)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec ('Gravitation.sci',-1)\n", +"//Given that\n", +"Vi = 1.2 * 10^4 //in m/sec\n", +"d = 10*Re;\n", +"m = 10 //let say it will mass cancel out later\n", +"//Sample Problem 5\n", +"printf('**Sample Problem 5**\n')\n", +"//we know that E(initial) = E(final)\n", +"//=> Ki + Ui = Kf + Uf\n", +"//K = .5*m*Vi*Vi (Kinetic Energy)\n", +"//U = gravitational potential (Potential Energy)\n", +"Ki = .5*m*Vi*Vi; \n", +"Ui = GPotential(m,Me,d);\n", +"Uf = GPotential(m,Me,Re);\n", +"Kf = Ki + Ui -Uf;\n", +"Vf = sqrt(2*Kf/m);\n", +"printf('The final velocity of the asteroid is equal to %e m/sec', Vf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec ('Gravitation.sci',-1)\n", +"\n", +"//Given that\n", +"T = 76 * 365 * 24 * 60 * 60 //time period in seconds (converting from years)\n", +"\n", +"//Sample Problem 6a\n", +"printf('**Sample Problem 6a**\n')\n", +"//We know that Ra + Re = 2*a\n", +"Rp = 8.9 * 10^10 //in meter\n", +"a = KeplerRadius(Ms,T)\n", +"//therefore\n", +"Ra = 2*a -Rp //in meter\n", +"printf('The Aphelion distance is equal to %em\n', Ra)\n", +"\n", +"//Sample Problem 6b\n", +"printf('\n**Sample Problem 6b**\n')\n", +"//we know that e*a = a - Rp\n", +"e = 1 - Rp/a\n", +"printf('The eccentricity of the path is %e ', e)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec ('Gravitation.sci',-1)\n", +"\n", +"//Given that\n", +"//Both the stars are moving around the centre of mass of the two particale system\n", +"//m1 = mass of visible star\n", +"//m2 = mass of invisible star\n", +"//r1 = distance of m1 from center of mass\n", +"//r2 = distance of m2 from center of mass\n", +"//r = r1+r2 distance between both the stars\n", +"//we have G*m1*m2/(r*r) = m1*v1*v1/r1 = m2*v2*v2/r2 ....1\n", +"v1 = 270*10^3 //in meter/sec\n", +"T = 1.7 * 24 * 60 * 60 //in s\n", +"m1 = 6* Ms \n", +"\n", +"//Sample Problem 7\n", +"printf('**Sample Problem 7**\n')\n", +"//m2 = ?\n", +"//using definition of center of mass\n", +"// we have r = r1 * (m1 + m2)/m2 ....2\n", +"//& 2*pi*r1/v1 = T ....3\n", +"//therefore\n", +"r1 = v1*T/(2*%pi); //from equation 3\n", +"//from equation 1 & 2\n", +"//G*(m2^3)/((r1*(m1+m2))^2) = v1*v1/r1\n", +"//we have a polynomial equation in order 3 \n", +"//(m2^3)/(m1+m2)^2 = v1*v1*r1/G\n", +"temp = v1*v1*r1/G; //say\n", +"//=> -m2^3 + temp*m2^2 + 2*m1*temp*m2+ m1*m1*temp\n", +"solpoly = (poly([-m1*m1,-2*m1,-1,1/temp],'x','c'));\n", +"sol = roots(solpoly,'e');\n", +"printf('The mass of the invisible star is equal to %e kg\n', sol(1))\n", +"printf('The mass of the invisible star is equal to %f times the mass of Sun', sol(1)/Ms)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec ('Gravitation.sci',-1)\n", +"\n", +"//Given that\n", +"m = 7.20 //in kg\n", +"h = 350 * 10^3 //altitude in meter\n", +"\n", +"//Sample Problem 8a\n", +"printf('**Sample Problem 8a**\n')\n", +"//mechanical energy E = K + U\n", +"//E = - G * M * m /(2* r)\n", +"E = .5*GPotential(m,Me,(h+Re))\n", +"printf('The total energy at the given altitude is %e joule\n',E)\n", +"\n", +"\n", +"//Sample Problem 8b\n", +"printf('\n**Sample Problem 8b**\n')\n", +"//here the k = 0\n", +"E0 = GPotential(m,Me,Re)\n", +"printf('The total energy on the launchpad is %e joule\n',E0)\n", +"deltaE = E - E0;\n", +"printf('The differencein both the energy %e joule',deltaE)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/15-Fluids.ipynb b/Fundamental_Of_Physics_by_D_Haliday/15-Fluids.ipynb new file mode 100644 index 0000000..68f90f2 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/15-Fluids.ipynb @@ -0,0 +1,278 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15: Fluids" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"l = 3.5 //in meter\n", +"b = 4.2 //in meter\n", +"h = 2.4 //in meter\n", +"g = 9.8 //in m/s^2\n", +"//Sample Problem 15-1a\n", +"printf('**Sample Problem 15-1a**\n')\n", +"P = 1.0*1.01*10^5 //in N/m^2\n", +"density_air = 1.21 //in kg/m^3 at 1atm pressure\n", +"V = l*b*h //Volume of the air\n", +"W = density_air*V*g //Weight of the air\n", +"printf('The weight of the air in the room is equal to %fkg\n', W)\n", +"//Sample Problem 15-1b\n", +"printf('\n**Sample Problem 15-1b**\n')\n", +"A = l*b\n", +"F = P*A\n", +"printf('The magnitude of atmosphere force on the floor of the room is %eN\n', F)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"deltaP = 9.3*10^3 //in N/m^2\n", +"density_water = 998 //in kg/m^3\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 15-2\n", +"printf('**Sample Problem 15-2**\n')\n", +"//deltaP = density_water*g*L\n", +"//therefore\n", +"L = deltaP/(density_water*g)\n", +"printf('The diver started at a depth of %fm', L)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"density_water = 998 //in kg/m^3\n", +"l = 135*10^-3 //in meter\n", +"d = 12.3*10^-3 //in meter\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 15-3\n", +"printf('**Sample Problem 15-3**\n')\n", +"d_oil = density_water*g*l/(g*(l+d)) //pressure at same height should be same\n", +"printf('The density of the oil is %fkg/m^3', d_oil)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"density_ice = 917 //in kg/m^3\n", +"density_seawater = 1024 //in kg/m^3\n", +"\n", +"//Sample Problem 15-4\n", +"printf('**Sample Problem 15-4**\n')\n", +"//Using archimedies principle\n", +"//density_seawater*Vs*g = density_ice*Vt*g\n", +"//fractoin of non-visible part 'f' is\n", +"f = density_ice/density_seawater //Vs/Vt\n", +"printf('The fraction of the visible part is equal to %f', (1-f))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 12.0 //in meter\n", +"m = 196 //in kg\n", +"density_He = 0.160 //in kg/m^3\n", +"density_air = 1.25 //in kg/m^3\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 15-5\n", +"printf('**Sample Problem 15-5**\n')\n", +"Vb = 4/3*%pi*R^3\n", +"PayLoad = (density_air*Vb*g - (density_He*Vb*g + m*g))/g\n", +"printf('The maximum mass attached can be %fkg', PayLoad)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Ao = 3*10^-4 //in m^2\n", +"Vo = 30*10^-2 //in m/s\n", +"A = 3*10^-7*10^-4 //in m^2\n", +"V = 0.05*10^-2 //in m/s\n", +"\n", +"//Sample Problem 15-6\n", +"printf('**Sample Problem 15-6**\n')\n", +"n = (Ao*Vo)/(A*V)\n", +"printf('Number of capillaries a person should have %e', n)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Bernauli.sci', -1)\n", +"\n", +"//Given that\n", +"Ao = 1.2*10^-4 //in m^2\n", +"A = 0.35*10^-4 //in m^2\n", +"h = 45*10^-3 //in m\n", +"density_water = 998 //in kg/m^3\n", +"\n", +"//Sample Problem 15-7\n", +"printf('**Sample Problem 15-7\n')\n", +"A = [A, Ao]\n", +"deltaP = 0 //in N/m^2\n", +"density = density_water\n", +"V = fsolve([0,0], Bernauli)\n", +"FlowRate = A(1)*V(1)\n", +"printf('The volume flow rate from the tap is equal to %fcm^3/s', FlowRate*10^6)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Bernauli.sci', -1)\n", +"\n", +"//Given that\n", +"density_Ethanol = 791 //in kg/m^3\n", +"A1 = 1.20*10^-3 //in m^2\n", +"A2 = A1/2\n", +"//Pressure in narrower section is lesser\n", +"deltaP = -4120 //in N/m^2\n", +"\n", +"//Sample Problem 15-8\n", +"printf('**Sample Problem 15-8**\n')\n", +"A = [A1, A2]\n", +"density = density_Ethanol\n", +"h = 0\n", +"V = fsolve([0, 1], Bernauli)\n", +"printf('The volume flow rate is %e m^3/s', abs(A1*V(1)))" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/16-Oscillation.ipynb b/Fundamental_Of_Physics_by_D_Haliday/16-Oscillation.ipynb new file mode 100644 index 0000000..cbb2e1f --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/16-Oscillation.ipynb @@ -0,0 +1,285 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16: Oscillation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 680*10^-3 //in kg\n", +"k = 65 //in N/m\n", +"x = 11*10^-2 //in meter\n", +"\n", +"//Sample Problem 16-1a\n", +"printf('**Sample Problem 16-1a**\n')\n", +"w = sqrt(k/m)\n", +"f = 2*%pi/w\n", +"T = 1/f\n", +"printf(' Angular Frequency - %f rad/s\n', w)\n", +"printf(' Frequency\t - %f Hz\n', f)\n", +"printf(' Time Period\t - %f s\n', T)\n", +"\n", +"//Sample Problem 16-1b\n", +"printf('\n**Sample Problem 16-1b**\n')\n", +"A = x\n", +"printf('The amplitude of oscillation is %fcm\n', A*100)\n", +"\n", +"//Sample Problem 16-1c\n", +"printf('\n**Sample Problem 16-1c**\n')\n", +"Vmax = A*w\n", +"printf('The maximum speed of the block is %fm/s\n', Vmax)\n", +"\n", +"//Sample Problem 16-1d\n", +"printf('\n**Sample Problem 16-1d**\n')\n", +"aMAX = Vmax*w\n", +"printf('The maximum acceleration of the block is %fm/s^2\n', aMAX)\n", +"\n", +"//Sample Problem 16-1e\n", +"printf('\n**Sample Problem 16-1e**\n')\n", +"//at t=0 x = A\n", +"phi = acos(x/A)\n", +"printf('The phase constant for the SHM x=Acos(w*t+phi) is 2*n*pi + %d where n is an integer', phi)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"x0 = -8.50*10^-2 //in m\n", +"v0 = -0.920 //in m/s\n", +"a0 = 47 //in m/s^2\n", +"exec('degree_rad.sci', -1)\n", +"\n", +"//Sample Problem 16-2a\n", +"printf('**Sample Problem 16-2a**\n')\n", +"w = sqrt(-a0/x0)\n", +"printf('The angular frequency of SHM is equal to %frad/s\n', w)\n", +"\n", +"//Sample Problem 16-2b\n", +"printf('\n**Sample Problem 16-2b**\n')\n", +"phi = atan(-(v0/x0)/w)\n", +"A = x0/cos(phi)\n", +"printf('The value of phi is %f degrees\n', rtod(phi))\n", +"printf('The maximum value of displacement is equal to %fcm\', A*100)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 16-3a\n", +"printf('**Sample Problem 16-3a**\n')\n", +"k = 65 //in N/m\n", +"A = 11*10^-2 //in meter\n", +"v = 0 //in m/s\n", +"E = 0.5*k*A^2 + 0\n", +"printf('The mechanical energy of the block is %fJ\n', E)\n", +"\n", +"//Sample Problem 16-3b\n", +"printf('\n**Sample Problem 16-3b**\n')\n", +"x = A/2\n", +"U = 0.5*k*x^2\n", +"K = E-U\n", +"printf('The potential energy at that position is %fJ\n', U)\n", +"printf('The kinetic energy of the block is %fJ', K)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 12.4*10^-2 // in meter\n", +"m = 135*10^-3 //in kg\n", +"Ta = 2.53 //in sec\n", +"Tb = 4.76 //in sec\n", +"\n", +"//Sample Problem 16-4\n", +"printf('**Sample Problem 16-4**\n')\n", +"//Time period is directly proportional to the squre root of rotational inertial\n", +"Ia = m*L^2/12\n", +"I = Ia*(Tb/Ta)^2\n", +"printf('The rotational inertail of X is %ekg.m^2', I)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 1.0 //in meter\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 16-5a\n", +"printf('**Sample Problem 16-5a**\n')\n", +"//(say)\n", +"m = 1\n", +"I = (1/3)*m*L^2\n", +"T = 2*%pi*sqrt(I/(m*g*(L/2)))\n", +"printf('The time period of SHm is %fs\n', T)\n", +"\n", +"//Sample Problem 16-5b\n", +"printf('\n**Sample Problem 16-5b**\n')\n", +"Lo = (T/(2*%pi))^2*g\n", +"printf('The distance of center of oscillation from the pivot is %fcm', Lo*100)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 2.0 //in meter\n", +"m = 12 //in kg\n", +"k = 1300 //in N/m\n", +"\n", +"//Sample Problem 16-6\n", +"printf('**Sample Problem 16-6**\n')\n", +"//I*a/L = -k*x*L\n", +"I = m*L^2/3\n", +"w = sqrt(k*L*L/I)\n", +"T = 2*%pi/w\n", +"printf('The time period of oscillation is %fs', T)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 250*10^-3 //in kg\n", +"k = 85 //in N/m\n", +"b = 70*10^-3 //in kg/s\\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 16-7a\n", +"printf('**Sample Problem 16-7a**\n')\n", +"T = 2*%pi*sqrt(m/k)\n", +"printf('The time period of oscillation is %fs\n', T)\n", +"\n", +"//Sample Problem 16-7b\n", +"printf('\n**Sample Problem 16-7b**\n')\n", +"//b*t/(2*m) = log(2)\n", +"t = log(2)*2*m/b\n", +"printf('The time taken to drop the amplitude half of its value is %fs\n', t)\n", +"\n", +"//Sample Problem 16-7c\n", +"printf('\nb**Sample Problem 16-7c**\n')\n", +"//amplitude should drop to A/sqrt(2) from A\n", +"tE = t/2\n", +"printf('The time taken to drop the mechanical energy half of its initial value is %fs', tE)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/17-Waves_l.ipynb b/Fundamental_Of_Physics_by_D_Haliday/17-Waves_l.ipynb new file mode 100644 index 0000000..31e6d51 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/17-Waves_l.ipynb @@ -0,0 +1,287 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 17: Waves l" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"funcprot(0)\n", +"deff('[y] = wave(t, x)', 'y = 0.00327*sin(72.1*x - 2.72*t)')\n", +"//Comparing the given equation with the standard wave equation\n", +"A = 0.00327 //in m\n", +"k = 72.1 //in rad/s\n", +"w = 2.72 //in rad/s\n", +"\n", +"//Sample Problem 17-1a\n", +"printf('**Sample Problem 17-1a**\n')\n", +"printf('The amplitude of the wave is %fm\n', A)\n", +"\n", +"//Sample Problem 17-1b\n", +"printf('\n**Sample Problem 17-1b**\n')\n", +"lambda = 2*%pi/k\n", +"printf(' wavelength = %fm\n', lambda)\n", +"T = 2*%pi/w\n", +"printf(' period = %fs\n',T)\n", +"f = 1/T\n", +"printf(' frequency = %fHz\n', f)\n", +"\n", +"//Sample Problem 17-1c\n", +"printf('\n**Sample Problem 17-1c**\n')\n", +"v = w/k\n", +"printf('The velocity of the wave is %fm/s\n', v)\n", +"\n", +"//Sample Problem 17-1d\n", +"printf('\n**Sample Problem 17-1d**\n')\n", +"y = wave(18.9, 22.5*10^-2)\n", +"printf('Displacement of the wave is %fm', y)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Example17_1.sce', -1)\n", +"clc\n", +"\n", +"//Sample Problem 17-2a\n", +"printf('\n\n**Sample Problem 17-2a**\n')\n", +"v = numdiff(list(wave, 22.5*10^-2), 18.9)\n", +"printf('The velocity of the particle is %fm/s\n', v)\n", +"\n", +"//Sample Problem 17-2b\n", +"printf('\n**Sample Problem 17-2b**\n')\n", +"dx = 0.001\n", +"a = (numdiff(list(wave, 22.5*10^-2), 18.9+dx) - numdiff(list(wave, 22.5*10^-2), 18.9))/dx\n", +"printf('The acceleration of the particle is %fm/s^2', a)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"u1 = 1.4*10^-4 //in kg/m\n", +"u2 = 2.8*10^-4 //in kg/m\n", +"L1 = 3.0 //in m\n", +"L2 = 2.0 //in m\n", +"T = 400 //in N\n", +"\n", +"//Sample Problem 17-3\n", +"printf('**Sample Problem 17-3**\n')\n", +"v1 = sqrt(T/u1)\n", +"v2 = sqrt(T/u2)\n", +"t1 = L1/v1\n", +"t2 = L2/v2\n", +"if t1<t2 then\n", +" printf('The pulse on string1 will reach the knot first')\n", +"else\n", +" printf('The pulse on string2 will reach the knot first')\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"u = 525*10^-3 //in kg/m\n", +"T = 45 //in N\n", +"f = 120 //in Hz\n", +"Ym = 8.5*10^-3 //in meter\n", +"\n", +"//Sample Problem 17-4\n", +"printf('**Sample Problem 17-4**\n')\n", +"v = sqrt(T/u)\n", +"w = 2*%pi*f\n", +"Pavg = 0.5*u*v*w^2*Ym^2\n", +"printf('The average power transfered is equal to %fW', Pavg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"Ym = 9.8*10^-3 //in meter\n", +"phi = dtor(100) //in degrees\n", +"\n", +"//Sample Problem 17-5a\n", +"printf('**Sample Problem 17-5a**\n')\n", +"Yn = 2*Ym*cos(phi/2)\n", +"printf('The amplitude of resultant wave is %fmm\n', Yn*10^3)\n", +"\n", +"//Sample Problem 17-5b\n", +"printf('\n**Sample Problem 17-5b**\n')\n", +"Y = 4.9*10^-3 //in meter\n", +"phi = 2*acos(Y/(2*Ym))\n", +"printf('The phase difference required is %frad', phi)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Ym1 = 4.0*10^-3 //in meter\n", +"Ym2 = 3.0*10^-3 //in meter\n", +"phi1 = 0 //in rad\n", +"phi2 = %pi/3 //in rad\n", +"\n", +"//Sample Problem 17-6\n", +"printf('**Sample Problem 17-6**\n')\n", +"//For horizontal component\n", +"Ymh = Ym1*cos(phi1) + Ym2*cos(phi2)\n", +"//For vertical component\n", +"Ymv = Ym1*sin(phi1) + Ym2*sin(phi2)\n", +"Ym = sqrt(Ymv^2 + Ymh^2)\n", +"beta = atan(Ymv/Ymh)\n", +"printf('The amplitude of the resultant wave is equal to %fmm\n', Ym*10^3)\n", +"printf('The phase constant for the wave is %frad\n', beta)\n", +"printf('The equation of wave is %f*sin(k*x - w*t + %frad)',Ym, beta)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 1.2 //in meter\n", +"u = 1.6*10^-3 //in kg/m\n", +"f = 120 //in Hz\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 17-7a\n", +"printf('**Sample Problem 17-7a**\n')\n", +"n = 4\n", +"//T = m*g\n", +"//v = sqrt(m*g/u)\n", +"//f = n*v/(2*L)\n", +"m = (2*L*f)^2*u/(g*n^2)\n", +"printf('The value of m should be %fkg\n', m)\n", +"\n", +"//Sample Problem 17-7b\n", +"printf('\n**Sample Problem 17-7b**\n')\n", +"M = 1 //in kg\n", +"n = sqrt((2*L*f)^2*u/(g*M))\n", +"if abs(n - round(n)) < 0.01 then\n", +" printf('The allowed wave mode is %d', n)\n", +"else\n", +" printf('No wave mode will be allowed')\n", +"end" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/18-Waves_ll.ipynb b/Fundamental_Of_Physics_by_D_Haliday/18-Waves_ll.ipynb new file mode 100644 index 0000000..296f7d0 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/18-Waves_ll.ipynb @@ -0,0 +1,251 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18: Waves ll" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"Vs = 1482 //in m/s\n", +"Vw = 343 //in m/s\n", +"\n", +"//Sample Problem 18-1\n", +"printf('**Sample Probelm 18-1**\n')\n", +"//deltaT = d/V = D*sin(theta)/V\n", +"//D*sin(90)/Vs = D*sin(theta)/Vw\n", +"theta = rtod(asin(Vw/Vs))\n", +"printf('The actual angle at which source is present, is %fdegree', theta)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"deltaPm = 28 //in N/m^2\n", +"density = 1.21 //in kg/m^3\n", +"f = 1000 //in Hz\n", +"v = 343 //in m/s\n", +"\n", +"//Sample Problem 18-2\n", +"printf('**Sample Problem 18-2**\n')\n", +"w = 2*%pi*f\n", +"Sm = deltaPm/(v*density*w)\n", +"printf('The amplitude of sound will be %em', Sm)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"D = 1.5 //times wavelength\n", +"\n", +"//Sample Problem 18-3a\n", +"printf('**Sample Problem 18-3a**\n')\n", +"deltaL = 0\n", +"printf('Waves undergo constructive interference at P1\n')\n", +"\n", +"//Sample Problem 18-3b\n", +"printf('\n**Sample Problem 18-3b**\n')\n", +"deltaL = D\n", +"deltaPhi = 1.5*2*%pi\n", +"if modulo((deltaPhi/%pi), 2)==0 then\n", +" printf('Waves undergo constructive interference at P2\n')\n", +"else\n", +" printf('Waves undergo desstructive interference at P2')\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 10 //in meter\n", +"Ps = 1.6*10^4\n", +"\n", +"//Sample Problem 18-4a\n", +"printf('**Sample Problem 18-4a**\n')\n", +"r = 12 //in meter\n", +"I = Ps/(2*%pi*r*L)\n", +"printf('The intensity of the sound at a distance %dm is equal to %fW/m^2\n', r, I)\n", +"\n", +"//Sapmle Problem 18-4b\n", +"printf('\n**Sample Problem 18-4b**\n')\n", +"Ad = 2*10^-4 //in m^2\n", +"Pd = I*Ad\n", +"printf('The sound energy intercepted by the acoustic detector is %eW', Pd)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"B2 = 120 //in dB\n", +"B1 = 92 //in dB\n", +"\n", +"//Sample Problem 18-5\n", +"printf('**Sample Problem 18-5**\n')\n", +"//B2 = 10 * log(I2/Io)\n", +"//B1 = 10 * log(I1/Io)\n", +"r = 10^((B2 - B1)/10) //The ratio\n", +"printf('The Ratio of the intensities is %f', r)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 67*10^-2 //in cm\n", +"Vs = 343 //in m/s\n", +"\n", +"//Sample Problem 18-6a\n", +"printf('**Sample Problem 18-6a**\n')\n", +"//Open Ends\n", +"f = Vs/(2*L)\n", +"printf('The frequency of sound in case of open end is %fHz\n', f)\n", +"\n", +"//Sample Problem 18-6b\n", +"printf('\n**Sample Problem 18-6b**\n')\n", +"//cloes end\n", +"fo = Vs/(4*L)\n", +"printf('The frequency of sound in case of open ends is %fHz', fo)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"v = 242 //in m/s\n", +"f = 1250 //in Hz\n", +"Vs = 343 //in m/s\n", +"\n", +"//Sample Problem 18-8a\n", +"printf('**Sample Problem 18-8a**\n')\n", +"F = (Vs/(Vs-v))*f\n", +"printf('The frequency measured by the detector on the pole is %fHz\n', F)\n", +"\n", +"//Sample Problem 18-8b\n", +"printf('\n**Sample Problem 18-8b**\n')\n", +"Fe = (Vs+v)/Vs*F\n", +"printf('The frequency measured by the detector on the rocket is %fHz', Fe)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/19-Temprature_Heat_and_the_First_Law_of_Thermodynamic.ipynb b/Fundamental_Of_Physics_by_D_Haliday/19-Temprature_Heat_and_the_First_Law_of_Thermodynamic.ipynb new file mode 100644 index 0000000..0e5503b --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/19-Temprature_Heat_and_the_First_Law_of_Thermodynamic.ipynb @@ -0,0 +1,266 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19: Temprature Heat and the First Law of Thermodynamic" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Bz = 65 //in degree Z\n", +"Fz = -14 //in degree Z\n", +"Tz = -98 //in degree Z\n", +"Bf = 212 //in degree F\n", +"Ff = 32 //in degree F\n", +"\n", +"//Sample Problem 19-1\n", +"printf('**Sample Problem 19-1**\n')\n", +"Tf = (Tz - Fz)/(Bz - Fz) * (Bf - Ff) + Ff\n", +"printf('-98 degree Z is equal to %f degree Forenheit', Tf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"V = 37*10^3 //in Litre\n", +"b = 9.50*10^-4 //in /degree C\n", +"deltaT = -23 //in degree C\n", +"\n", +"//Sample Problem 19-2\n", +"printf('**Sample Problem 19-2**\n')\n", +"deltaV = V* b* deltaT\n", +"Vd = V + deltaV\n", +"printf('The amount of oil delievered is %dL', Vd)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 720*10^-3 //in kg\n", +"Ti = -10 //in degree C\n", +"Tw = 15 //in degree C\n", +"Si = 2220 //in J/kg.K\n", +"L = 333*10^3 //in J/Kg\n", +"Sw = 4190 //in J/kg.K\n", +"//Sample Problem 19-3a\n", +"printf('**Sample Problem 19-3a**\n');\n", +"//When ice comes to temprature 0 degree celcius\n", +"deltaTi = 0 - Ti\n", +"Q1 = m*Si*deltaTi\n", +"//When ice melts to water\n", +"Q2 = m*L\n", +"//when water comes to temprature 15 degree celcius\n", +"deltaTw = Tw - 0\n", +"Q3 = m*Sw*deltaTw\n", +"Q = Q1 + Q2 + Q3\n", +"printf('The heat absorbed by the ice is %fKJ\n', Q*10^-3)\n", +"//Sample Problem 19-3b\n", +"printf('\n**Sample Problem 19-3b**\n')\n", +"Qg = 210*10^3 //in J\n", +"//Energy left after the ice comes to 0 degree C\n", +"Qleft = Qg - Q1\n", +"//the mass of ice left aftr all the heat is consumed\n", +"Mleft = Qleft/L\n", +"printf('The mass of ice left is equal to %fg', Mleft*10^3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Mc = 75 //in g\n", +"To = 312 //in degree C\n", +"Mw = 220 //in g\n", +"Cb = 45 //in cal/K\n", +"Ti = 12 //in degree C\n", +"Sc = 0.0923 //in cal/g.K\n", +"Cb = 45 //in cal/K\n", +"Sw = 1 //in cal/g.K\n", +"//Sample Problem 19-4\n", +"printf('**Sample Problem 19-4**\n')\n", +"Tf = (Sc*Mc*To + Cb*Ti + Sw*Mw*Ti)/(Sw*Mw + Cb + Sc*Mc)\n", +"printf('The final temprature of the system is %f degree C', Tf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 1.00 //in kg\n", +"T = 100 //in degree C\n", +"P = 1.01*10^5 //in Pascal\n", +"Vi = 1.00*10^-3 //in m^3\n", +"Vf = 1.671 //in m^3\n", +"Lv = 2256*10^3 //in J/kg\n", +"//Sample Problem 19-5a\n", +"printf('**Sample Problem 19-5a**\n')\n", +"W = P*(Vf - Vi)\n", +"printf('The work done by the system during the process is %fkJ\n', W*10^-3)\n", +"//Sample Problem 19-5b\n", +"printf('\n**Sample Problem 19-5b**\n')\n", +"Q = Lv*M\n", +"printf('The heat supplied to the system is equal to %eJ\n', Q)\n", +"//Sample Problem 19-5c\n", +"printf('\n**Sample Problem 19-5c**\n')\n", +"deltaE = Q - W\n", +"printf('The change in internal energy is equal to %eJ', deltaE)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"La = 1 //(say)\n", +"Ld = 2.0*La\n", +"ka = 1 //(say)\n", +"kd = 5.0*ka\n", +"T1 = 25 //in degree C\n", +"T2 = 20 //in degree C\n", +"T5 = -10 //in degree C\n", +"//Sample Problem 19-6\n", +"printf('**Sample Problem 19-6**\n')\n", +"//The rate of thermal conduction will be same through (a) & (d)\n", +"//ka*A*(T2-T1)/La = kd*A*(T5-T4)/Ld\n", +"T4 = T5 - ka*(T2-T1)/La * (Ld/kd)\n", +"printf('The T4 is equal to %d degree C', T4)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 2*10^-2 //in meter\n", +"n = 500 //number of bees\n", +"t = 20*60 //in sec\n", +"e = .80\n", +"T1 = 35 + 273 //in K\n", +"T2 = 47 + 273 //in K\n", +"sigma = 5.67*10^-8 //in W/m^2\n", +"\n", +"//Sample Problem 19-7\n", +"printf('**Sample Problem 19-7**\n')\n", +"A = 4*%pi*R^2\n", +"Eaddi = sigma*e*A*(T2^4 - T1^4)*t //for n bees\n", +"E = Eaddi/n\n", +"printf('The additional energy produced by each bee is equal to %fJ', E)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/2-Motion_Along_a_Straight_Line.ipynb b/Fundamental_Of_Physics_by_D_Haliday/2-Motion_Along_a_Straight_Line.ipynb new file mode 100644 index 0000000..f21bfb6 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/2-Motion_Along_a_Straight_Line.ipynb @@ -0,0 +1,416 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Motion Along a Straight Line" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1_a: Sample_Problem_1a.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"velocity = 70 //in km/h\n", +"distance_covered = 8.4 //in km\n", +"next_time = 30 //in min\n", +"next_walk = 2 //in km\n", +"\n", +"//Sample Problem 2-1a\n", +"printf('**Sample Problem 2-1a**\n')\n", +"overall_displacement = distance_covered + next_walk\n", +"printf('Overall displacement from begining of the drive to the station is %f km', overall_displacement)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1_b: Sample_Problem_1b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Example2_1a.sce', -1)\n", +"clc\n", +"\n", +"//Sample Problem 2-1b\n", +"printf('\n**Sample Problem 2-1b**\n')\n", +"time = distance_covered / velocity //in hr\n", +"delta_t = time + next_time /60 //in hr\n", +"printf('Time interval from the begining of the drive to the arrival at the station is %f hr', delta_t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1_c: Sample_Problem_1c.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Example2_1a.sce', -1)\n", +"clc\n", +"\n", +"//Sample Problem 2-1c\n", +"printf('\n**Sample Problem 2-1c**\n')\n", +"average_velocity = overall_displacement/delta_t\n", +"printf('The average velocity over the whole journey is %f km/h\n', average_velocity)\n", +"\n", +"//from position v/s time graph\n", +"xset('window',1)\n", +"xtitle ('position v/s time','time(hr)','position(Km)');\n", +"//drawing reference lines\n", +"plot(linspace(delta_t,delta_t,10),linspace(0,overall_displacement,10),'--.x')\n", +"plot(linspace(0,delta_t,10),linspace(overall_displacement,overall_displacement,10),'--.o')\n", +"plot(linspace(time,time,10),linspace(0,distance_covered,10),'--')\n", +"plot(linspace(0,time,10),linspace(distance_covered,distance_covered,10),'--')\n", +"//position v/s time graph\n", +"x = linspace(0,time,10);\n", +"y = linspace(0,distance_covered,10);\n", +"plot(x,y,'r');\n", +"//average graph\n", +"x = linspace(time,delta_t,10);\n", +"y = linspace(distance_covered,overall_displacement,10);\n", +"plot(x,y,'r');\n", +"//slope of this line will give us the average velocity\n", +"x = linspace(0,delta_t,10);\n", +"y = linspace(0,overall_displacement,10);\n", +"plot(x,y,'m');\n", +"legend('$\delta x=10.4 km$','time interval=.62hr')\n", +"printf('The average velocity from the graph is %f km/h', 10.4/delta_t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1_d: Sample_Problem_1d.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"exec('Example2_1a.sce', -1)\n", +"exec('Example2_1a.sce', -1)\n", +"clc\n", +"\n", +"//Sample Problem 2-1d\n", +"printf('\n**Sample Probelm 2-1d**\n')\n", +"time_station = 45 //in min\n", +"//he trevels 2 km back to the truck\n", +"final_displacement = overall_displacement + 2 //in km\n", +"final_average_velocity = final_displacement /(delta_t + time_station /60)\n", +"printf('Average velocity from the begining of the drive till reaching back to the truck is %f km/h', final_average_velocity)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 2-2\n", +"printf('**Sample Problem 2-2**\n')\n", +"//we know that v=dx/dt\n", +"xset('window',2)\n", +"//velocity v/s time graph\n", +"subplot(2,1,1);\n", +"xtitle ('VELOCITY v/s TIME','time (sec)','velocity (m/s)');\n", +"plot2d(linspace(0,0,10),linspace(0,0,10),style=3,rect=[0,0,10,5]);\n", +"//drawing the graph\n", +"plot(linspace(0,1,10),linspace(0,0,10),'m');\n", +"plot(linspace(1,3,10),linspace(0,4,10),'m');\n", +"plot(linspace(3,8,10),linspace(4,4,10),'m');\n", +"plot(linspace(8,9,10),linspace(4,0,10),'m');\n", +"plot(linspace(9,10,10),linspace(0,0,10),'m');\n", +"//dotted lines\n", +"plot(linspace(0,3,5),linspace(4,4,5),'--');\n", +"plot(linspace(3,3,5),linspace(0,4,5),'--');\n", +"plot(linspace(8,8,5),linspace(0,4,5),'--');\n", +"\n", +"//acceleration v/s time graph\n", +"subplot(2,1,2);\n", +"xtitle('ACCELERATION v/s TIME','time (sec)', 'acceleration s(m/s^2)')\n", +"plot2d(linspace(0,0,10),linspace(0,0,10),style=0,rect=[0,-5,10,5]);\n", +"//drawing the graph\n", +"plot(linspace(0,1,5),linspace(0,0,5),'m');\n", +"plot(linspace(1,1,5),linspace(0,2,5),'m');\n", +"plot(linspace(1,3,5),linspace(2,2,5),'m');\n", +"plot(linspace(3,3,5),linspace(2,0,5),'m');\n", +"plot(linspace(3,8,5),linspace(0,0,5),'m');\n", +"plot(linspace(8,8,5),linspace(0,-4,5),'m');\n", +"plot(linspace(8,9,5),linspace(-4,-4,5),'m');\n", +"plot(linspace(9,9,5),linspace(-4,0,5),'m');\n", +"plot(linspace(9,10,5),linspace(0,0,5),'m');\n", +"//dotted lines\n", +"plot(linspace(1,1,5),linspace(-5,0,5),'--');\n", +"plot(linspace(3,3,5),linspace(-5,0,5),'--');\n", +"plot(linspace(8,8,5),linspace(-5,-4,5),'-.');\n", +"plot(linspace(9,9,5),linspace(-5,-4,5),'-.');\n", +"plot(linspace(0,1,5),linspace(2,2,5),'--');\n", +"plot(linspace(0,8,5),linspace(-4,-4,5),'--');\n", +"plot(linspace(1,3,5),linspace(0,0,5),'--');\n", +"plot(linspace(8,9,5),linspace(0,0,5),'--');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"t = poly(0,'t');\n", +"x = 7.8 + 9.2 * t - 2.1 * t^3;\n", +"\n", +"//Sample Problem 2-3\n", +"printf('**Sample Problem 2-3**\n')\n", +"v = derivat(x)\n", +"velocity = horner(v,3.5)\n", +"printf('Velocity at t=3.5 second is equal to %f m/s', velocity)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"t = poly(0,'t')\n", +"x = t^3 - 27 * t + 4\n", +"\n", +"//Sample Problem 2-4a\n", +"printf('**Sample Problem 2-4a**\n')\n", +"velocity = derivat(x)\n", +"acceleration = derivat(velocity)\n", +"printf('The velocity as a function of time in m/s is')\n", +"disp(velocity)\n", +"printf('The acceleration as a function of time in m/s^2 is')\n", +"disp(acceleration)\n", +"\n", +"//Sample Problem 2-4b\n", +"printf('\n**Sample Problem 2-4b**\n')\n", +"v0 = roots(velocity)\n", +"printf('velocity of the particle is zero at times in sec')\n", +"disp(v0)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"conv = 5/18 //converts velocity from km/h to in m/s\n", +"speed_initial = 100 * conv //in km/h\n", +"speed_final = 80 * conv //in km/h\n", +"displacement = 88 //in meter\n", +"\n", +"//Sample Problem 2-5a\n", +"printf('**Sample Problem 2-5a\n')\n", +"//using newton's 3rd equation of motion\n", +"acceleration = (speed_final^2 - speed_initial^2)/(2 * displacement)\n", +"printf('The acceleration is equal to %f m/sec^2\n', acceleration)\n", +"\n", +"//Sample Problem2-5b\n", +"printf('\n**Sample Problem 2-5a**\n')\n", +"//using newton's first equation of motion\n", +"time = (speed_final - speed_initial)/acceleration\n", +"printf('The time taken to decrease the speed is %f sec', time)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = -9.8 //in m/sec^2\n", +"displacement = -48 //in meter\n", +"\n", +"//Sample Problem 2-6a\n", +"printf('**Sample Problem 2-6a**\n')\n", +"//using newton's equation of motion\n", +"//displacement = 0 * t + .5 * g * t * t\n", +"//displacement = .5 * g * t * t\n", +"time = sqrt(displacement/(.5 * g))\n", +"printf('The time taken to reach at the ground is %f sec\n', time)\n", +"\n", +"//Sample Problem 2-6b\n", +"printf('\n**Sample Problem 2-6b**\n')\n", +"t = poly(0,'t');\n", +"dis_t = 0 * t + .5 * g *t * t\n", +"ds = horner(dis_t,[1,2,3])\n", +"printf('The displacements at times 1,2,3 sec in meter is')\n", +"disp(ds)\n", +"\n", +"//Sample Problem 2-6c\n", +"printf('\n**Sample Problem 2-6c**\n')\n", +"//using newton's first equation of motion v = u + a*t\n", +"velocity = 0 + g * time\n", +"printf('The velocity at water surface is equal to %f km/h\n', velocity)\n", +"\n", +"//Sample Problem 2-6d\n", +"printf('\n**Sample Problem 2-6d**\n')\n", +"//using newton's first equation of motion\n", +"v_at_time_t = 0 + g * t\n", +"velocities = horner(v_at_time_t,[1,2,3])\n", +"printf('The velocitis at times 1,2,3 sec in m/s is')\n", +"disp(velocities)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = -9.8 //in m/sec^2\n", +"v_initial = 12 //in m/s\n", +"v_final = 0 //at maximum height velocity equal to zero\n", +"\n", +"//Sample Problem 2-7a\n", +"printf('**Sample Problem 2-7a**\n')\n", +"//using newton'd first equation of motion\n", +"// v_final = v_initial + g * t\n", +"h_max_time = (v_final - v_initial)/g\n", +"printf('After %f sec, the ball will attain its maximum height\n', h_max_time)\n", +"\n", +"//Sample Problem 2-7b\n", +"printf('\n**Sample Problem 2-7b**\n')\n", +"//using newton's second equation of motion\n", +"h_max = (v_final^2 - v_initial^2)/(2 * g)\n", +"printf('The maximum height reached by the baseball is %f m\n', h_max)\n", +"\n", +"//Sample Problem 2-7c\n", +"printf('\n**Sample Problem 2-7c**\n')\n", +"displacement = 5\n", +"t = poly(0,'t')\n", +"quad_t = v_initial * t + .5 * g * t * t - displacement\n", +"t_5 = roots(quad_t)\n", +"printf('At following times in sec, the ball will be at height 5m')\n", +"disp(t_5)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/20-The_Kinetic_Theory_of_Gases.ipynb b/Fundamental_Of_Physics_by_D_Haliday/20-The_Kinetic_Theory_of_Gases.ipynb new file mode 100644 index 0000000..54f3fdb --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/20-The_Kinetic_Theory_of_Gases.ipynb @@ -0,0 +1,303 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20: The Kinetic Theory of Gases" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Vi = 12 //in L\n", +"Ti = 20+273 //in K\n", +"Pi = 15 //in atm\n", +"Tf = 35+273 //in K\n", +"Vf = 8.5 //in L\n", +"R = .0821 //in atm.lit/(mol.K)\n", +"\n", +"//Sample Problem 20-1\n", +"printf('**Sample Problem 20-1**\n')\n", +"Pf = (Pi*Vi/(R*Ti))/(Vf/(R*Tf))\n", +"printf('The final pressure of the gas is %fatm', Pf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 1 //in mole\n", +"T = 310 //in K\n", +"Vi = 12 //in L\n", +"Vf = 19 //in L\n", +"R = .0821 //in atm.lit/(mol.K)\n", +"\n", +"//Sample Probelm 20-2\n", +"printf('**Sample Problem 20-2**\n')\n", +"W = n*R*T*log(Vf/Vi) //in atm.lit\n", +"printf('The work done by the gas is equal to %fJ', W*1.0125*10^5*10^-3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = [5, 11, 32, 67, 89]\n", +"\n", +"//Sample Probelm 20-3a\n", +"printf('**Sample Problem 20-3a**\n')\n", +"navg = sum(n)/length(n)\n", +"printf('The average value of numbers is %2.2f\n', navg)\n", +"\n", +"//Sample Probelm 20-3b\n", +"printf('\n**Sample Problem 20-3b**\n')\n", +"nrms = sqrt(sum(n.*n)/5)\n", +"printf('The rms value of numbers is %2.2f', nrms)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 8.31 //in J/mol.K\n", +"Na = 6.023*10^23 //in /mol\n", +"K = R/Na\n", +"\n", +"//Sample Problem 20-4a\n", +"printf('**Sample Problem 20-4a**\n')\n", +"T = 300 //in K\n", +"P = 1*1.0125*10^5 //in pa\n", +"d = 290*10^-12 //in meter\n", +"lambda = K*T/(sqrt(2)*%pi*d^2*P)\n", +"printf('The mean free path is equal to %em\n', lambda)\n", +"\n", +"//Sample Problem 20-4b\n", +"printf('\n**Sample Problem 20-4b**\n')\n", +"v = 450 //in m/s\n", +"t = lambda/v\n", +"printf('The time between two collisions is %esec\n', t)\n", +"f = 1/t\n", +"printf('The frequency of collision is %eHz', f)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"T = 300 //in K\n", +"v1 = 599 //m/s\n", +"v2 = 601 //in m/s\n", +"M = 0.0320 //in kg/mol\n", +"R = 8.31 //J/(mol.K)\n", +"\n", +"//Sample Problem 20-5\n", +"printf('**Sample Problem 20-5**\n')\n", +"v = (v1+v2)/2\n", +"deltav = v2 - v1\n", +"f = 4*%pi*(M/(2*%pi*R*T))^(3/2)*v^2*%e^(-M*v^2/2/R/T)*deltav\n", +"printf('The fraction of molecules having speed in given range is %e', f)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 0.0320 //in kg/mol\n", +"T = 300 //in K\n", +"R = 8.31 //in J/mol.cal\n", +"\n", +"//Sample Problem 20-6a\n", +"printf('**Sample Problem 20-6a**\n')\n", +"Vavg = sqrt(8*R*T/%pi/M)\n", +"printf('The average speed of molecule is %fm/s\n', Vavg)\n", +"\n", +"//Sample Problem 20-6b\n", +"printf('\n**Sample Problem 20-6b**\n')\n", +"Vrms = sqrt(3*R*T/M)\n", +"printf('The rms speed of molecule is %fm/s\n', Vrms)\n", +"\n", +"//Sample Problem 20-6c\n", +"printf('\n**Sample Problem 20-6c**\n')\n", +"Vmps = sqrt(2*R*T/M)\n", +"printf('The most probable speed of molecule is %fm/s', Vmps)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 5.00 //in mol\n", +"deltaT = 20\n", +"R = 8.31 //in J/mol.K\n", +"\n", +"//Sample Problem 20-7a\n", +"printf('**Sample Problem 20-7a**\n')\n", +"//For Helium\n", +"Cv = 3/2*R\n", +"Cp = Cv + R\n", +"Q = n*Cp*deltaT\n", +"printf('The heat added to the bubble is equal to %fJ\n', Q)\n", +"\n", +"//Sample Problem 20-7b\n", +"printf('\n**Sample Problem 20-7b**\n')\n", +"deltaEint = n*Cv*deltaT\n", +"printf('The change in internal energy is equal to %fJ\n', deltaEint)\n", +"\n", +"//Sample Problem 20-7c\n", +"printf('\n**Sample Problem 20-7c**\n')\n", +"W = Q - deltaEint\n", +"printf('The work done by the system is equal to %fJ', W)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 20.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 1 //in mol\n", +"Ti = 310 //in K\n", +"Vi = 12 //in L\n", +"Vf = 19 //in L\n", +"\n", +"//Sample Probelm 20-9a\n", +"printf('**Sample Problem 20-9a**\n')\n", +"gama = 7/5 //for diatomic gases\n", +"Tf = Ti*(Vi/Vf)^(gama-1)\n", +"printf('The final temprature of the gas is %fK\n', Tf)\n", +"\n", +"//Sample Probelm 20-9b\n", +"printf('\n**Sample Problem 20-9b**\n')\n", +"Tf = Ti //Temprature does not change in free expansion\n", +"Pi = 2 //in Pa\n", +"Pf = Pi*Vi/Vf\n", +"printf('The final pressure of the gas is %fPa\n', Pf)\n", +"printf('The final temprature of the gas is %fK', Tf)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/21-Entropy_and_the_Second_Law_of_Thermodynamics.ipynb b/Fundamental_Of_Physics_by_D_Haliday/21-Entropy_and_the_Second_Law_of_Thermodynamics.ipynb new file mode 100644 index 0000000..b5d8a6f --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/21-Entropy_and_the_Second_Law_of_Thermodynamics.ipynb @@ -0,0 +1,205 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21: Entropy and the Second Law of Thermodynamics" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 1 //in mol\n", +"Vi = 1 //(say)\n", +"Vf = 2*Vi\n", +"R = 8.31 //in J/mol.K\n", +"T = 1 //(say)\n", +"\n", +"//Sample Problem 21-1\n", +"printf('**Sample Problem 21-1**\n')\n", +"Q = n*R*T*log(Vf/Vi)\n", +"deltaS = Q/T\n", +"printf('The change in entropy for the irreversible process is equal to %fJ/K', deltaS)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 1.5 //in kg\n", +"TiL = 60 + 273 //in K\n", +"TiR = 20 + 273 //in K\n", +"Tf = 40 + 273 //in K\n", +"Sc = 386 //in J/kg.K\n", +"\n", +"//Sample Problem 21-2\n", +"printf('**Sample Problem 21-2**\n')\n", +"SL = m*Sc*integrate('1/T', 'T', TiL, Tf)\n", +"SR = m*Sc*integrate('1/T', 'T', TiR, Tf)\n", +"Srev = SR + SL\n", +"printf('The net entropy change in the reversible process is equal to %fJ/K', Srev)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"TH = 850 //in K\n", +"TL = 300 //in K\n", +"W = 1200 //in J\n", +"t = 0.25 //in sec\n", +"\n", +"//Sample Problem 21-3a\n", +"printf('**Sample Problem 21-3a**\n')\n", +"eta = 1 - (TL/TH)\n", +"printf('The efficiency of the cycle is equal to %f\n', eta)\n", +"\n", +"//Sample Problem 21-3b\n", +"printf('\n**Sample Problem 21-3b**\n')\n", +"P = W/t\n", +"printf('The average power of the cycle is %fW\n', P)\n", +"\n", +"//Sample Problem 21-3c\n", +"printf('\n**Sample Problem 21-3c**\n')\n", +"QH = W/eta\n", +"printf('The heat extracted from the reservoir is equal to %fJ\n', QH)\n", +"\n", +"//Sample Problem 21-3d\n", +"printf('\n**Sample Problem 21-3d**\n')\n", +"QL = W - QH\n", +"printf('The heat delivered to the reservoir is equal to %fJ\n', QL)\n", +"\n", +"//Sample Problem 21-3e\n", +"printf('\n**Sample Problem 21-3e**\n')\n", +"S = QH/TH + QL/TL\n", +"printf('The net entropy change for the cycle is %fJ/k', S)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"eta = 0.75\n", +"TH = 100 + 273\n", +"TL = 0 + 273\n", +"\n", +"//Sample Problem 21-5\n", +"printf('**Sample Problem 21-5**\n')\n", +"etaMAX = 1 - TL/TH\n", +"if eta < etaMAX then\n", +" printf('It is possible')\n", +"else \n", +" printf('It is not possible')\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 21.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 100\n", +"\n", +"//Sample Problem 21-5a\n", +"printf('**Sample Problem 21-5a**\n')\n", +"n1 = 50\n", +"n2 = 50\n", +"W = factorial(n)/(factorial(n1)*factorial(n2))\n", +"printf('The total number of possible configuration is %e\n', W)\n", +"\n", +"//Sample Problem 21-5b\n", +"printf('\n**Sample Problem 21-5b**\n')\n", +"n1 = 100\n", +"n2 = 0\n", +"W = factorial(n)/(factorial(n1)*factorial(n2))\n", +"printf('The total number of possible configuration is %e', W)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/22-Electric_Charge.ipynb b/Fundamental_Of_Physics_by_D_Haliday/22-Electric_Charge.ipynb new file mode 100644 index 0000000..2d4e947 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/22-Electric_Charge.ipynb @@ -0,0 +1,188 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22: Electric Charge" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"q1 = 1.60*10^-19 //in C\n", +"q2 = 3.20*10^-19 //in C\n", +"R = 0.0200 //in meter\n", +"\n", +"//Sample Problem 22-1a\n", +"printf('**Sample Problem 22-1a**\n')\n", +"F12 = coulomb(q1, q2, R)\n", +"printf('The coulombic force between the charged particle is %eN\n', F12)\n", +"\n", +"//Sample Problem 22-1b\n", +"printf('\n**Sample Problem 22-1b**\n')\n", +"q3 = -3.20*10^-19 //in C\n", +"R13 = 3/4*R\n", +"F1 = coulomb(q1, q3, R13) + F12\n", +"printf('The net force on particle 1 is equal to %eN\n', abs(F1))\n", +"\n", +"//Sample Problem 22-1c\n", +"printf('\n**Sample Problem 22-1c**\n')\n", +"q4 = -3.20*10^-19 //in C\n", +"R14 = 3/4*R\n", +"theta = dtor(60)\n", +"F14 = coulomb(q1, q4, R14)\n", +"F1net = [F12+F14*cos(theta), F14*sin(theta)]\n", +"printf('The net force on particle 1 is equal to %eN', norm(F1net))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"q = 1 //(say)\n", +"q1 = 8*q\n", +"q2 = -2*q\n", +"L = 1 //(say)\n", +"x = L\n", +"\n", +"//Sample Problem 22-2\n", +"printf('**Sample Problem 22-2**\n')\n", +"//let the distance of proton from q1 is y\n", +"//We know that y>L\n", +"y = poly(0, 'y')\n", +"F1 = coulomb(q1, e, y)\n", +"F2 = coulomb(q2, e, y-L)\n", +"p = F1 + F2\n", +"r = roots(denom(inv(p)))\n", +"printf('the proton should be placed at a distance %1.1fL from q1', r(2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"Q = 1 //(say)\n", +"q1 = Q\n", +"q2 = 0\n", +"a = 1 //(say)\n", +"\n", +"//Sample Problem 22-3a\n", +"printf('**Sample Problem 22-3a**\n')\n", +"q1 = (q1 + q2)/2\n", +"q2 = q1\n", +"F = coulomb(q1, q2, a)\n", +"printf('The net force between the two sphere is %eQ^2/a^2\n', F)\n", +"\n", +"\n", +"//Sample Problem 22-3b\n", +"printf('\n**Sample Problem 22-3b**\n')\n", +"q1 = 0\n", +"F = coulomb(q1, q2, a)\n", +"printf('The net force between the two sphere is %fN', F)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 22.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"exec('gravitation.sci', -1)\n", +"\n", +"//Given that\n", +"r = 4*10^-15 //in meter\n", +"n = 26\n", +"mp = 1.67*10^-27 //in kg\n", +"\n", +"//Sample Problem 22-4a\n", +"printf('**Sample Problem 22-4a**\n')\n", +"Fpp = coulomb(e, e, r)\n", +"printf('The force of repulsion is %eN\n', Fpp)\n", +"\n", +"//Sample Problem 22-4b\n", +"printf('\n**Sample Problem 22-4b**\n')\n", +"Fg = GForce(mp, mp, r)\n", +"printf('The magnitude of gravitational force between two proton is %eN', Fg)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/23-Electric_Fields.ipynb b/Fundamental_Of_Physics_by_D_Haliday/23-Electric_Fields.ipynb new file mode 100644 index 0000000..dfb1f7e --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/23-Electric_Fields.ipynb @@ -0,0 +1,169 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23: Electric Fields" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"Q = 1 //(say)\n", +"q1 = +2*Q\n", +"q2 = -2*Q\n", +"q3 = -4*Q\n", +"d = 1 //(say)\n", +"theta = dtor(30)\n", +"\n", +"//Sample Problem 23-2\n", +"printf('**Sample Problem 23-2**\n')\n", +"E1 = coulomb(q1, 1, d)*[cos(theta), sin(theta)]\n", +"E2 = coulomb(q2, 1, d)*[-cos(theta), sin(theta)]\n", +"E3 = coulomb(q3, 1, d)*[-cos(theta), -sin(theta)]\n", +"E = E1 + E2 + E3\n", +"printf('The net electric field at origin is equal to %eQ/d^2 N/C', norm(E))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"Q = 1 //(say)\n", +"A = dtor(120)\n", +"r = 1 //(say)\n", +"\n", +"//Sample Problem 23-3\n", +"printf('**Sample Problem 23-3**\n')\n", +"Eunit_angle = coulomb(Q/A, 1, r)\n", +"//overall only x-component will survive\n", +"E = integrate('Eunit_angle*cos(theta)', 'theta', -A/2, A/2)\n", +"printf('The electric field due to arc at point P is equal to %eQ/r^2 N/C', E)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 1.3*10^-10 //in kg\n", +"Q = 1.5*10^-13 //in C\n", +"Vx = 18 //in m/s\n", +"L = 1.6*10^-2 //in meter\n", +"E = 1.4*10^6 //in N/C\n", +"\n", +"//Sample Problem 23-4\n", +"printf('**Sample Problem 23-4**\n')\n", +"a = E*Q/m\n", +"t = L/Vx\n", +"dv = 0.5*a*t^2 //vertical drop\n", +"printf('The vertical drop of the drop is equal to %fmm', dv*10^3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 23.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"P = 6.2*10^-30 //in C.m\n", +"\n", +"//Sample Problem 23-5a\n", +"printf('**Sample Problem 23-5a**\n')\n", +"q = 10*e\n", +"l = P/q\n", +"printf('The effective ditance between the positive & negative center is %em\n', l)\n", +"\n", +"//Sample Problem 23-5b\n", +"printf('\n**Sample Problem 23-5b**\n')\n", +"E = 1.5*10^4 //in N/C\n", +"T = 2*E*q*l/2\n", +"printf('The net torque on the dipole is %eN.m\n', T)\n", +"\n", +"//Sample Problem 23-5c\n", +"printf('\n**Sample Problem 23-5c**\n')\n", +"W = -(P*E*(cos(dtor(180)) - cos(0)))\n", +"printf('The work done by the external agent is equal to %eJ', W)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/24-Gauss_Law.ipynb b/Fundamental_Of_Physics_by_D_Haliday/24-Gauss_Law.ipynb new file mode 100644 index 0000000..6ab88a7 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/24-Gauss_Law.ipynb @@ -0,0 +1,229 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 24: Gauss Law" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"R = 1 //(say)\n", +"E = 1 //(say)\n", +"A = 1 //cuve surface area of cylinder(say)\n", +"\n", +"//Sample Problem 24-1\n", +"printf('**Sample Problem 24-1**\n')\n", +"flux = E*A + (-E*A) + E*A*cos(dtor(90))\n", +"printf('The net flux passing through the cylinder is equal to %fN.m^2/C', flux)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"x = poly(0, 'x')\n", +"E = [3*x, 4, 0]\n", +"x1 = 1.0 //in m\n", +"x2 = 3.0 //in m\n", +"y1 = 0.0 //in m\n", +"y2 = 2.0 //in m\n", +"z1 = 0.0 //in m\n", +"z2 = 2.0 //in m\n", +"\n", +"//Sample Problem 24-2\n", +"printf('**Sample Problem 24-2**\n')\n", +"//top face\n", +"A = [0; 2; 0] //area per unit x\n", +"Phi_top = integrate('[3*x, 4, 0]*A', 'x', x1, x2)\n", +"printf('The flux through the top face is equal to %fN.m^2/C\n', Phi_top)\n", +"//left face\n", +"A = [-2*2; 0; 0]\n", +"Phi_left = horner(E, x1)*A\n", +"printf('The flux through the left face is equal to %fN.m^2/C\n', Phi_left)\n", +"//Right face\n", +"A = [2*2; 0; 0]\n", +"Phi_right = horner(E, x2)*A\n", +"printf('The flux through the right face is equal to %fN.m^2/C', Phi_right)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"q1 = +3.1*10^-9 //in C\n", +"q4 = q1\n", +"q2 = -5.9*10^-9 //in C\n", +"q5 = q2\n", +"q3 = -3.1*10^-9 //in C\n", +"\n", +"//Sample Problem 24-3\n", +"printf('**Sample Problem 24-3**\n')\n", +"//Using gauss law\n", +"flux = (q1+q2+q3)/Eo\n", +"printf('The flux through the surface is equal to %fN.m^2/C', flux)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"q = -5 //in micro coulomb\n", +"\n", +"//Sample Problem 24-4\n", +"printf('**Sample Problem 24-4**\n')\n", +"qin = -q\n", +"qout = -qin\n", +"printf('Charge on the inner surface is equal to %dmicroCoulomb\n', qin)\n", +"printf('Charge on the outer surface is equal to %dmicroCoulom', qout)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"lambda = -1*10^-3 //in C/m\n", +"Eb = 3*10^6 //in N/C\n", +"\n", +"//Sample Problem 24-5\n", +"printf('**Sample Problem 24-5**\n')\n", +"r = lambda/(2*%pi*Eo*Eb)\n", +"printf('The radius of the column is equal to %fm', abs(r))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 24.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"sigmaP = 6.8*10^-6 //in C.m^2\n", +"sigmaN = 4.3*10^-6 //in C.m^2\n", +"\n", +"Ep = sigmaP/(2*Eo) //field due to positive plate\n", +"En = sigmaN/(2*Eo) //field due to negative plate\n", +"//Sample Problem 24-6a\n", +"printf('**Sample Problem 24-6a**\n')\n", +"El = En - Ep\n", +"printf('Electric field on the left of the sheets is equal to %eN/C\n', El)\n", +"\n", +"//Sample Problem 24-6b\n", +"printf('\n**Sample Problem 24-6b**\n')\n", +"Eb = En + Ep\n", +"printf('Field in between is equal to %eN/C\n', Eb)\n", +"\n", +"//Sample Problem 24-6c\n", +"printf('\n**Sample Problem 24-6c**\n')\n", +"Er = -En + Ep\n", +"printf('Field in between is equal to %eN/C', Er)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/25-Electric_Potential.ipynb b/Fundamental_Of_Physics_by_D_Haliday/25-Electric_Potential.ipynb new file mode 100644 index 0000000..10795ad --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/25-Electric_Potential.ipynb @@ -0,0 +1,155 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25: Electric Potential" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"E = 150 //in N/C\n", +"d = -520 //in meter\n", +"\n", +"//Sample Problem 25-1\n", +"printf('**Sample Problem 25-1**\n')\n", +"W = E*(-e)*d\n", +"deltaU = -W\n", +"printf('The change in potential energy is %eJ', deltaU)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"d = 1.3 //in meter\n", +"q1 = 12*10^-9 //in C\n", +"q2 = -24*10^-9 //in C\n", +"q3 = +31*10^-9 //in C\n", +"q4 = 17*10^-9 //in C\n", +"\n", +"//Sample Problem 25-3\n", +"printf('**Sample Problem 25-3**\n')\n", +"V = EPotential(q1+q2+q3+q4, d/sqrt(2))\n", +"printf('The electric potential at point P is equal to %eV', V)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"n = 12\n", +"qT = -n*e\n", +"R = 1 //(say)\n", +"\n", +"//Sample Problem 25-4a\n", +"printf('**Sample Problem 25-4a**\n')\n", +"V = EPotential(qT, R)\n", +"printf('The electric potential at the center is equal to %e/R Volts', V)\n", +"\n", +"//Sample Problem 25-4b\n", +"printf('**Sample Problem 25-4b**\n')\n", +"printf('It does not change in this configuration')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"d = 12*10^-2 //in cm\n", +"q = 150*10^-9 //in C\n", +"q1 = +q\n", +"q2 = -4*q\n", +"q3 = +2*q\n", +"\n", +"//Sample Problem 25-6\n", +"printf('**Sample Problem 25-6**\n')\n", +"U = EPotential(q1, d)*q2 + EPotential(q2, d)*q3 + EPotential(q3, d)*q1\n", +"printf('The electric potential energy of the system is equal to %eJ', U)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/26-Capacitance.ipynb b/Fundamental_Of_Physics_by_D_Haliday/26-Capacitance.ipynb new file mode 100644 index 0000000..a2d747a --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/26-Capacitance.ipynb @@ -0,0 +1,251 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 26: Capacitance" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"C = 55*10^-15 //in F\n", +"V = 5.3 //in V\n", +"\n", +"//Sample Problem 26-1\n", +"printf('**Sample Problem 26-1**\n')\n", +"Q = C*V\n", +"n = Q/e\n", +"printf('The number of excess electron is equal to %e', n)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"C1 = 12 //in uF\n", +"C2 = 5.30 //in uF\n", +"C3 = 4.50 //in uF\n", +"V = 12.5 //in Volts\n", +"\n", +"//Sample Problem 26-2a\n", +"printf('**Sample Problem 26-2a**\n')\n", +"C12 = C1 + C2 //in series\n", +"C123 = C12*C3/(C12 + C3) //in parallel\n", +"printf('The equivalent capacitance for the given circuit is %fuF\n', C123)\n", +"\n", +"//Sample Problem 26-2b\n", +"printf('\n**Sample Problem 26-2b**\n')\n", +"Q123 = C123*V\n", +"Q12 = Q123 //in series\n", +"Q1 = Q12*C1/(C1+C2)\n", +"printf('The charge on the capacitor C1 is equal to %fuC', Q1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"C1 = 3.55 //in uF\n", +"Vo = 6.30 //in Volts\n", +"C2 = 8.95 //in uF\n", +"\n", +"//Sample Problem 26-3\n", +"printf('**Sample Problem 26-3**\n')\n", +"qT = C1*Vo //Total charge\n", +"q1 = qT*C1/(C1+C2) //in parallel\n", +"V = q1/C1\n", +"printf('The final potential difference between each capacitor is equal to %fV', V)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"R = 6.85*10^-2 //in m\n", +"q = 1.25*10^-9 //in C\n", +"\n", +"//Sample Problem 26-4a\n", +"printf('**Sample Problem 26-4a**\n')\n", +"C = 4*%pi*Eo*R\n", +"U = q^2/(2*C)\n", +"printf('The electric energy stored is equal to %eJ\n', U)\n", +"\n", +"//Sample Problem 26-4b\n", +"printf('\n**Sample Problem 26-4b**\n')\n", +"E = coulomb(q, 1, R)\n", +"u = 1/2*Eo*E^2\n", +"printf('The energy density is equal to %eJ/m^3', u)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"C = 13.5*10^-12 //in F\n", +"V = 12.5 //in Volts\n", +"x = 6.50\n", +"\n", +"//Sample Problem 26-5\n", +"printf('**Sample Problem 26-5**\n')\n", +"q = C*V\n", +"Ui = q^2/(2*C)\n", +"printf('The initial stored energy is equal to %eJ\n', Ui)\n", +"C = x*C\n", +"Uf = q^2/(2*C)\n", +"printf('The energy stored after the slab is inserted is equal to %eJ', Uf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 26.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('electrostatics.sci', -1)\n", +"\n", +"//Given that\n", +"A = 115*10^-4 //in m^2\n", +"d = 1.24*10^-2 //in meter\n", +"Vo = 85.5 //in Volts\n", +"b = 0.780*10^-2 //in meter\n", +"x = 2.61\n", +"\n", +"//Sample Problem 26-6a\n", +"printf('**Sample Problem 26-6a**\n')\n", +"Co = A*Eo/d\n", +"printf('The capacitance of the plates before the dielectric slab is inserted is equal to %fpF\n', Co*10^12)\n", +"\n", +"//Sample Problem 26-6b\n", +"printf('\n**Sample Problem 26-6b**\n')\n", +"Q = Co*Vo\n", +"printf('Free charge on the plates is equal to%fpC\n', Q*10^12)\n", +"\n", +"//Sample Problem 26-6c\n", +"printf('\n**Sample Problem 26-6c**\n')\n", +"E = Q/(A*Eo)\n", +"printf('The electric field is equal to %fV/m\n', E)\n", +"\n", +"//Sample Problem 26-6d\n", +"printf('\n**Sample Problem 26-6d**\n')\n", +"E1 = Q/(A*Eo*x)\n", +"printf('The electric field in dielectric slab is equal to %fV/m\n', E1)\n", +"\n", +"//Sample Problem 26-6e\n", +"printf('\n**Sample Problem 26-6e**\n')\n", +"V = E*(d-b) + E1*b\n", +"printf('The new potential difference is equal to %fV\n', V)\n", +"\n", +"//Sample Problem 26-6f\n", +"printf('\n**Sample Problem 26-6f**\n')\n", +"C = Q/V\n", +"printf('The new capacitance is equal to %fpF', C*10^12)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/27-Current_and_Resistance.ipynb b/Fundamental_Of_Physics_by_D_Haliday/27-Current_and_Resistance.ipynb new file mode 100644 index 0000000..29da3f6 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/27-Current_and_Resistance.ipynb @@ -0,0 +1,224 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 27: Current and Resistance" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"rate = 450*10^-6 //in m^3/s\n", +"e = 1.6*10^-19\n", +"Na = 6.023*10^23\n", +"M = 18*10^-3 //in kg/mol\n", +"density = 1000 //in kg/m^3\n", +"\n", +"//Sample Problem 27-1\n", +"printf('**Sample Problem 27-1**\n')\n", +"n = 10\n", +"i = e*n*Na/M*density*rate\n", +"printf('The current of negative charge is equal to %eA', i)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 2*10^-3 //in meter\n", +"J = 2*10^5 //in A/m^2\n", +"\n", +"//Sample Problem 27-2a\n", +"printf('**Sample Problem 27-2a**\n')\n", +"//As current density is uniform\n", +"A = %pi*(R^2 - (R/2)^2)\n", +"I = J*A\n", +"printf('The current flowing through the outer portion is %fA\n', I)\n", +"\n", +"//Sample Problem 27-2b\n", +"printf('\n**Sample Problem 27-2b**\n')\n", +"a = 3*10^11 //in SI unit\n", +"Iv = integrate('a*r^2*2*%pi*r', 'r', R/2, R)\n", +"printf('Now the current will be %fA', Iv)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r = 900*10^-6 //in m\n", +"i = 17*10^-3 //in A\n", +"e = 1.6*10^-19 //in C\n", +"densityCopper = 8.96*10^3 //in kg/m^3\n", +"M = 63.54*10^-3 //in kg/mol\n", +"Na = 6.023*10^23\n", +"\n", +"//Sample Problem 27-3\n", +"printf('**Sample Problem 27-3**\n')\n", +"A = %pi*r^2\n", +"J = i/A\n", +"n = densityCopper/M*Na\n", +"Vd = J/(n*e)\n", +"printf('The drift speed is %em/s', Vd)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"l = 1.2*10^-2 //in meter\n", +"b = 1.2*10^-2 //in meter\n", +"h = 15*10^-2 //in meter\n", +"resistivityIron = 9.68*10^-8 //in ohm.m\n", +"\n", +"//Sample Problem 27-4(1)\n", +"printf('**Sample Problem 27-4(1)**\n')\n", +"R1 = resistivityIron*h/(l*b)\n", +"printf('The resistance of the block is equal to %eOhm\n', R1)\n", +"\n", +"//Sample Problem 27-4(2)\n", +"printf('\n**Sample Problem 27-4(2)**\n')\n", +"R2 = resistivityIron*l/(b*h)\n", +"printf('The resistance of the block is equal to %eOhm', R2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"e = 1.6*10^-19 //in C\n", +"Re = 1.69*10^-8 //in Ohm.m\n", +"n = 8.49*10^28\n", +"m = 9.1*10^-31 //mass of electron in kg\n", +"Veff = 1.6*10^6 //in m/s\n", +"\n", +"//Sample Problem 27-5a\n", +"printf('**Sample Problem 27-5a**\n')\n", +"//resistivity = m/(n*e^2*t)\n", +"t = m/(n*e^2*Re)\n", +"printf('The mean free time between two collision is %es\n', t)\n", +"\n", +"//Sample Problem 27-5b\n", +"printf('\n**Sample Problem 27-5b**\n')\n", +"lambda = Veff*t\n", +"printf('The mean free path is equal to %em', lambda)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 27.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 72 //in Ohm\n", +"V = 120 //in volts\n", +"\n", +"//Sample Problem 27-6\n", +"printf('**Sample Problem 27-6**\n')\n", +"H1 = V^2/R\n", +"printf('The Power dissipated in first case is equal to %dW\n', H1)\n", +"H2 = V^2/(R/2) * 2\n", +"printf('The Power dissipated in second case is equal to %dW\n', H2)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/28-Circuits.ipynb b/Fundamental_Of_Physics_by_D_Haliday/28-Circuits.ipynb new file mode 100644 index 0000000..dde1bf5 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/28-Circuits.ipynb @@ -0,0 +1,218 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 28: Circuits" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"E1 = 4.4 //in Volts\n", +"E2 = 2.1 //in olts\n", +"r1 = 2.3 //in Ohm\n", +"r2 = 1.8 //in Ohm\n", +"R = 5.5 //in Ohm\n", +"\n", +"//Sample Problem 28-1a\n", +"printf('**Sample Problem 28-1a**\n')\n", +"i = poly(0, 'i')\n", +"p = E1 - E2 - i*r2 - i*R - i*r1\n", +"i = roots(p)\n", +"printf('The current in the circuit is equal to %fA\n', i)\n", +"\n", +"//Sample Problem 28-1b\n", +"printf('\n**Sample Problem 28-1b**\n')\n", +"Ve1 = E1 - i*r1\n", +"printf('The potential difference between the terminal of the battery is equal to %fV', Ve1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R1 = 20 //in Ohm\n", +"R2 = 20 //in Ohm\n", +"R3 = 30 //in Ohm\n", +"R4 = 8 //in Ohm\n", +"E = 12 //in Volts\n", +"\n", +"//Sample Problem 28-2a\n", +"printf('**Sample Problem 28-2a**\n')\n", +"R23 = R2*R3/(R2+R3)\n", +"Req = R1 + R23 + R4\n", +"i = poly(0, 'i')\n", +"i = E/Req\n", +"printf('The current through the battery is %fA\n', i)\n", +"\n", +"//Sample Problem 28-2b\n", +"printf('\n**Sample Problem 28-2b**\n')\n", +"i2 = i*R23/R2\n", +"printf('The current through R2 is %fA\n', i2)\n", +"\n", +"//Sample Problem 28-2c\n", +"printf('\n**Sample Problem 28-2c**\n')\n", +"i3 = i2*R2/R3\n", +"printf('The current through R3 is %fA', i3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"E1 = 3.0 //in Volts\n", +"E2 = 6.0 //in Volts\n", +"R1 = 2.0 //in Ohm\n", +"R2 = 4.0 //in Ohm\n", +"\n", +"//Sample Problem 28-3\n", +"printf('**Sample Problem 28-3**\n')\n", +"function [f] = circuit(i)\n", +" f = zeros(2, 1)\n", +" //Using KVL in both the loops\n", +" f(1) = -i(1)*R1 - E1 - i(1)*R1 + E2 + i(2)*R2\n", +" f(2) = E2 + i(2)*R2 + (i(1)+i(2))*R1 - E2 + (i(1)+i(2))*R1\n", +"endfunction\n", +"i = fsolve([0,0], circuit)\n", +"printf('i1 = %fA\n', i(1))\n", +"printf('i2 = %fA\n', i(2))\n", +"printf('i3 = %fA', sum(i))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"rows = 140\n", +"n = 5000\n", +"E = 0.15 //in Volts\n", +"r = 0.25 //in Ohm\n", +"Rw = 800 //in Ohm\n", +"\n", +"//Sample Problem 28-4a\n", +"printf('**Sample Problem 28-4a**\n')\n", +"Rrow = n*r\n", +"Req = Rrow/rows\n", +"Erow = n*E\n", +"Eeq = (Erow/Rrow * rows)/(rows/Rrow)\n", +"I = Eeq/(Rw + Req)\n", +"printf('The magnitude of current produced in the water is %fA\n', i(1))\n", +"\n", +"//Sample Problem 28-4b\n", +"printf('\n**Sample Problem 28-4b**\n')\n", +"Irow = I/rows\n", +"printf('Current in each row is equal to %fA', Irow)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 28.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"C = 1 //(say)\n", +"R = 1 //(say)\n", +"\n", +"//Sample Problem 28-5a\n", +"printf('**Sample Problem 28-5a**\n')\n", +"//q = qo*e^(-t/(R*C))\n", +"//q = qo/2 when t = Tq\n", +"Tq = log(2)*R*C\n", +"printf('At t=%fT, the capacitor will be half charged\n', Tq)\n", +"\n", +"//Sample Problem 28-5b\n", +"printf('\n**Sample Problem 28-5b**\n')\n", +"//U = Uo*e^(-2*t(R*C))\n", +"//U = Uo/2 When t = Tu\n", +"Tu = log(2)/2*R*C\n", +"printf('At t=%fT, the enrgy stored will be half of its MAX value', Tu)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/29-Magnetic_fields.ipynb b/Fundamental_Of_Physics_by_D_Haliday/29-Magnetic_fields.ipynb new file mode 100644 index 0000000..0fbfebe --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/29-Magnetic_fields.ipynb @@ -0,0 +1,287 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 29: Magnetic fields" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"B = 1.2*10^-3 //in T\n", +"e = 1.6*10^-19 //in C\n", +"K = 5.3*10^6*e //in J\n", +"m = 1.6*10^-27 //in kg\n", +"\n", +"//Sample Problem 29-1\n", +"printf('**Sample Problem 29-1**\n')\n", +"v = sqrt(2*K/m)\n", +"F = e*v*B\n", +"printf('The magnitude of magnetic force acting on the proton is %eN', F)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"d = 1.5*10^-2 //in meter\n", +"v = 4.0 //in m/s\n", +"B = 0.050 //in T\n", +"\n", +"//Sample Problem 29-2a\n", +"printf('**Sample Problem 29-2a**\n')\n", +"//force is in right direction\n", +"printf('The right surface will be at high potential\n')\n", +"\n", +"//Sample Problem 29-2b\n", +"printf('\n**Sample Problem 29-2b**\n')\n", +"F = v*B //force per unit charge\n", +"deltaU = d*F //energy per unit charge = potential difference\n", +"printf('The potential difference between the two surface is %1.1eV', deltaU)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"B = 80*10^-3 //in T\n", +"V = 1000.0 //in V\n", +"q = 1.6022*10^-19 //in C\n", +"x = 1.6254 //in m\n", +"conv = 1.6605*10^-27 //in kg/u\n", +"\n", +"//Sample Problem 29-3\n", +"printf('**Sample Problem 29-3**\n')\n", +"m = poly(0, 'm')\n", +"r = x/2\n", +"//r = m*v/(q*B)\n", +"//v = q*B*r/m\n", +"m = 0.5*(q*r*B)^2/V/q\n", +"printf('The mass of the particle is %ekg', m)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"e = 1.6*10^-19\n", +"K = 22.5*e //in J\n", +"B = 4.55*10^-4 //in T\n", +"theta = dtor(65.5)\n", +"m = 9.11*10^-31 //in kg\n", +"\n", +"//Sample Problem 29-4\n", +"printf('**Sample Problem 29-4**\n')\n", +"q = e\n", +"v = sqrt(2*K/m)\n", +"r = m*v*sin(theta)/(q*B)\n", +"T = 2*%pi*r/(v*sin(theta))\n", +"p = v*cos(theta)*T\n", +"printf('The pitch of the electron is equal to %fm', p)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"f = 12*10^6 //in Hz\n", +"R = 53*10^-2 //in m\n", +"q = 1.6*10^-19 //in C\n", +"m = 3.34*10^-27 //in kg\n", +"\n", +"//Sample Problem 29-5a\n", +"printf('**Sample Problem 29-5a**\n')\n", +"B = 2*%pi*m*f/q\n", +"printf('The magnitude of magnetic field should be %fT\n', B)\n", +"\n", +"//Sample Problem 29-5b\n", +"printf('\n**Sample Problem 29-5b**\n')\n", +"v = q*B*R/m\n", +"K = 0.5*m*v^2\n", +"printf('The kinetic energy of the deuteron will be %eJ', K)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"i = 28 //in A\n", +"LD = 46.6*10^-3 //in kg/m\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 29-6\n", +"printf('**Sample Problem 29-6**\n')\n", +"L = 1 //(say)\n", +"m = LD*L\n", +"B = m*g/i/L\n", +"printf('The minimum magnetic field required to suspend the wire is equal to %eT', B)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"h = 2.1*10^-2 //in m\n", +"w = 1.2*10^-2 //in m\n", +"n = 250\n", +"B = 0.23 //in T\n", +"i = 100*10^-6 //in A\n", +"theta = 28 //in degree\n", +"\n", +"//Sample Problem 29-7\n", +"printf('**Sample Problem 29-7**\n')\n", +"A = h*w\n", +"T = B*i*n*A\n", +"k = T/theta\n", +"printf('The value of constant k is equal to %eN.m/degree', k)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 29.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 250\n", +"A = 2.52*10^-4 //in m^2\n", +"i = 100*10^-6 //in A\n", +"B = 0.85 //in T\n", +"\n", +"//Sample Problem 29-8\n", +"printf('**Sample Problem 29-8**\n')\n", +"mu = n*i*A\n", +"U1 = -mu*B*cos(%pi/2)\n", +"U2 = -mu*B*cos(0)\n", +"W = U1 - U2\n", +"printf('Work done is equal to %eJ', W)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/3-Vectors.ipynb b/Fundamental_Of_Physics_by_D_Haliday/3-Vectors.ipynb new file mode 100644 index 0000000..cf6af25 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/3-Vectors.ipynb @@ -0,0 +1,275 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: Vectors" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"a = [2,0]\n", +"b = [2 *cos(dtor(30)),2 *sin(dtor(30))]\n", +"c = [-1,0]\n", +"\n", +"//Sample Problem 3-1\n", +"printf('**Sample Problem 3-1**\n')\n", +"poss = [norm(a+b+c) norm(a-b+c), norm(a+b-c), norm(a-b-c)]\n", +"max_norm = 0\n", +"for v = poss\n", +" if v > max_norm then max_norm = v \n", +" end\n", +"end\n", +"printf('The maximum possible value is %f m', max_norm)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"dis = 215 //in km\n", +"position = [dis * cos(dtor(22)), dis * sin(dtor(22))]\n", +"\n", +"//Sample Problem 3-2\n", +"printf('**Sample Problem 3-2**\n')\n", +"printf('The plane is %f km in the north & %f in the east', position(1),position(2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"displacement_vector = [-2.6,-3.9,.025] //each in km\n", +"\n", +"//Sample Problem 3-3\n", +"printf('**Sample Problem 3-3**\n')\n", +"mag = norm(displacement_vector)\n", +"sw_angle = atan(displacement_vector(2)/displacement_vector(1))\n", +"up_angle = displacement_vector(3)/norm(displacement_vector)\n", +"printf('The team displacement vector had a magnitude %f km,\n and was at an angle of %d south of west and\n at an angle of %f upward', mag, rtod(sw_angle), rtod(up_angle))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"a = [4.2,-1.5]\n", +"b = [-1.6,2.9]\n", +"c = [0,-3.7]\n", +"\n", +"//Sample Problem 3-4\n", +"printf('**Sample Problem 3-4**\n')\n", +"r = a + b + c\n", +"magnitude = norm(r)\n", +"angle = rtod(atan(r(2)/r(1)))\n", +"printf('The magnitude of the vector is %f m & the angle measured from the x axis is %f', magnitude, (angle) )" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"a = [36,0] //in km\n", +"c = [25 *cos(dtor(135)), 25 *sin(dtor(135))] //in km\n", +"d_mag = 62 //in km\n", +"\n", +"//Sample Problem 3-5\n", +"printf('**Sample Problem 3-5**\n')\n", +"//we have a + b + c = d\n", +"//therefore ax = bx + cx + dx\n", +"// bx = 0\n", +"d_x = a(1) + c(1)\n", +"d_y = d_mag * sqrt(1 - (d_x/d_mag)^2)\n", +"d = [d_x, d_y]\n", +"b = d(2) - a(2) - c(2)\n", +"printf('The magnitude of b is equal to %f km', b)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"a = [3,-4,0]\n", +"b = [-2,0,3]\n", +"\n", +"//Sample Problem 3-6\n", +"printf('**Sample Problem 3-6**\n')\n", +"angle_ab = acos(-norm(a*b')/(norm(a) * norm(b)))\n", +"printf('The angle between given vectors is %f degress', rtod(angle_ab))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"exec('cross_product.sci',-1)\n", +"\n", +"//Given that\n", +"a = [18 * cos(dtor(250)), 18 * sin(dtor(250)),0]\n", +"b = [0,0,12]\n", +"\n", +"//Sample Problem 3-7\n", +"printf('**Sample Problem 3-7**\n')\n", +"cross_ab = crossproduct(a,b)\n", +"angle_x = acos(cross_ab(1)/norm(cross_ab))\n", +"printf('The magnitude of cross product of given vectors is %f \n and angle with the x axis in degrees is %f', norm(cross_ab),rtod(angle_x))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"exec('cross_product.sci',-1)\n", +"\n", +"//Given that\n", +"a = [3,-4,0]\n", +"b = [-2,0,3]\n", +"\n", +"//Sample Problem 3-8\n", +"printf('**Sample Problem 3-8**\n')\n", +"cross_ab = crossproduct(a,b)\n", +"printf('The cross product of given vectors is ')\n", +"disp(cross_ab)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/30-Magnetic_fields_due_to_Current.ipynb b/Fundamental_Of_Physics_by_D_Haliday/30-Magnetic_fields_due_to_Current.ipynb new file mode 100644 index 0000000..ea971f3 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/30-Magnetic_fields_due_to_Current.ipynb @@ -0,0 +1,128 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 30: Magnetic fields due to Current" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"i1 = 15 //in A\n", +"i2 = 32 //in A\n", +"d = 5.3*10^-2 //in m\n", +"uo = 4*%pi*10^-7 //in SI unit\n", +"\n", +"//Sample Problem 30-2\n", +"printf('**Sample Problem 30-2**\n')\n", +"R = d/sqrt(2)\n", +"B1 = uo*i1/(2*%pi*R)\n", +"B2 = uo*i2/(2*%pi*R)\n", +"theta = atan(B1/B2)\n", +"B = sqrt(B1^2 + B2^2)\n", +"printf('The magnitude of net magnetic field is %eT\n', B)\n", +"printf('The angle made by the net magnetic field by the positive x-axis is %fdegrees', rtod(theta)+45)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"a = 2*10^-2 //in meter\n", +"b = 4*10^-2 //in meter\n", +"r = 3*10^-2 //in meter\n", +"c = 3*10^6 //in A/m^4\n", +"uo = 4*%pi*10^-7 //in SI unit \n", +"\n", +"//Sample Problem 30-3\n", +"printf('**Sample Problem 30-3**\n')\n", +"//Using gauss law\n", +"//B*L = uo*I\n", +"Ienc = integrate('c*x^2*2*%pi*x', 'x', a, r)\n", +"L = 2*%pi*r\n", +"B = uo*Ienc/L\n", +"printf('The magnetic field at x=r is %eT', B)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 30.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 1.23 //in meter\n", +"d = 3.55*10^-2 //in meter\n", +"i = 5.57 //in A\n", +"n = 850*5\n", +"uo = 4*%pi*10^-7 //in SI unit\n", +"\n", +"//Sample Problem 30-4\n", +"printf('**Sample Problem 30-4**\n')\n", +"B = uo*n/L*i\n", +"printf('The magnetic field inside the solenoid is %eT', B)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/31-Induction_and_Inductance.ipynb b/Fundamental_Of_Physics_by_D_Haliday/31-Induction_and_Inductance.ipynb new file mode 100644 index 0000000..d79f531 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/31-Induction_and_Inductance.ipynb @@ -0,0 +1,333 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 31: Induction and Inductance" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"i = 1.5 //in A\n", +"D = 3.2*10^-2 //in meter\n", +"N = 220/10^-2 //in turns/m\n", +"n = 130\n", +"d = 2.1*10^-2 //in meter\n", +"deltaT = 25*10^-3 //in s\n", +"uo = 4*%pi*10^-7 //in SI unit\n", +"\n", +"//Sample Problem 31-1\n", +"printf('**Sample Problem 31-1**\n')\n", +"A = %pi*(d/2)^2\n", +"deltaPhi = uo*N*i*A\n", +"E = n*deltaPhi/deltaT\n", +"printf('The emf induced is equal to %eV', E)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r = 0.20 //in meter\n", +"t = poly(0, 't')\n", +"B = 4.0*t^2 + 2.0*t + 3.0\n", +"E = 2.0 //in Volts\n", +"R = 2 //in Ohm\n", +"\n", +"//Sample Problem 31-2a\n", +"printf('**Sample Problem 31-2a**\n')\n", +"t = 10 //in sec\n", +"flux = B*%pi*r^2/2\n", +"Et = derivat(flux)\n", +"E1 = horner(Et, t)\n", +"printf('The Emf induced is equal to %fV\n', E1)\n", +"\n", +"//Sample Problem 31-2b\n", +"printf('\n**Sample Problem 31-2b**\n')\n", +"I = (E1-E)/R\n", +"printf('The induced current is equal to %fA', I)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"t = poly(0, 't')\n", +"//B = 4*t^2*x^2\n", +"W = 3.0 //in meter\n", +"H = 2.0 //in meter\n", +"t1 = 0.10 //in sec\n", +"\n", +"//Sample Problem 31-3\n", +"printf('**Sample Problem 31-3**\n')\n", +"flux = integrate('4*x^2*H', 'x', 0, W)\n", +"E = derivat(flux*t^2)\n", +"E1 = horner(E, t1)\n", +"printf('The induced emf is equal to %fV', E1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 8.5*10^-2 //in meter\n", +"Rb = 0.13 //in T/s\n", +"r = 5.2*10^-2 //in meter\n", +"\n", +"//Sample Problem 31-4a\n", +"printf('**Sample Problem 31-4a**\n')\n", +"//Using Faraday's law\n", +"Rf = Rb*%pi*r^2\n", +"E = Rf/(2*%pi*r)\n", +"printf('The induced electric field is equal to %eV/m\n', E)\n", +"\n", +"//Sample Problem 31-4b\n", +"printf('\n**Sample Problem 31-4b**\n')\n", +"r = 12.5*10^-2 //in meter\n", +"Rf = Rb*%pi*R^2\n", +"E = Rf/(2*%pi*r)\n", +"printf('The induced electric field is equal to %eV/m', E)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 9.0 //in Ohm\n", +"L = 2*10^-3 //in Henery\n", +"E = 18 //in Volts\n", +"\n", +"//Sample Problem 31-5a\n", +"printf('**Sample Problem 31-5a**\n')\n", +"//As soon as switch is closed the inductor will act like current barrier\n", +"Io = E/R\n", +"printf('The current as soon as qwitch is closed is equal to %1.2fA\n', Io)\n", +"\n", +"//Sample Problem 31-5b\n", +"printf('\n**Sample Problem 31-5b**\n')\n", +"//After long time inductor will act like short circuit\n", +"Req = R/3\n", +"If = E/(R/3)\n", +"printf('The current through the battery after long time will be %1.2fA', If)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 53*10^-3 //in H\n", +"R = 0.37 //in Ohm\n", +"\n", +"//Sample Problem 31-6\n", +"printf('**Sample Problem 31-6**\n')\n", +"//i = io(1-e^(t/T))\n", +"//ln2 = t/T\n", +"T = L/R\n", +"t = T*log(2)\n", +"printf('The time taken to rach the current to half of its stedy state value is %fs', t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 53*10^-3 //in H\n", +"R = 0.35 //in Ohm\n", +"V = 12 //in Volts\n", +"\n", +"//Sample Problem 31-7a\n", +"printf('**Sample Problem 31-7a**\n')\n", +"i = V/R //in steady state\n", +"E = 1/2*L*i^2\n", +"printf('The Energy stored in the inductor in steady state is %fJ\n', E)\n", +"\n", +"//Sample Problem 31-7b\n", +"printf('\n**Sample Problem 31-7b**\n')\n", +"Et = E/2\n", +"//hence It = Io/sqrt(2)\n", +"f = log(1-1/sqrt(2)) //the number of times of time constant\n", +"printf('After t=%1.1fT, the energy stored in the inductor will be half of tis steady state value', f)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"a = 1.2*10^-3 //in meter\n", +"b = 3.5*10^-3 //in meter\n", +"i = 2.7 //in Amp\n", +"l = 1 //in meter(say)\n", +"uo = 4*%pi*10^-7\n", +"\n", +"//Sample Problem 31-8\n", +"printf('**Sample Problem 31-8**\n')\n", +"B = uo*i/(2*%pi) //divided by r\n", +"Ul = B^2/(2*uo) //divided by r^2\n", +"//Energy as a funtion of r\n", +"U = Ul*2*%pi*l //divided by r by r\n", +"Energy = integrate('U/r', 'r', a, b)\n", +"printf('Energy per unit length is equal to %1.2eJ/m', Energy)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"N1 = 1200 //turns\n", +"N2 = N1\n", +"R2 = 1.1*10^-2 //in meter\n", +"R1 = 15*10^-2 //in meter\n", +"uo = 4*%pi*10^-7\n", +"\n", +"//Sample Problem 31-9\n", +"printf('**Sample Problem 31-9**\n')\n", +"//let's assume\n", +"i = 1 //in amp\n", +"B1 = uo*N1*i/(2*R1)\n", +"phi2 = B1*%pi*R2^2*N2\n", +"M = phi2/i\n", +"printf('The mutual inductance of the two coil is equal to %1.2eH', M)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/32-Magnetism_of_Matter_Maxwell_Equation.ipynb b/Fundamental_Of_Physics_by_D_Haliday/32-Magnetism_of_Matter_Maxwell_Equation.ipynb new file mode 100644 index 0000000..ec8a78f --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/32-Magnetism_of_Matter_Maxwell_Equation.ipynb @@ -0,0 +1,126 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 32: Magnetism of Matter Maxwell Equation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"T = 300 //in K\n", +"B = 1.5 //in T\n", +"ub = 9.27*10^-24 //in J/T\n", +"mu = 1.0*ub\n", +"K = 1.38*10^-23 //in J/K\n", +"e = 1.6*10^-19 //in coulomb\n", +"\n", +"//Sample Problem 32-1\n", +"printf('**Sample Problem 32-1**\n')\n", +"K = 3/2*K*T\n", +"deltaU = 2*ub*B\n", +"printf('The average translation kinetic energy of the atoms is %1.2eeV\n', K/e)\n", +"printf('The difference between the energy of the two arrangement is %1.2eeV', deltaU/e)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"density = 7900 //in kg/m^3\n", +"L = 3*10^-2 //in meter\n", +"w = 1*10^-3 //in meter\n", +"t = 0.50*10^-3 //in meter\n", +"MFe = 2.1*10^-23 //in J/T\n", +"f = 10/100\n", +"M = 55.847*10^-3 //in kg/mol\n", +"Na = 6.023*10^23 //in /mol\n", +"\n", +"//Sample Problem 32-2\n", +"printf('**Sample Problem 32-2**\n')\n", +"N = density*L*w*t/M * Na\n", +"MD = N*f*MFe\n", +"printf('The needles magnetic dipole moment is %1.2eJ/T', MD)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 32.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r = 11.0*10^-3 //in meter\n", +"R = 5*r\n", +"Edot = 1.50*10^12 //in V/m.s\n", +"uo = 4*%pi*10^-7\n", +"Eo = 8.85*10^-12 //in C^2/N.m^2\n", +"\n", +"//Sample Problem 32-3\n", +"printf('**Sample Problem 32-3**\n')\n", +"//for r=R/5\n", +"B = uo*Eo*R^2/(2*r)*Edot\n", +"printf('The magnetic field is at r=R/5 is equal to %1.2eT', B)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/33-Electromagnetic_Oscillations_and_Alternating_Current.ipynb b/Fundamental_Of_Physics_by_D_Haliday/33-Electromagnetic_Oscillations_and_Alternating_Current.ipynb new file mode 100644 index 0000000..536f30a --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/33-Electromagnetic_Oscillations_and_Alternating_Current.ipynb @@ -0,0 +1,353 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 33: Electromagnetic Oscillations and Alternating Current" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"C = 1.5*10^-6 //in F\n", +"V = 57 //in volts\n", +"L = 12*10^-3 //in H\n", +"\n", +"//Sample Problem 33-1\n", +"printf('**Sample Problem 33-1**\n')\n", +"Imax = V*sqrt(C/L)\n", +"printf('The maximum current in the circuit is %1.2eA', Imax)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"C = 1.5*10^-6 //in F\n", +"V = 57 //in volts\n", +"L = 12*10^-3 //in H\n", +"\n", +"//Sample Problem 33-2a\n", +"printf('**Sample Problem 33-2a**\n')\n", +"//V(accross Inductor) = V(accross Capacitor)\n", +"//-L*(dI/dt) = V\n", +"//I = C*(dV/dt)\n", +"//L*C*(d^2V/dt^2) = -V\n", +"//at t=0, Potential difference = V\n", +"w = 1/sqrt(L*C)\n", +"printf('The potential defference accross the inductor is V=%d*cos(%d*t)\n', V, w)\n", +"\n", +"//Sample Problem 33-2b\n", +"printf('\n**Sample Problem 33-2b**\n')\n", +"MaxRate = abs(-V/L)\n", +"printf('The maximum rate of change in current is %1.2famp/s', MaxRate)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 12*10^-3 //in H\n", +"C = 1.6*10^-6 //in F\n", +"R = 1.5 //in ohm\n", +"\n", +"//Sample Problem 33-3a\n", +"printf('**Sample Probelm 33-3a**\n')\n", +"//Q/2 = Q*e^(-R*t/(2*L))\n", +"t = -2*L/R*log(0.50)\n", +"printf('At time t=%1.2esec, the amplitude of charge oscillation is half of the maximum value\n', t)\n", +"\n", +"//Sample Problem 33-3b\n", +"printf('\n**Sample Probelm 33-3b**\n')\n", +"w = 1/sqrt(L*C)\n", +"T = (2*%pi)/w\n", +"n = t/T\n", +"printf('The number of oscillation are %1.2f till t=%1.2e', n, T)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 200 //in Ohm\n", +"Em = 36 //in volts\n", +"fd = 60 //in Hz\n", +"t = poly(0, 't')\n", +"w = 2*%pi*fd\n", +"//V = Em*sin(w*t)\n", +"\n", +"//Sample Problem 33-4a\n", +"printf('**Sample Problem 33-4a**\n')\n", +"//Vr = Emax*sin(w*t)\n", +"printf('The voltage drop across the resistor is Vr=%1.2f*sin(%1.2f*t)\n', Em, w)\n", +"\n", +"//Sample Problem 33-4b\n", +"printf('\n**Sample Problem 33-4b**\n')\n", +"Ir = Em/R\n", +"printf('The current in the resistor as a function of time is Ir=%1.2f*sin(%1.2f*t)', Ir, w)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"C = 15*10^-6 //in Farad\n", +"Em = 36.0 //in volts\n", +"fd = 60.0 //in Hz\n", +"\n", +"//Sample Problem 33-5a\n", +"printf('**Sample Problem 33-5a**\n')\n", +"//Vc = Emax*sin(w*t)\n", +"printf('The voltage drop across the capacitor is Vc=%1.2f*sin(%1.2f*t)\n', Em, w)\n", +"\n", +"//Sample Problem 33-5b\n", +"printf('\n**Sample Problem 33-5b**\n')\n", +"//I = -C*(dV/dt)\n", +"IcMAX = abs(w*C*Em)\n", +"printf('The current in the capacitor as a function of time is Ic=%1.2f*cos(%1.2f*t)', IcMAX, w)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 230*10^-3 //in Farad\n", +"Em = 36.0 //in volts\n", +"fd = 60.0 //in Hz\n", +"\n", +"//Sample Problem 33-6a\n", +"printf('**Sample Problem 33-6a**\n')\n", +"//Vc = Emax*sin(w*t)\n", +"printf('The voltage drop across the inductor is Vi=%1.2f*sin(%1.2f*t)\n', Em, w)\n", +"\n", +"//Sample Problem 33-6b\n", +"printf('\n**Sample Problem 33-6b**\n')\n", +"//V = -L*(dI/dt)\n", +"IcMAX = abs(Em/(w*L))\n", +"printf('The current in the inductor as a function of time is Ic=-%1.2f*cos(%1.2f*t)', IcMAX, w)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 200 //in ohm\n", +"C = 15*10^-6 //in F\n", +"L = 230*10^-3 //in H\n", +"Em = 36.0 //in volts\n", +"fd = 60.0 //in Hz\n", +"\n", +"//Sample Problem 33-7a\n", +"printf('**Sample Problem 33-7a**\n')\n", +"w = 2*%pi*fd\n", +"Xl = w*L\n", +"Xc = 1/(w*C)\n", +"Z = sqrt(R^2 + (Xl - Xc)^2)\n", +"Imax = Em/Z\n", +"printf('The amplitude of current in the circuit is %1.2fA, Imax\n', Imax)\n", +"\n", +"//Sample Problem 33-7b\n", +"printf('\n**Sample Problem 33-7a**\n')\n", +"phi = atan((Xl-Xc)/R)\n", +"printf('The phase constant is equal to %fdegrees', phi)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Erms = 120 //in volts\n", +"fd = 60 //in Hz\n", +"R = 200 //in ohm\n", +"Xl = 80.0 //in ohm\n", +"Xc = 150 //in ohm\n", +"\n", +"//Sample Problem 33-8a\n", +"printf('**Sample Problem 33-8a**\n')\n", +"Z = sqrt(R^2 + (Xl - Xc)^2)\n", +"pf = R/Z\n", +"printf('The power factor for the circuit is %.3f\n', pf)\n", +"\n", +"//Sample Problem 33-8b\n", +"printf('\n**Sample Problem 33-8b**\n')\n", +"Irms = Erms/R\n", +"Pavg = Erms*Irms*pf\n", +"printf('The average rate of disscipation of energy is equal to %1.2fW\n', Pavg)\n", +"\n", +"//Sample Problem 33-8c\n", +"printf('\n**Sample Problem 33-8c**\n')\n", +"Xc = Xl\n", +"w = 2*%pi*fd\n", +"Cnew = 1/Xc/w\n", +"printf('The new capacitance should be %1.2eF', Cnew)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 33.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Vp = 8.5*10^3 //in Volts\n", +"Vs = 120 //in volts\n", +"P = 78*10^3 //in W\n", +"\n", +"//Sample Problem 33-9a\n", +"printf('**Sample Problem 33-9a**\n')\n", +"ratio = Vp/Vs\n", +"printf('The turn ratio is equal to %.3f\n', ratio)\n", +"\n", +"//Sample Problem 33-9b\n", +"printf('\n**Sample Problem 33-9b**\n')\n", +"Is = P/Vs\n", +"Ip = P/Vp\n", +"printf('The current in primary circuit is %1.2eA\n', Ip)\n", +"printf('The current in secondary circuit is %1.2eA\n', Is)\n", +"\n", +"//Sample Problem 33-9c\n", +"printf('\n**Sample Problem 33-9c**\n')\n", +"Rs = Vs/Is\n", +"Rp = Vp/Ip\n", +"printf('The resistance in primary circuit is %1.2eA\n', Rp)\n", +"printf('The resistance in secondary circuit is %1.2eA\n', Rs)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/34-Electromagnetic_Waves.ipynb b/Fundamental_Of_Physics_by_D_Haliday/34-Electromagnetic_Waves.ipynb new file mode 100644 index 0000000..00e1c52 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/34-Electromagnetic_Waves.ipynb @@ -0,0 +1,196 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 34: Electromagnetic Waves" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"d = 1.8 //in meter\n", +"P = 250 //in W\n", +"c = 3*10^8 //in m/s\n", +"mu = 4*%pi*10^-7 //in SI unit\n", +"\n", +"//Sample Problem 34-1\n", +"printf('**Sample Problem 34-1**\n')\n", +"Erms = sqrt(P*c*mu/(4*%pi*d^2))\n", +"Brms = Erms/c\n", +"printf('The rms value of electric field is equal to %1.2eV/m\n', Erms)\n", +"Brms = printf('The rms value of magnetic field is equal to %1.2eT', Brms)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Gravitation.sci', -1)\n", +"\n", +"//Given that\n", +"density = 3.5*10^3 //in kg/m^3\n", +"c = 3*10^8 //in m/s\n", +"d = 1 //(say)\n", +"Ps = 3.9*10^26 //in W\n", +"\n", +"//Sample Problem 34-2\n", +"printf('**Sample Problem 34-2**\n')\n", +"R = poly(0, 'R')\n", +"A = %pi*R^2\n", +"Ad = 4*%pi*d^2\n", +"I = Ps/Ad\n", +"Fr = I*A/c\n", +"V = 4/3*%pi*R^3\n", +"m = density*V\n", +"Fg = GForce(m, Ms, d)\n", +"R = roots(Fr-Fg)\n", +"printf('The radius of the dust particle is %1.3em', R(1))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"theta1 = dtor(60)\n", +"theta2 = dtor(90-60)\n", +"I = 1 //(say)\n", +"\n", +"//Sample Problem 34-3\n", +"printf('**Sample Problem 34-3**\n')\n", +"//half of the original intensity, from the one-half rule\n", +"I1 = I/2 \n", +"I2 = I1*cos(theta1)^2\n", +"I3 = I2*cos(theta2)^2\n", +"printf('The ratio of the initial inensity to the final intensity of the light is %.4f', I3)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"n1 = 1.33\n", +"n2 = 1.77\n", +"n3 = 1.00\n", +"theta1 = 50 //in degrees\n", +"\n", +"//Sample Problem 34-4a\n", +"printf('**Sample Problem 34-4a**\n')\n", +"AORl = 90 - theta1\n", +"AORr = rtod(asin(n1/n2*sin(dtor(AORl))))\n", +"printf('The angle of reflection is %1.2fdegrees\n', AORl)\n", +"printf('The angle of refraction is %1.2fdegrees\n', AORr)\n", +"\n", +"//Sample Problem 34-4b\n", +"printf('\n**Sample Problem 34-4b**\n')\n", +"Af = rtod(asin(n2/n3*sin(dtor(AORr))))\n", +"printf('The final angle of refraction is %1.2fdegrees', Af)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 34.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"e = 45 //in degrees\n", +"\n", +"//Sample Problem 34-5\n", +"printf('**Sample Problem 34-5**\n')\n", +"//For extrem case\n", +"n = 1/sin(dtor(e))\n", +"printf('The index of refraction should be at least %1.1f', n)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/35-Images.ipynb b/Fundamental_Of_Physics_by_D_Haliday/35-Images.ipynb new file mode 100644 index 0000000..21ec4d7 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/35-Images.ipynb @@ -0,0 +1,156 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 35: Images" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"h = 1 //(say)\n", +"f = 40 //in cm\n", +"hdash = .20*h\n", +"\n", +"//Sample Problem 35-1a\n", +"printf('**Sample Problem 35-1a**\n')\n", +"printf('The image is virtual & on the opposite side of mirror because of having same orientation\n')\n", +"\n", +"//Sample Problem 35-1b\n", +"printf('\n**Sample Problem 35-1b**\n')\n", +"printf('The height of image is smaller than the object. Therefore, the mirror is concave\n')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n1 = 1.6\n", +"n2 = 1.00\n", +"R = -3.0 //in mm\n", +"i = -5.0 //in mm\n", +"\n", +"//Sample Problem 35-2\n", +"printf('**Sample Problem 35-2**\n')\n", +"//n1/d + n2/i = (n2-n1)/R\n", +"d = n1/(- n2/i + (n2-n1)/R)\n", +"printf('The real depth of the mosquito is %1.2fmm', d)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Xo = -20 //in cm\n", +"m = -0.25\n", +"n = 1.65\n", +"\n", +"//Sample Problem 35-3a\n", +"printf('**Sample Problem 35-3a**\n')\n", +"printf('The image is real real because m<0 as well as m<1\n')\n", +"printf('The mens is converging because magnification is negative for real image\n')\n", +"printf('The object is outside the focal length because m<1\n')\n", +"printf('The image is on the opposite side of the image from the lens\n')\n", +"printf('The image is erect\n')\n", +"\n", +"//Sample Problem 35-3b\n", +"printf('\n**Sample Problem 35-3b**\n')\n", +"f = Xo*m*Xo/(Xo-m*Xo)\n", +"R = (n-1)*2*f\n", +"printf('The radius of curvature of the lens is %1.2fcm', f)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 35.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L= 10 //in cm\n", +"f1 = 24 //in cm\n", +"f2 = 9 //in cm\n", +"xo = -6 //in cm\n", +"\n", +"//Sample Problem 35-4\n", +"printf('**Sample Problem 35-4**\n')\n", +"xi1 = xo*f1/(xo+f1)\n", +"xo2 = xi1 - L\n", +"xi2 = xo2*f2/(xo2+f2)\n", +"printf('The final image will be at a distance of %dcm from the second mirror', xi2)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/36-Interference.ipynb b/Fundamental_Of_Physics_by_D_Haliday/36-Interference.ipynb new file mode 100644 index 0000000..7b1d07c --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/36-Interference.ipynb @@ -0,0 +1,220 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 36: Interference" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"l = 550*10^-9 //in meter\n", +"n2 = 1.60\n", +"n1 = 1.00\n", +"t = 2.6*10^-6 //in meter\n", +"\n", +"//Sample Problem 36-1a\n", +"printf('**Sample Problem 36-1a**\n')\n", +"deltaPHI = t/l*(n2 - n1)*360\n", +"printf('The phase difference is equal to %1.2fdegrees\n', deltaPHI)\n", +"\n", +"//Sample Problem 36-1b\n", +"printf('\n**Sample Problem 36-1b**\n')\n", +"printf('The interference produced would be constructive')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"l = 546*10^-9 //in meter\n", +"d = 12*10^-5 //in meter\n", +"D = 55*10^-2 //in meter\n", +"\n", +"//Sample Problem 36-2\n", +"printf('**Sample Problem 36-2**\n')\n", +"beeta = l*D/d\n", +"printf('The difference between two adjacent maxima is %1.2em', beeta)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"Eo = 1 //(say)\n", +"E1 = Eo\n", +"E2 = Eo\n", +"E3 = Eo\n", +"phi1 = dtor(0)\n", +"phi2 = dtor(60)\n", +"phi3 = dtor(-30)\n", +"\n", +"//Sample Problem 36-3\n", +"printf('**Sample Problem 36-3**\n')\n", +"Eh = E1*cos(phi1) + E2*cos(phi2) + E3*cos(phi3)\n", +"Ev = E1*sin(phi1) + E2*sin(phi2) + E3*sin(phi3)\n", +"Er = sqrt(Ev^2 + Eh^2)\n", +"theta = rtod(atan(Ev/Eh))\n", +"printf('The resultant electric field is E=%1.2f*Eo*sin(w*t + %1.2f)', Er, theta)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"lmin = 400 //in nm\n", +"lmax = 690 //in nm\n", +"n2 = 1.33\n", +"L = 320 //in nm\n", +"\n", +"//Sample Problem 36-4\n", +"printf('**Sample Problem 36-4**\n')\n", +"flag = 1\n", +"odd_number = 1\n", +"while flag == 1\n", +" lambda = 4*L*n2/odd_number\n", +" if lambda > lmin & lambda < lmax then\n", +" flag = 0\n", +" end\n", +" odd_number = odd_number + 2\n", +"end\n", +"printf('The wavelength of the light is %1.2enm', lambda)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n1 = 1.38\n", +"n2 = 1.50\n", +"lambda = 550 //in nm\n", +"\n", +"//Sample Problem 36-5\n", +"printf('**Sample Problem 36-5**\n')\n", +"Lmin = lambda/4/n1\n", +"printf('The minimum value of wavelength possible is %1.2fnm', Lmin)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 36.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"lambda = 632.8*10^-9 //in meter\n", +"i = 0 //in rad\n", +"dFringes = 6\n", +"bFringes = 5\n", +"\n", +"//Sample Problem 36-6\n", +"printf('**Sample Problem 36-6**\n')\n", +"//Assume the difference in thickness is 't'\n", +"//then the path difference will be 2*L\n", +"//hence\n", +"deltaL = bFringes/2*lambda\n", +"printf('The difference in thickness is equal to %fnm', deltaL*10^9)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/37-Diffraction.ipynb b/Fundamental_Of_Physics_by_D_Haliday/37-Diffraction.ipynb new file mode 100644 index 0000000..ded6c99 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/37-Diffraction.ipynb @@ -0,0 +1,208 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 37: Diffraction" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 37.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"lambda = 650*10^-9 //in meter\n", +"theta = dtor(15) //in radians\n", +"\n", +"//Sample Problem 37-1a\n", +"printf('**Sample Problem 37-1a**\n')\n", +"//We know that [a*sin(theta) = m*lambda] for m=Integer\n", +"m = 1\n", +"a = m*lambda/sin(theta)\n", +"printf('The slit width is equal to %fnm\n', a*10^9)\n", +"\n", +"//Sample Problem 37-1b\n", +"printf('\n**Sample Problem 37-1b**\n')\n", +"m = 3/2 //for first side maxima\n", +"lambdaDESH = a*sin(theta)/ m\n", +"printf('The wavelength of the light is equal to %fnm', lambdaDESH*10^9)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 37.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 37-2\n", +"printf('**Sample Problem 37-2**\n')\n", +"I = [];\n", +"for m = 1:3\n", +" theta = (m+.5)*%pi\n", +" I = [I, (sin(theta)/theta)^2]\n", +"end\n", +"printf('The intensities of secondary maximas relative to intensity of CBF is-\n')\n", +"printf('\tI1/Im = %.4f\n', I(1))\n", +"printf('\tI2/Im = %.4f\n', I(2))\n", +"printf('\tI3/Im = %.4f', I(3))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 37.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"d = 32*10^-3 //in meter\n", +"f= 24*10^-2 //in meter\n", +"lam = 550*10^-9 //in meter\n", +"\n", +"//Sample Problem 37-3a\n", +"printf('**Sample Problem 37-3a**\n')\n", +"theta = 1.22*lam/d\n", +"printf('Angular sepration should be equal to %erad\n', theta)\n", +"\n", +"//Sample Problem 37-3b\n", +"printf('\n**Sample Problem 37-3b**\n')\n", +"deltaX = f*theta\n", +"printf('The sepration between the two images is %em', deltaX)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 37.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"l = 405*10^-9 //in meter\n", +"d = 19.44*10^-6 //in meter\n", +"a = 4.050*10^-6 //in meter\n", +"\n", +"//Sample Problem 37-4a\n", +"printf('**Sample Problem 37-4a**\n')\n", +"n = floor(d/a)\n", +"printf('The number of bright fringes are %d\n', 2*n+1)\n", +"\n", +"//Sample Problem 37-4b\n", +"printf('\n**Sample Problem 37-4b**\n')\n", +"num = ceil(2*d/a)\n", +"printf('The number of bright fringes within either of the first side peak is %d', num/2 - 1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 37.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"N = 1.26*10^4\n", +"w = 25.4*10^-3 //in meter\n", +"l1 = 589*10^-9 //in meter\n", +"l2 = 589.59*10^-9 //in meter\n", +"\n", +"//Sample Problem 37-5a\n", +"printf('**Sample Problem 37-5a**\n')\n", +"d = w/N\n", +"m = 1\n", +"theta = asin(m*l1/d)\n", +"theta = rtod(theta)\n", +"printf('The first order maxima occurs at an angle of %fdegree from the center\n', theta)\n", +"\n", +"//Sample Problem 37-5b\n", +"printf('\n**Sample Problem 37-5b**\n')\n", +"theta = rtod(theta)\n", +"D = m/(d*cos(theta))\n", +"deltaL = l2-l1\n", +"deltaTHETA = D*deltaL\n", +"printf('The anglar sepration between the two first orderlines is %erad\n', deltaTHETA)\n", +"\n", +"//Sample Problem 37-5c\n", +"printf('\n**Sample Problem 37-5c**\n')\n", +"Lavg = (l1+l2)/2\n", +"R = Lavg/(m*deltaL)\n", +"N = R/m\n", +"printf('The least number of rulings a grating can have is %d', N)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/38-Relativity.ipynb b/Fundamental_Of_Physics_by_D_Haliday/38-Relativity.ipynb new file mode 100644 index 0000000..c8644a6 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/38-Relativity.ipynb @@ -0,0 +1,267 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 38: Relativity" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 38.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r = 0.9990\n", +"t = 10 //in years\n", +"\n", +"//Sample Problem 38-1\n", +"printf('**Sample Problem 38-1**\n')\n", +"y = 1/sqrt(1-r^2)\n", +"tEarth = t*y\n", +"T = 2*tEarth\n", +"printf('The time as measured from the earth is %1.2fy', T)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 38.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Tl = 0.1237*10^-6 //in sec\n", +"c = 3*10^8 //in m/s\n", +"r = 0.990\n", +"\n", +"//Sample Problem 38-2\n", +"printf('**Sample Problem 38-2**\n')\n", +"y = 1/sqrt(1-r^2)\n", +"Tb = Tl*y //in laboratory frame\n", +"v = r*c\n", +"d = v*Tb\n", +"printf('The kaon can go till %dm', d)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 38.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Lp = 230 ///in meter\n", +"t = 3.57*10^-6 //in meter\n", +"c = 3*10^8 //in m/s\n", +"\n", +"//Sample Problem 38-3\n", +"printf('**Sample Problem 38-3**\n')\n", +"//y = 1/sqrt(1-r^2)\n", +"//L = Lp/y\n", +"//L = r*c*t\n", +"//solving -\n", +"r = Lp/sqrt((c*t)^2 + Lp^2)\n", +"printf('The relative velocoty is equal to %.3fc m/s', r)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 38.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"deltaT = 1.10 //in sec\n", +"x = 4.00*10^8 //in meter\n", +"c = 3*10^8 //in m/s\n", +"r = 0.980\n", +"\n", +"//Sample Problem 38-4\n", +"printf('**Sample Problem 38-4**\n')\n", +"y = 1/sqrt(1-r^2)\n", +"Xe = y*(x - r*c*deltaT)\n", +"printf('The distance in earth frame is %1.2em\n', Xe)\n", +"Te = y*(deltaT - r*x/c)\n", +"printf('The time interval in earth frame is %1.2f', Te)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 38.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"lp1 = 499.8*10^-9 //in meter\n", +"lp2 = 501.6*10^-9 //in meter\n", +"c = 3*10^8 //in m/s\n", +"Ms = 1.99*10^30 //in kg\n", +"G = 6.67*10^-11 //in SI unit\n", +"R = 100 //in light year\n", +"conv = 9.46*10^15 //conversion factor from light year to sec\n", +"\n", +"//Sample Problem 38-5a\n", +"printf('**Sample Problem 38-5a**\n')\n", +"lo = lp1 + lp2\n", +"lo = lo/2\n", +"deltaL = abs(lp1 - lo)\n", +"v = deltaL/lo * c\n", +"printf('The speed of gas relative to us is %1.2em/s\n', v)\n", +"\n", +"//Sample Problem 38-5b\n", +"printf('\n**Sample Problem 38-5b**\n')\n", +"//G*M*m/r^2 = m*v^2/r\n", +"r = R*conv\n", +"M = v^2*r/G\n", +"ratio = M/Ms\n", +"printf('The mass of galaxy is %1.2e*Ms', ratio)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 38.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"K = 2.53 //in Mev\n", +"Me = 9.109*10^-31 //in kg\n", +"c = 3*10^8 //in m/s\n", +"conv = 1.6*10^-19*10^6 //Mev to joule conversion factor\n", +"\n", +"//Sample Problem 38-6a\n", +"printf('**Sample Problem 38-6a**\n')\n", +"Eactual = Me*c^2/conv + K\n", +"printf('The actual energy of the elctron is %1.2fMev\n', Eactual)\n", +"\n", +"//Sample Problem 38-6b\n", +"printf('\n**Sample Problem 38-6b**\n')\n", +"p = sqrt(Eactual^2 - (Me*c^2))\n", +"printf('The momentum of the electron is %1.2fMev/c', p)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 38.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"K = 3.0*10^20*1.6*10^-19 //in J\n", +"Mp = 1.67*10^-27 //in kg\n", +"c = 3*10^8 //in m/s\n", +"conv = 9.46*10^15 //conversion factor from light year to sec\n", +"D = 9.8*10^4 //in light year\n", +"\n", +"//Sample Problem 38-7a\n", +"printf('**Sample Problem 38-7a**\n')\n", +"Erm = (Mp*c^2)\n", +"y = (K + Erm)/Erm\n", +"r = sqrt(1 - (1/y)^2)\n", +"printf('The velocity is approximately equal to %1.2f*c\n', r)\n", +"\n", +"//Sample Problem 38-7b\n", +"printf('\n**Sample Problem 38-7b**\n')\n", +"deltaT = D //in year\n", +"printf('The time taken is %1.1ey\n', deltaT)\n", +"\n", +"//Sample Problem 38-7c\n", +"printf('\n**Sample Problem 38-7c**\n')\n", +"deltaTp = deltaT/y * 365*24*3600\n", +"printf('The time taken in reference frame of proton is %1.2fs', deltaTp)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/39-Photons_and_Matter_Waves.ipynb b/Fundamental_Of_Physics_by_D_Haliday/39-Photons_and_Matter_Waves.ipynb new file mode 100644 index 0000000..791939a --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/39-Photons_and_Matter_Waves.ipynb @@ -0,0 +1,246 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 39: Photons and Matter Waves" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"P = 100 //in W\n", +"lambda = 590*10^-9 //in meter\n", +"h = 6.62*10^-34 //in J.s\n", +"c = 3*10^8 //in m/s\n", +"\n", +"//Sample Problem 39-1\n", +"printf('**Sample Problem 39-1**\n')\n", +"Ep = h*c/lambda //Energy of each photon\n", +"N = P/Ep\n", +"printf('The rate at which photons are absorbed is %1.2e/s', N)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r = 3.5 //in meter\n", +"P = 1.5 //in W\n", +"phi = 2.2 //in ev\n", +"conv = 1.6*10^-19 //ev to Joule to conversion factor\n", +"R = 5.0*10^-11 //in meter\n", +"\n", +"//Sample Problem 39-2\n", +"printf('**Sample Problem 39-2**\n')\n", +"I = P/(4*%pi*r^2)\n", +"A = %pi*R^2\n", +"deltaT = phi*conv/(I*A)\n", +"printf('The time taken in ejecting electron is %ds', deltaT)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"h = 6.62*10^-34 //in J.s\n", +"fo = 5.5*10^14 //in Hz\n", +"conv = 1.6*10^-19 //ev to J conversion factor\n", +"\n", +"//Sample Problem 39-3\n", +"printf('**Sample Problem 39-3**\n')\n", +"phi = h*fo/conv //in ev\n", +"printf('The work function of sodium is %1.2fev', phi)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"lambda = 22*10^-12 //in meter\n", +"conv = 1.6*10^-19 //ev to J conversion factor\n", +"E = 56*10^3*conv\n", +"theta = dtor(85) //in rad\n", +"h = 6.62*10^-34 //in J.s\n", +"Me = 9.1*10^-31 //in kg\n", +"c = 3*10^8 //in m/s\n", +"\n", +"//Sample Problem 39-4a\n", +"printf('**Sample Problem 39-4a**\n')\n", +"deltaL = h/(Me*c)*(1 - cos(theta))\n", +"printf('compton shift is equal to %1.2fpm\n', deltaL*10^12)\n", +"\n", +"//Sample Problem 39-4b\n", +"printf('\n**Sample Problem 39-4b**\n')\n", +"frac = deltaL/(lambda + deltaL)\n", +"printf('The fraction of energy transfered is %1.3f', frac)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"K = 120*1.6*10^-19 //in J\n", +"Me = 9.11*10^-31 //in kg\n", +"h = 6.62*10^-34 //in J.s\n", +"\n", +"//Sample Problem 39-5\n", +"printf('**Sample Problem 39-5**\n')\n", +"p = sqrt(2*K*Me)\n", +"lambda = h/p\n", +"printf('The wavelength of the electon is %dpm', lambda*10^12)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"v = 2.05*10^6 //in m/s\n", +"h = 6.62*10^-34 //in J.s\n", +"hC = h/(2*%pi)\n", +"precision = .50/100\n", +"Me = 9.109*10^-31 //in kg\n", +"\n", +"//Sample Problem 39-6\n", +"printf('**Sample Probelm 39-6**\n')\n", +"Px = Me*v\n", +"deltaPx = precision*Px\n", +"deltaX = hC/deltaPx\n", +"printf('The error in measuring x is %dnm', deltaX*10^9)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 39.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"conv = 1.6*10^-19 //ev to J conversion factor\n", +"E = 5.1*conv //in ev\n", +"Uo = 6.8*conv //in ev\n", +"L = 750*10^-12 //in m\n", +"h = 6.62*10^-34 //in J.s\n", +"Me = 9.11*10^-31 //in kg\n", +"\n", +"//Sample Problem 39-7a\n", +"printf('**Sample Problem 39-7a**\n')\n", +"k = sqrt(8*%pi^2*Me*(Uo-E)/h^2)\n", +"T = %e^(-2*k*L)\n", +"printf('The transmission coefficient is %e', T)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/4-Motion_in_Two_and_Three_Dimesions.ipynb b/Fundamental_Of_Physics_by_D_Haliday/4-Motion_in_Two_and_Three_Dimesions.ipynb new file mode 100644 index 0000000..5cbd6c6 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/4-Motion_in_Two_and_Three_Dimesions.ipynb @@ -0,0 +1,451 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Motion in Two and Three Dimesions" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.10: Sample_Problem_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//To convert velocity m/s from km/h\n", +"conv = 5/18\n", +"\n", +"//Given that\n", +"v_BA = 52 //in km/hr\n", +"v_PA = -78 //in km/hr\n", +"\n", +"//Sample Problem 4-10a\n", +"printf('**Sample Problem 4-10a**\n')\n", +"//using concept of relative velocity\n", +"v_PB = v_PA - v_BA\n", +"printf('The velocity of P as measured by Barbara is %d km/hr\n',v_PB)\n", +"\n", +"//Sample Problem 4-10b\n", +"printf('\n**Sample Problem 4-10b**\n')\n", +"//In frame of Alex\n", +"delta_t = 10 //in sec\n", +"a_PA = (0 - v_PA)* conv/delta_t\n", +"printf('The accleration of P in frame of Alex is %f m/s^2\n', a_PA)\n", +"\n", +"//Sample Problem 4-10c\n", +"printf('\n**Sample Problem 4-10c**\n')\n", +"a_BA = 0\n", +"a_PB = a_PA - a_BA\n", +"printf('The acceleration of P as measured by B is %f m/s^2', a_PB)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.11: Sample_Problem_11.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"v_WG = [65 * sin(dtor(20)),65 * cos(dtor(20))] //in km/h\n", +"v_PG_y = 0\n", +"v_PW_mag = 215 //in km/h\n", +"\n", +"//Sample Problem 4-11\n", +"printf('**Sample Problem 4-11**\n')\n", +"//therefore in direction\n", +"v_PW_y = v_PG_y - v_WG(2)\n", +"v_PW_x = sqrt(v_PW_mag^2 - v_PW_y^2)\n", +"v_PG_x = v_PW_x + v_WG(1)\n", +"printf('The magnitude of velocity of plane relative to ground is %f km/h', v_PG_x)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r_initial = [-3,2,5] //in meter\n", +"r_final = [9,2,8] //in meter\n", +"//Sample Problem 4-1\n", +"printf('**Sample Problem 4-1**\n')\n", +"dis_v = r_final - r_initial\n", +"printf('The displacement vector of the particle in meter is')\n", +"disp(dis_v)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2_a: Sample_Problem_2a.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"t = poly(0,'t')\n", +"x = -0.31 *t^2 + 7.2 *t +28 //in meter\n", +"y = 0.22 *t^2 - 9.1 *t + 30 //in meter\n", +"\n", +"//Sample Problem 4-2a\n", +"printf('**Sample Problem 4-2a**\n')\n", +"time_t =15 //in sec\n", +"position_r = [horner(x,time_t),horner(y,time_t)]\n", +"printf('The position vector of the rabbit at t=15sec in meter is')\n", +"disp(position_r)\n", +"printf('The magnitude of the position vector is %f m\n', norm(position_r))\n", +"printf('The angle made by the position vector with the x axis in degrees at the same time %f', rtod(atan(position_r(2)/position_r(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2_b: Sample_Problem_2b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Example4_2a.sce',-1)\n", +"clc\n", +"\n", +"//Sample Problem 4-2b\n", +"printf('**Sample Problem 4-2b**\n')\n", +"xx = horner(x, [0:2:25])\n", +"yy = horner(y, [0:2:25])\n", +"xset('window',3)\n", +"xtitle ('Y v/s X [t=0sec to t=25sec]','X (m)','Y (m)');\n", +"plot2d(linspace(0,0,10),linspace(0,0,10),style=3,rect=[0,-80,80,40]);\n", +"//plotting grid\n", +"plot(linspace(10,10,5),linspace(-80,40,5),'--')\n", +"plot(linspace(20,20,5),linspace(-80,40,5),'--')\n", +"plot(linspace(30,30,5),linspace(-80,40,5),'--')\n", +"plot(linspace(40,40,5),linspace(-80,40,5),'--')\n", +"plot(linspace(50,50,5),linspace(-80,40,5),'--')\n", +"plot(linspace(60,60,5),linspace(-80,40,5),'--')\n", +"plot(linspace(70,70,5),linspace(-80,40,5),'--')\n", +"plot(linspace(80,80,5),linspace(-80,40,5),'--')\n", +"plot(linspace(0,80,5),linspace(-60,-60,5),'--')\n", +"plot(linspace(0,80,5),linspace(-40,-40,5),'--')\n", +"plot(linspace(0,80,5),linspace(-20,-20,5),'--')\n", +"plot(linspace(0,80,5),linspace(0,0,5),'--')\n", +"plot(linspace(0,80,5),linspace(20,20,5),'--')\n", +"plot(linspace(0,80,5),linspace(40,40,5),'--')\n", +"//plotting graph\n", +"plot(xx,yy,'m.<-')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Example4_2a.sce',-1)\n", +"clc\n", +"\n", +"//Sample Problem 4-3\n", +"printf('\n**Sample Problem 4-3**\n')\n", +"velocity_v_x = derivat(x)\n", +"velocity_v_y = derivat(y)\n", +"v_time_t = [horner(velocity_v_x,time_t),horner(velocity_v_y,time_t)]\n", +"printf('The velocity vector of the rabbit at t=15sec in m/s is')\n", +"disp(v_time_t)\n", +"printf('The magnitude of the velocity vector is %f m/s\n', norm(v_time_t))\n", +"printf('The angle made by the velocity vector with the x axis in degrees at the same time %f', rtod(atan(v_time_t(2)/v_time_t(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('Example4_3.sce',-1)\n", +"clc\n", +"\n", +"//Sample Problem 4-4\n", +"printf('\n**Sample Problem 4-4**\n')\n", +"acceler_x = derivat(velocity_v_x)\n", +"acceler_y = derivat(velocity_v_y)\n", +"a_time_t = [horner(acceler_x,time_t),horner(acceler_y,time_t)]\n", +"printf('The acceleration vector of the rabbit at t=15sec in m/sec^2 is')\n", +"disp(a_time_t)\n", +"printf('The magnitude of the acceleration vector is %f m/sec^2\n', norm(a_time_t))\n", +"printf('The angle made by the acceleration vector with the x axis in degrees at the same time %f', rtod(atan(a_time_t(2)/a_time_t(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"velocity_v0 = [-2,4] //in m/s\n", +"acceler_a = [3 *cos(dtor(130)), 3 *sin(dtor(130))] //in m/sec^2\n", +"time_t = 5 //in sec\n", +"\n", +"//Sample Problem 4-5\n", +"printf('**Sample Problem 4-5**\n')\n", +"//using newton's first equation of motion v = u + a *t\n", +"velocity_t = velocity_v0 + acceler_a * time_t\n", +"printf('The velocity vector of the particle at t=5sec in m/s is')\n", +"disp(velocity_t)\n", +"printf('The magnitude of the velocity vector is %f m/s\n', norm(velocity_t))\n", +"printf('The angle made by the velocity vector with the x axis in degrees at the same time %f', rtod(atan(velocity_t(2)/velocity_t(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"v_rescue = [55,0] //in m/s\n", +"dis_y = -500 //in m\n", +"g = -9.8 //in m/s^2\n", +"\n", +"//Sample Problem 4-6a\n", +"printf('**Sample Problem 4-6a**\n')\n", +"//using newton's second equation of motion\n", +"time = sqrt(2 *dis_y /g)\n", +"dis_x = v_rescue(1) *time\n", +"printf('The angle of the piliots line of sight to the victim %f degrees\n',rtod(atan(-dis_x/dis_y)))\n", +"\n", +"//Sample Problem 4-6b\n", +"printf('\n**Sample Problem 4-6b**\n')\n", +"u_initial = v_rescue\n", +"//using newton's first equation of motion\n", +"v_final = u_initial + [0,g] * time\n", +"printf('The velocity vector of the capsule near water in m/s is')\n", +"disp(v_final)\n", +"printf('The magnitude of the velocity vector is %f m/s\n', norm(v_final))\n", +"printf('The angle made by the velocity vector with the x axis in degrees at the same time %f', rtod(atan(v_final(2)/v_final(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"range_x = 560 //in m\n", +"v0_mag = 82 //in m/sec\n", +"g = -9.8 //in m/s^2\n", +"\n", +"//Sample Problem 4-7a\n", +"printf('**Sample Problem 4-7a**\n')\n", +"theta = .5 *asin(-g* range_x/v0_mag^2)\n", +"printf('The angle at which the ball be fired to hit the ship is %f degrees or %f\n', rtod(theta), (90 - rtod(theta)))\n", +"\n", +"//Sample Problem 4-7b\n", +"printf('\n**Sample Problem 4-7b**\n')\n", +"//Range is maximum when theta = 45 degree\n", +"R_max = -v0_mag^2/g\n", +"printf('The maximum possible range is %f m', R_max)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"gr_height = 3 //in m\n", +"theta = dtor(53)\n", +"g = -9.8 //in m/s^2\n", +"v0 = 26.5 //in m/s\n", +"tower_height = 18 //in m\n", +"\n", +"//Sample Problem 4-8a\n", +"printf('**Sample Problem 4-8a**\n')\n", +"x = poly(0,'x')\n", +"y = x * tan(theta) + g * x * x /(2* v0^2) * sec(theta)^2\n", +"y_tower1 = horner(y,23)\n", +"if y_tower1<0 then printf('No, It does not clear the first Ferris wheel\n')\n", +" else printf('Yes, It clears the first Ferris wheel\n')\n", +"end\n", +"\n", +"//Sample Proble, 4-8b\n", +"printf('\n**Sample Problem 4-8b**\n')\n", +"y_max = horner(y,34.5)\n", +"printf('The balls clearance above middle tower is %f m\n', y_max + gr_height - tower_height)\n", +"\n", +"//Sample Problem 4-8c\n", +"printf('\n**Sample Problem 4-8c**\n')\n", +"Range = -v0^2 * sin(2*theta)/g\n", +"printf('The centre of the net should be placed at a diastance of %f m form the cannon', Range)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"v = 694 //in m/s\n", +"r = 5800 //in m\n", +"\n", +"//Sample Problem 4-9\n", +"printf('**Sample Problem 4-9**\n')\n", +"cent_a = v^2 / (r *g)\n", +"printf('Centripetal acceleration of the pilot is %f*g m/s^2', cent_a)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/40-More_About_Matter_waves.ipynb b/Fundamental_Of_Physics_by_D_Haliday/40-More_About_Matter_waves.ipynb new file mode 100644 index 0000000..87f2b64 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/40-More_About_Matter_waves.ipynb @@ -0,0 +1,211 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 40: More About Matter waves" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('quantum.sci', -1)\n", +"\n", +"//Given that\n", +"L = 100*10^-12 //in m\n", +"\n", +"//Sample Problem 40-1a\n", +"printf('**Sample Prblem 40-1a**\n')\n", +"n = 1 //for min energy\n", +"Emin = Ediff(n, 0, Me, L)\n", +"printf('The least possible energy is %1.2fev\n', Emin/conv)\n", +"\n", +"//Sample Problem 40-1b\n", +"printf('\n**Sample Prblem 40-1b**\n')\n", +"n2 = 3\n", +"n1 = 1\n", +"deltaE13 = Ediff(n2, n1, Me, L)\n", +"printf('The energy to be transferred is %1.2fev\n', deltaE13/conv)\n", +"\n", +"//Sample Problem 40-1c\n", +"printf('\n**Sample Prblem 40-1c**\n')\n", +"lambda = wavelength(deltaE13)\n", +"printf('The wavelength of photon is %1.2fnm\n', lambda*10^9)\n", +"\n", +"//Sample Problem 40-1d\n", +"printf('\n**Sample Prblem 40-1d**\n')\n", +"deltaE12 = Ediff(2, 1, Me, L)\n", +"lambda1 = wavelength(deltaE12)\n", +"deltaE23 = Ediff(3, 2, Me, L)\n", +"lambda2 = wavelength(deltaE23)\n", +"printf('The possible wavelength of photon is :\n')\n", +"printf('\t %1.2fnm\n', lambda*10^9)\n", +"printf('\t %1.2fnm\n', lambda1*10^9)\n", +"printf('\t %1.2fnm', lambda2*10^9)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 100*10^-12 //in m\n", +"\n", +"//Sample Problem 40-3a\n", +"printf('**Sample Prblem 40-3a**\n')\n", +"P = integrate('2/L*(sin(%pi/L*x))^2', 'x', 0, L/3)\n", +"printf('The probability is equal to %1.2f\n', P)\n", +"\n", +"//Sample Problem 40-3b\n", +"printf('\n**Sample Prblem 40-3b**\n')\n", +"P = integrate('2/L*(sin(%pi/L*x))^2', 'x', L/3, 2*L/3)\n", +"printf('The probability is equal to %1.2f\n', P)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('quantum.sci', -1)\n", +"\n", +"//Given that\n", +"L = 100*10^-12 //in m\n", +"Uo = 450*conv //in ev\n", +"l = 2*10^-9 //in m\n", +"\n", +"//Sample Problem 40-4a\n", +"printf('**Sample Prblem 40-4a**\n')\n", +"E1 = Ediff(1, 0, Me, L)\n", +"E = Uo - E1\n", +"lambda = wavelength(E)\n", +"printf('The wavelength of the free electron is %1.2fnm\n', lambda*10^9)\n", +"\n", +"//Sample Problem 40-4b\n", +"printf('\n**Sample Prblem 40-4b**\n')\n", +"E2 = Energy(l)\n", +"K = E2 - E\n", +"printf('The electron energy is %1.2fev', K/conv)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('quantum.sci', -1)\n", +"\n", +"//Sample Problem 40-6a\n", +"printf('**Sample Prblem 40-6a**\n')\n", +"//lyman series : least energetic photon => 2 to 1 transition\n", +"deltaE21 = -(13.6) * (1/2^2 - 1/1^2)*conv\n", +"lambda = wavelength(deltaE21)\n", +"printf('The wavelength of the least energetic photon in lyman series is %1.2fnm\n', lambda*10^9)\n", +"\n", +"//Sample Problem 40-6b\n", +"printf('\n**Sample Prblem 40-6b**\n')\n", +"//lyman series limit => 1 to infinity transition\n", +"deltaE = -(13.6) * (0 - 1/1^2)*conv\n", +"lambda = wavelength(deltaE)\n", +"printf('The wavelength of the lyman series limit is %1.2fnm\n', lambda*10^9)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"p = 0.90\n", +"\n", +"//Sample Problem 40-8\n", +"printf('**Sample Prblem 40-8**\n')\n", +"funcprot(0)\n", +"function [P] = f(x)\n", +" P = 1 - (%e^(-2*x))*(1 + 2*x + 2*x^2) - p\n", +"endfunction\n", +"p = fsolve(1, f)\n", +"printf('The possible value of radius is %1.2f*a', p)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/41-All_About_Atoms.ipynb b/Fundamental_Of_Physics_by_D_Haliday/41-All_About_Atoms.ipynb new file mode 100644 index 0000000..302c81a --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/41-All_About_Atoms.ipynb @@ -0,0 +1,205 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 41: All About Atoms" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Bdot = 1.4/10^-3 //in T/m\n", +"w = 3.5*10^-2 //in m\n", +"v = 750 //in m/s\n", +"M = 1.8*10^-25 //in kg\n", +"u = 9.27*10^-24 //in J/T\n", +"\n", +"//Sample Problem 41-1\n", +"printf('**Sample Problem 41-1**\n')\n", +"Fz = u*Bdot\n", +"a = Fz/M\n", +"t = w/v\n", +"d = 0.5*a*t^2\n", +"printf('The atoms have been deflected %e m', d)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"B = 1.80 //in T\n", +"Uz =1.41*10^-26 //in J/T\n", +"h = 6.62*10^-34 //in J-s\n", +"c = 3*10^8 //in m/s\n", +"\n", +"//Sample Problem 42-2\n", +"printf('**Sample Problem 42-2**\n')\n", +"f = 2*Uz*B/h\n", +"printf('The frequency of the alternating field is %e Hz\n', f)\n", +"lambda = c/f\n", +"printf('The wavelength of the field is %fm', lambda)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 41_3\n", +"printf('Sample Problem 42_3')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"K = 35*10^3 //in ev\n", +"e = 1.6*10^-19 //in coulomb\n", +"h = 6.62*10^-34 //in J-s\n", +"c = 3*10^8 //in m/s\n", +"\n", +"//Sample Problem 41-4\n", +"printf('**Sample Problem 41-4**\n')\n", +"lambdaMin = h*c/(K*e)\n", +"printf('The cutoff wavelength is %em', lambdaMin)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Kalpha = 178.9*10^-12 //in m\n", +"Kimpure = 143.5*10^-12 //in m\n", +"Z = 27\n", +"\n", +"//sample Problem 41-5\n", +"printf('**Sample Problem 41-5**\n')\n", +"Zx = sqrt(Kalpha/Kimpure)*(Z-1) + 1\n", +"printf('The proton number of the impurity is %d', Zx)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 41.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"lambda = 550*10^-9 //in m\n", +"T = 300 //in K room temprature\n", +"h = 6.62*10^-34 //in J-s\n", +"c = 3*10^8 //in m/s\n", +"e = 1.6*10^-19\n", +"K = 8.62*10^-5*e\n", +"\n", +"//Sample Problem 41-6a\n", +"printf('**Sample Problem 41-6a**\n')\n", +"deltaE = h*c/lambda\n", +"ratio = %e^(-(deltaE)/(K*T))\n", +"printf('The ratio is equal to %e\n', ratio)\n", +"\n", +"//Sample Problem 41-6b\n", +"printf('\n**Sample Problem 41-6b**\n')\n", +"ratio = 0.5\n", +"T = -deltaE/K/log(ratio)\n", +"printf('The temprature required for the given ratio is equal to %fK', T)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/42-Conduction_of_Electricity_in_Solids.ipynb b/Fundamental_Of_Physics_by_D_Haliday/42-Conduction_of_Electricity_in_Solids.ipynb new file mode 100644 index 0000000..b3f12dd --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/42-Conduction_of_Electricity_in_Solids.ipynb @@ -0,0 +1,248 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 42: Conduction of Electricity in Solids" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"T = 300 //in K\n", +"e = 1.6*10^-19 //in coulomb\n", +"Eg = 5.5*e //in J\n", +"K = 8.62*10^-5*e //in J/K\n", +"\n", +"//Sample Problem 42-1\n", +"printf('**Sample Problem 42-1**\n')\n", +"P = %e^(-Eg/(K*T))\n", +"printf('The probability is equal to %e', P)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 2\n", +"V = 2*10^-6 //in m^3\n", +"density = 1.738*10^3 //in kg/m^3\n", +"M = 24.312*10^-3 //in kg/m^3\n", +"Na = 6.023*10^23\n", +"\n", +"//Sample Problem 42-2\n", +"printf('**Sample Problem 42-2**\n')\n", +"num = density*V*Na/M\n", +"electrons = n*num\n", +"printf('The number of electrons is equal to %e', electrons)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"E = 7 //in ev\n", +"V = 2*10^-9 //in m^3\n", +"density = 2*10^28 //in m^3/ev\n", +"deltaE = 3*10^-3 //in ev\n", +"\n", +"//Sample Problem 42-3a\n", +"printf('**Sample Problem 42-3a**\n')\n", +"n = density*V\n", +"printf('The number of states are equal to %1.2e per ev\n', n)\n", +"\n", +"//Sample Problem 42-3b\n", +"printf('\n**Sample Problem 42-3b**\n')\n", +"n = n*deltaE\n", +"printf('The number of states are equal to %1.2e per ev\n', n)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"E = 0.10 //in ev\n", +"T = 800 //in K\n", +"k = 8.62*10^-5 //Boltzman constant\n", +"\n", +"//Sample Problem 42-4a\n", +"txt = mopen('Example42_4_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 42-4a**\n')\n", +"expo = E/(k*T)\n", +"P = 1/(%e^expo + 1)\n", +"mfprintf(txt, 'The probability of occupying the given energy state is equal to %f\n', P)\n", +"\n", +"//Sample Problem 42-4b\n", +"Pbelow = 1/(1 + %e^-expo)\n", +"mfprintf(txt, '\n**Sample Problem 42-4**\n')\n", +"mfprintf(txt, 'The probability of occupying the given energy state is equal to %f', Pbelow)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"E = 7.0 //in ev\n", +"density = 2*10^28 //density of states \n", +"V = 2*10^-9 //in m^3\n", +"\n", +"//Sample Problem 42-5\n", +"txt = mopen('Example42_5_result.txt', 'wt')\n", +"mfprintf(txt, '**Sample Problem 42-5**\n')\n", +"P = 0.50\n", +"No = density * P * V\n", +"mfprintf(txt, 'Number of occupied states per eV at 7 ev is equal to %e', No)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"No = 10^16 //number per m^3\n", +"T = 298 //in K\n", +"fac = 10^6\n", +"density = 2330 //in kg/m^3\n", +"Na = 6.023*10^23\n", +"M = 28.1*10^-3 //in kg/mol\n", +"\n", +"//Sample Problem 42-6\n", +"pt = mopen('Example42_6_result.txt', 'wt')\n", +"mfprintf(pt, '**Sample Problem 42-6**\n')\n", +"Np = fac*No + No\n", +"NSi = density*Na/M\n", +"fraction = Np/NSi\n", +"mfprintf(pt, 'The fraction of Phosphorus atoms with Silicon atoms is equal to %e', fraction)\n", +"mclose(pt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 42.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Egap = 1.9 //in eV\n", +"\n", +"//Sample Problem 42-7\n", +"pt = mopen('Example42_7_result.txt', 'wt')\n", +"mfprintf(pt, '**Sample Problem 42-7**\n')\n", +"lambada = 1242/Egap\n", +"mfprintf(pt, 'The wavelength emitted is equal to %dnm', lambada)\n", +"mclose(pt)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/43-Nuclear_Physics.ipynb b/Fundamental_Of_Physics_by_D_Haliday/43-Nuclear_Physics.ipynb new file mode 100644 index 0000000..fbfc8ef --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/43-Nuclear_Physics.ipynb @@ -0,0 +1,341 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 43: Nuclear Physics" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.10: Sample_Problem_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"e = 1.6*10^-19 //conversion from electron volt to Joule\n", +"deltaE = 0.20*e\n", +"h = 6.62*10^-34 //in J.s\n", +"\n", +"//Sample Problem 43-10\n", +"txt = mopen('Example43_10_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-9**\n')\n", +"Tavg = h/(2*%pi)/deltaE\n", +"mfprintf(txt, 'The average life of the compound is %esec', Tavg)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"e = 1.6*10^-19 //ev to joule conversion\n", +"E = 5.30*10^6*e //in Joules\n", +"n = 79 //number of protons\n", +"\n", +"//Sample Problem 43-1\n", +"txt = mopen('Example43_1_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-1**\n')\n", +"K = 9*10^9 //in SI unit\n", +"q1 = 2*e\n", +"q2 = n*e\n", +"//K*q1*q2/d = E\n", +"d = K*q1*q2/E\n", +"mfprintf(txt, 'The distance of the alpha particles from gold nucleus is equal to %em', d)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 43-2\n", +"txt = mopen('Example43_2_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-2**\n')\n", +"A = 1 //say for the purpose of calculation\n", +"Mp = 1.67*10^-27\n", +"Mass = A*Mp\n", +"Rnot = 1.2*10^-15\n", +"r = Rnot*A^(1/3)\n", +"Volume = 4/3*%pi*r^3\n", +"density = Mass/Volume\n", +"mfprintf(txt, 'The density of nucleus is %eKg/m^3', density)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Nn = 70\n", +"Np = 50\n", +"A = Nn+Np\n", +"\n", +"//Sample Problem 43-3\n", +"txt = mopen('Example43_3_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-3**\n')\n", +"Msn = 119.902199 //in Atomic mass unit\n", +"uCsqure = 931.5 //in Mev\n", +"Mp = 1.007825 //in Atomic mass unit\n", +"Mn = 1.008625 //in Atomic mass unit\n", +"deltaE = (Np*Mp + Nn*Mn - Msn)*uCsqure/A\n", +"mfprintf(txt, 'The binding energy per nucleon of Sn is %fMev/nucleon', deltaE)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 43-4\n", +"txt = mopen('Example43_4_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-4**\n')\n", +"slope = (0-6.2)/(225-0)\n", +"lambda = -slope\n", +"mfprintf(txt, 'The disintegration constant for the radionuclide is %fmin-1\n', lambda)\n", +"Th = log(2)/lambda\n", +"mfprintf(txt, 'The half life is equal to %dmin', Th)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 2.71 //in g\n", +"R = 4490 //in Bq\n", +"fraction = 1.17/100\n", +"Mo = 74.555 //gm/mol\n", +"Na = 6.023*10^23 //n /mol\n", +"\n", +"//Sample Problem 43-5\n", +"txt = mopen('Example43_5_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-5**\n')\n", +"Nk = Na * M * fraction/Mo\n", +"Th = log(2)*Nk/R\n", +"mfprintf(txt, 'The half life of the substance is %eSec', Th)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"uCsqure = 931.5 //in Mev\n", +"M_H = 1.007825 //in Atomic mass unit\n", +"M_U = 238.05079 //in Atomic mass unit\n", +"M_Th = 234.04363 //in Atomic mass unit\n", +"M_He = 4.00260 //in Atomic mass unit\n", +"M_Pa = 237.05121 //in Atomic mass unit\n", +"\n", +"txt = mopen('Example43_6_result.txt','wt')\n", +"//Sample Problem 43-6a\n", +"mfprintf(txt, '**Sample Problem 43-6a**\n')\n", +"Q = (M_U - (M_Th + M_He)) * uCsqure\n", +"mfprintf(txt, 'Energy released during alpha decay of uranium is %fMev\n', Q)\n", +"\n", +"//Sample Problem 43-6b\n", +"mfprintf(txt, '\n**Sample Problem 43-6b**\n')\n", +"Q = (M_U - (M_Pa + M_H)) * uCsqure\n", +"if(Q<0) \n", +" mfprintf(txt, 'It cannot emit a proton spontaneously')\n", +"else\n", +" mfprintf(txt, 'It can emit proton spontaneously')\n", +"end\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M_P = 31.97391 //in u\n", +"M_S = 31.97207 //in u\n", +"uCsqure = 931.5 //in Mev\n", +"\n", +"//Sample Problem 43-7\n", +"txt = mopen('Example43_7_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-7**\n')\n", +"Q = -(M_S - M_P)*uCsqure\n", +"mfprintf(txt, 'The disintegration energy for the beta decay of Phosphorus is %fMev', Q)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"ratio = 10.3\n", +"Th = 1.25*10^9 //in years\n", +"\n", +"//Sample Problem 43-8\n", +"txt = mopen('Example43_8_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-8**\n')\n", +"t = Th * log(1 + ratio)/log(2) \n", +"mfprintf(txt, 'The life of rock is %eyears', t)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 43.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"dose = 3 //in J/kg\n", +"c = 4180 //in J.kg/K\n", +"m = 1 //(say)\n", +"\n", +"//Sample problem 43-9\n", +"txt = mopen('Example43_9_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 43-9**\n')\n", +"deltaT = (dose/m)/c\n", +"mfprintf(txt, 'The change in temprature of the body is %eK', deltaT)\n", +"mclose(txt)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/44-Energy_from_the_Nucleus.ipynb b/Fundamental_Of_Physics_by_D_Haliday/44-Energy_from_the_Nucleus.ipynb new file mode 100644 index 0000000..1a5e55d --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/44-Energy_from_the_Nucleus.ipynb @@ -0,0 +1,249 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 44: Energy from the Nucleus" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M_U = 235.0439 //in Atomic mass unit\n", +"M_Ce = 139.9054 //in Atomic mass unit\n", +"M_Zr = 93.9063 //in Atomic mass unit\n", +"M_n = 1.00867 //in Atomic mass unit\n", +"uCsqure = 931.5 //in Mev\n", +"\n", +"//Sample Problem 44-1\n", +"txt = mopen('Example44_1_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 44-1**\n')\n", +"Q = -((M_Ce + M_Zr + M_n) - M_U) * uCsqure\n", +"mfprintf(txt, 'The disintegration energy is %dMev', Q)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"c = 3*10^8 //in m/s\n", +"e = 1.6*10^-19\n", +"conv = 3600*24 //day to sec conversion\n", +"Pgen = 3400*10^6 //in W\n", +"Pused = 1100*10^6 //in W\n", +"fuel = 8.60*10^4 //in kg\n", +"Q = 200*10^6*e //in J\n", +"Uinitial = 8.6*10^4 //in kg\n", +"M_u = 3.90*10^-25 //in kg/atom\n", +"N = 5.70*10^4\n", +"p = 3/100\n", +"\n", +"txt = mopen('Example44_2_result.txt','wt')\n", +"//Sample Problem 44-2a\n", +"mfprintf(txt, '**Sample Problem 44-2a**\n')\n", +"eff = Pused/Pgen*100\n", +"mfprintf(txt, 'The efficiency of the power plant is %d\%\n', eff)\n", +"\n", +"//Sample Problem 44-2b\n", +"mfprintf(txt, '\n**Sample Problem 44-2b**\n')\n", +"R = Pgen/Q\n", +"mfprintf(txt, 'The fissions in the reactor per second is %e\n', R)\n", +"\n", +"//Sample Proble 44-2c\n", +"mfprintf(txt, '\n**Sample Problem 44-2c**\n')\n", +"RateDay = (1+0.25)*R*M_u*conv\n", +"mfprintf(txt, 'The uranium use in a day is equal to %fkg/day\n', RateDay)\n", +"\n", +"//Sample Problem 44-2d\n", +"mfprintf(txt, '\n**Sample Problem 44-2d**\n')\n", +"T = fuel*p/RateDay\n", +"mfprintf(txt, 'The U will long for %ddays\n', T)\n", +"\n", +"//Sample Problem 44-2e\n", +"mfprintf(txt, '\n**Sample Problem 44-2e**\n')\n", +"MassConvRate = Pgen/c^2\n", +"mfprintf(txt, 'The mass conversion rate is %ekg/s', MassConvRate)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"ratio = 0.0072\n", +"T = 2.0*10^9 //in years\n", +"Th1 = 7.04*10^8 //in years\n", +"Th2 = 44.7*10^8 //in years\n", +"\n", +"//Sample Problem 44-3\n", +"txt = mopen('Example44_3_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 44-3**\n')\n", +"l1 = log(2)/Th1\n", +"l2 = log(2)/Th2\n", +"ratioEarlier = ratio*%e^((l1 - l2)*T)\n", +"mfprintf(txt, 'The earlier ratio is equal to %f', ratioEarlier)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 10^-15 //in meter\n", +"e = 1.6*10^-19 //in coloumb\n", +"q1 = e\n", +"k = 9*10^9 //in SI unit\n", +"B = 1.38*10^-23 //in J/K\n", +"\n", +"txt = mopen('Example44_4_result.txt','wt')\n", +"//Sample Problem 44-4a\n", +"mfprintf(txt, '**Sample Problem 44-4a**\n')\n", +"K = k*q1^2/(2*R)/2\n", +"mfprintf(txt, 'The initial kinetic energy is equal to %dKev\n', K/e/10^3)\n", +"\n", +"//Sample Problem 44-4b\n", +"mfprintf(txt, '\n**Sample Problem 44-4b**\n')\n", +"T = 2*K/(3*B) //B is Boltzman constant\n", +"mfprintf(txt, 'The temprature required to achieve that energy equal to %eK', T)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 44-5\n", +"txt = mopen('Example44_5_result.txt','wt')\n", +"mfprintf(txt, '**Sample Problem 44-5**\n')\n", +"MassRate = 4*1.67*10^-27/(4.20*10^-12) //mass of proton required to produce 1 unit of energy \n", +"Ps = 3.90*10^26 //in W\n", +"Rate = MassRate*Ps\n", +"mfprintf(txt, 'The rate at which hydrogen is consumed is %ekg/s', Rate)\n", +"mclose(txt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 44.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"d = 200 //in kg/m^3\n", +"fac = 10^3\n", +"Na = 6.023*10^23\n", +"Mt = 3*10^-3 //in kg/mol\n", +"Md = 2*10^-3 //in kg/mol\n", +"\n", +"txt = mopen('Example44_6_result.txt','wt')\n", +"//Sample Problem 44-6a\n", +"mfprintf(txt, '**Sample Problem 44-6a**\n')\n", +"n = 2*fac *d *Na /(Mt + Md)\n", +"mfprintf(txt, 'The number of particle in unit volume is %em^-3\n', n)\n", +"\n", +"//Sample Problem 44-6b\n", +"mfprintf(txt, '\n**Sample Problem 44-6b**\n')\n", +"TauMin = 10^20/n\n", +"mfprintf(txt, 'The duration of time, pallet can maintain is of the order of %esec', TauMin)\n", +"mclose(txt)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/45-Quarks_Leptons_and_the_Big_Bang.ipynb b/Fundamental_Of_Physics_by_D_Haliday/45-Quarks_Leptons_and_the_Big_Bang.ipynb new file mode 100644 index 0000000..2a3b31f --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/45-Quarks_Leptons_and_the_Big_Bang.ipynb @@ -0,0 +1,199 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 45: Quarks Leptons and the Big Bang" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Rpi = 139.6 //in Mev\n", +"Rmu = 105.7 //in Mev\n", +"c = 3*10^8 //in m/s\n", +"\n", +"//Sample Problem 45-1\n", +"pt = mopen('Example45_1_result.txt', 'wt')\n", +"mfprintf(pt, '**Sample Problem 45-1**\n')\n", +"Kmu = poly(0, 'Kmu')\n", +"//CONSERVATION OF ENERGY\n", +"//Rpi + Kpi = Rmu + Kmu + Rv + Kv\n", +"//putting Rv(as mass is 0) & Kpi equal to 0\n", +"//Rpi = Rmu + Kmu + Kv\n", +"SUM = Rpi - Rmu //sum of Kmu & Kv\n", +"Kv = SUM - Kmu\n", +"//for neutrino\n", +"Pv = Kv/c\n", +"//CONSERVATION OF LINEAR MOMENTUM\n", +"//Ppi = Pmu + Pv\n", +"//putting Ppi equal to 0\n", +"Pmu = - Pv\n", +"//for pion\n", +"P = Kmu + 2*Kmu*Rmu - (Pmu*c)^2\n", +"Kmu = roots(P)\n", +"Kmu = Kmu(2)\n", +"mfprintf(pt, 'The kinetic energy of the antimuon is %fMev\n', Kmu)\n", +"mfprintf(pt, 'The kinetic energy of the neutrino is %fMev\n', SUM - Kmu)\n", +"mclose(pt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.2: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Epi = 139.6 //in Mev\n", +"Ek = 493.7 //in Mev\n", +"Ep = 983.3 //in Mev\n", +"Es = 1189.4 //in Mev\n", +"\n", +"//Sample Problem 45-2\n", +"pt = mopen('Example45_2_result.txt', 'wt')\n", +"mfprintf(pt, '**Sample Problem 45-2**\n')\n", +"Q = Epi + Ep - Ek - Es\n", +"mfprintf(pt, 'The energy of the reaction is %dMev', Q)\n", +"mclose(pt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Ep = 938.3 //in Mev\n", +"Epi = 135.0 //in Mev\n", +"Epip = 139.6 //in Mev\n", +"\n", +"//Sample Problem 45-3\n", +"pt = mopen('Example45_3_result.txt', 'wt')\n", +"mfprintf(pt, '**Sample Problem 45-3**\n')\n", +"Q = Ep - Epi - Epip\n", +"if (Q > 0) then\n", +" mfprintf(pt, 'The proton can decay according to given scheme')\n", +"else\n", +" mfprintf(pt, 'The proton cannot decay according to the given scheme')\n", +"end\n", +"mclose(pt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"v = 2.8*10^8 //in m/s\n", +"H = 19.3*10^-3 //in m/s.ly\n", +"\n", +"//Sample Problem 45-6\n", +"pt = mopen('Example45_6_result.txt', 'wt')\n", +"mfprintf(pt, '**Sample Problem 45-6**\n')\n", +"r = v/H\n", +"mfprintf(pt, 'The quasar is approx at a distance of %ely', r)\n", +"mclose(pt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 45.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"w = 1 //(say)\n", +"W = 1.1*w\n", +"c = 3*10^8 //in m/s\n", +"H = 19.3*10^-3 //in m/s.ly\n", +"\n", +"//Sample Problem 45-7\n", +"pt = mopen('Example45_7_result.txt', 'wt')\n", +"mfprintf(pt, '**Sample Problem 45-7**\n')\n", +"deltaW = W - w\n", +"r = c/H * deltaW/w\n", +"mfprintf(pt, 'The galaxy is at a distance of %ely', r)\n", +"mclose(pt)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/5-Force_and_Motion_l.ipynb b/Fundamental_Of_Physics_by_D_Haliday/5-Force_and_Motion_l.ipynb new file mode 100644 index 0000000..2683c61 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/5-Force_and_Motion_l.ipynb @@ -0,0 +1,351 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Force and Motion l" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"m = .20 //in kg\n", +"F_1 = 4 * [1,0]\n", +"F_2 = 2 * [-1,0]\n", +"F_3 = 1 * [cos(dtor(30)),sin(dtor(30))]\n", +"\n", +"//Sample Problebb nmkn nm 5-1\n", +"printf('**Sample Problem 5-1**\n')\n", +"acceleration_a = F_1(1)/m\n", +"acceleration_b = F_2(1)/m\n", +"acceleration_c = (F_2(1) + F_3(1))/m\n", +"printf('The acceleration of puck in case a is %d m/s^2\n',acceleration_a)\n", +"printf('The acceleration of puck in case b is %d m/s^2\n',acceleration_b)\n", +"printf('The acceleration of puck in case c is %f m/s^2\n',acceleration_c)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"mass = 2 //in kg\n", +"acceleration = 3 * [cos(dtor(50)),sin(dtor(50))] //in m/s^2\n", +"F1 = 10 * [cos(dtor(180+30)),sin(dtor(180+30))] //in N\n", +"F2 = 20 * [0,1] //in N\n", +"\n", +"//Sample Problem 5-2\n", +"printf('**Sample Problem 5-2**\n')\n", +"//from newton's first law\n", +"//F1 + F2 + F3 = mass * acceleration\n", +"F3 = mass * acceleration - F2 - F1\n", +"printf('The third Force vector F3 in N is')\n", +"disp(F3)\n", +"printf('The magnitude F3 is %f m/s\n', norm(F3))\n", +"printf('The angle made by F3 with the x axis in degrees %f', rtod(atan(F3(2)/F3(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//GIven that\n", +"F_A = 220 * [cos(dtor(180-47)),sin(dtor(180-47))] //in N\n", +"F_B_dir = [0,-1]\n", +"F_C_mag = 170 //in N\n", +"\n", +"//Sample Problem 5-3\n", +"printf('**Sample Problem 5-3**\n')\n", +"//net sum of three forces must be zero\n", +"phi = acos(- F_A(1) / F_C_mag)\n", +"F_B_mag = F_C_mag * sin(phi) + F_A(2)\n", +"printf('The magnitude of Bettys force is %f N', F_B_mag)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"g = 10 //in m/s^2\n", +"mass = 80 //in kg\n", +"theta = 30 //in degrees\n", +"Force_applied = 2.5 * mass * g * [cos(dtor(theta)),sin(dtor(theta))]\n", +"W_car = 7 * 10^5 //in N\n", +"distance = 1 //in meter\n", +"\n", +"//Sample Problem 5-4\n", +"printf('**Sample Problem 5-4**\n')\n", +"//using Newtons first law\n", +"acceleration = Force_applied(1) / (W_car /g)\n", +"//using newtons third equation of motion\n", +"v_final = sqrt(2 * acceleration * distance)\n", +"printf('The velocity after trevelling 1 m distance is %f m/s', v_final)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"M = 3.3 //in kg\n", +"m = 2.1 //in kg\n", +"\n", +"//Sample Problem 5-5\n", +"printf('**Sample Problem 5-5**\n')\n", +"//from FBD1\n", +"//both will have common acceleration\n", +"//mg - T = ma\n", +"//T = Ma\n", +"//adding -> mg = (M+m)a\n", +"a = m * g /(M + m)\n", +"T = m *g - m *a\n", +"printf('The acceleration of both the blocks is %f m/s^2\n', a)\n", +"printf('The tension in the string is %f N', T)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"m = 15 //in kg\n", +"//from FBD\n", +"T = m *g\n", +"\n", +"//Sample Problem 5-6\n", +"printf('**Sample Problem 5-6**\n')\n", +"//we have-\n", +"//T1cos(28) - T2cos(47) = 0\n", +"//T1sin(28) + T2sin(47) = T\n", +"//therefore\n", +"mat_1 = [cos(dtor(28)),-cos(dtor(47));sin(dtor(28)),sin(dtor(47))]\n", +"mat_2 = [0 ; T]\n", +"//wr have -> mat_1 * ans = mat_2\n", +"mat_ans = inv(mat_1) * mat_2\n", +"printf('The tension in the first chord is %f N\n', mat_ans(1,1))\n", +"printf('The tension in the second chord is %f N\n', mat_ans(2,1))\n", +"printf('The tension in the third chord is %f N', T)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci', -1)\n", +"\n", +"//Given that\n", +"m = 15 //in kg\n", +"g = 9.8 //in m/s^2\n", +"T = m* g* sin(dtor(27))\n", +"N = m* g* cos(dtor(27))\n", +"\n", +"//Sample Problem 5-7a\n", +"printf('**Sample Problem 5-7a**\n')\n", +"printf('The tension in the chord is %f N\n', T)\n", +"printf('The Normal force is %f N\n', N)\n", +"\n", +"//Sample Problem 5-7b\n", +"printf('\n**Sample Problem 5-7b**\n')\n", +"a = g * sin(dtor(27))\n", +"printf('The acceleration of block after cutting the chord is %f m/s^2', a)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"m = 72.2 //in kg\n", +"\n", +"//Sample Problem 5-8a\n", +"printf('**Sample Problem 5-8a**\n')\n", +"scale_read_b = m *g\n", +"printf('The reading of the scale in case a is %f ', scale_read_b)\n", +"\n", +"//Sample Problem 5-8b\n", +"printf('**Sample Problem 5-8b**\n')\n", +"//N - mg = ma\n", +"a_U = 3.2 //in m/s^2\n", +"scale_read_c_U = m * (g + a_U)\n", +"printf('The reading of the scale in case b if cab moves upaward is %f\n', scale_read_c_U)\n", +"a_D = -3.2 //in m/s^2\n", +"scale_read_c_D = m * (g + a_D)\n", +"printf('The reading of the scale in case b if cab moves downaward is %f\n', scale_read_c_D)\n", +"\n", +"//Sample Problem 5-8b\n", +"printf('\n**Sample Problem 5-8b**\n')\n", +"//using newtons second law => net force = mass * acceleration\n", +"F_net = m *a_U\n", +"printf('The net force on passenger during upward journey is %f N\n', F_net)\n", +"printf('The acceleration of passenger in the frame if cab is 0')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 5-9a\n", +"F_ap = 20 //in N\n", +"m_A = 4 //in kg\n", +"m_B = 6 //in kg\n", +"\n", +"//Sample Problem 5-9a\n", +"printf('**Sample Problem 5-9a**\n')\n", +"ac = F_ap /(m_A + m_B)\n", +"printf('The comon acceleration of the blocks is %f m/s^2\n', ac)\n", +"\n", +"//Sample Problem 5-9b\n", +"printf('\n**Sample Problem 5-9b**\n')\n", +"//from FBD of A\n", +"//F - F_AB = m_A * a\n", +"F_AB = F_ap - m_A *ac\n", +"printf('The force on block B by Block A is %f N', F_AB)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/6-Force_and_Motion_ll.ipynb b/Fundamental_Of_Physics_by_D_Haliday/6-Force_and_Motion_ll.ipynb new file mode 100644 index 0000000..3d03e2b --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/6-Force_and_Motion_ll.ipynb @@ -0,0 +1,326 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Force and Motion ll" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"mu_k = 0.6\n", +"d = 290 //in meter\n", +"g = 9.8 //in m/s^2\n", +"v_final = 0\n", +"\n", +"//Sample Problem 6-1\n", +"printf('**Sample Problem 6-1**\n')\n", +"//using newton's 3rd equation of motion\n", +"ac = - mu_k * g //due to friction hence a negative sign\n", +"v_initial = sqrt(v_final^2 - 2 * ac * d)\n", +"printf('The inital velocity of that car would have been %f m/s', v_initial)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//GIven that\n", +"g = 9.8 //in /s^2\n", +"mass = 75 //in kg\n", +"mu_k = 0.10\n", +"phi = dtor(42)\n", +"\n", +"//Sample Problem 6-2a\n", +"printf('**Sample Problem 6-2a**\n')\n", +"//T * cos(phi)- mu_k * N = 0\n", +"//T * sin(phi) + N = mass * g\n", +"mat_A = [cos(phi),-mu_k;sin(phi),1]\n", +"mat_B = [0 ;mass * g]\n", +"F = inv(mat_A) * mat_B\n", +"printf('The Tension in the string is %f N\n', F(1))\n", +"\n", +"//Sample Problem 6-2b\n", +"printf('\n**Sample Problem 6-2b**\n')\n", +"printf('The force of friction will not change')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"theta = dtor(13)\n", +"\n", +"//Sample Problem 6-3\n", +"printf('**Sample Problem 6-3**\n')\n", +"//N = mg cos(theta)\n", +"//f_s = mg sin(theta)\n", +"//dividing->\n", +"//f_s/N = tan(theta)\n", +"mu_s = tan(theta)\n", +"printf('The value of coefficient of static friction is %f', mu_s)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"v1 = 97 //in km/hr\n", +"//Assuming\n", +"A1 = 1 \n", +"//therefore\n", +"A2 = 2\n", +"\n", +"//Sample Problem 6-4\n", +"printf('**Sample Problem 6-4**\n')\n", +"//the terminal velocity is inversly proportional to squre root of area\n", +"v2 = v1 * sqrt(A1/A2)\n", +"printf('The final velocity of cat will be %f km/hr', v2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"Radius = 1.5 * 10^-3 //in meter\n", +"height = 1200 //in meter\n", +"C_drag = 0.60\n", +"density_water = 1000 //in kg/m^3\n", +"density_air = 1.2 //in kg/m^3\n", +"\n", +"//Sample Problem 6-5a\n", +"printf('**Sample Problem 6-5a**\n')\n", +"//v_t = sqrt(2*F_g/(C*density*A))\n", +"volume_drop = 4/3*%pi*Radius^3\n", +"mass_drop = density_water *volume_drop\n", +"Area_drop = %pi *Radius^2\n", +"v_terminal = sqrt(2*mass_drop*g/(C_drag*density_air*Area_drop))\n", +"printf('The terminal velocity will be %f m/s\n', v_terminal)\n", +"\n", +"//Sample Problem 6-5b\n", +"printf('\n**Sample Problem 6-5b**\n')\n", +"v_without_drag = sqrt(2 *g * height)\n", +"printf('The velocity just before the impact if there were no drag force would be %f m/s', v_without_drag)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Radius_earth = 6.37 * 10^6\n", +"h_alti = 520 * 10^3 //in meter\n", +"velocity = 7.6 * 10^3 //in m/s converted from km/s\n", +"mass = 79 //in kg\n", +"\n", +"//Sample Problem 6-6a\n", +"printf('**Sample Problem 6-6a**\n')\n", +"acce = velocity^2/(h_alti + Radius_earth)\n", +"printf('The acceleration is equal to %f m/s^2\n', acce)\n", +"\n", +"//Sample Problem 6-5b\n", +"printf('\n**Sample Problem 6-6b**\n')\n", +"Force_total = mass * acce\n", +"printf('The net force is equal ro %f N', Force_total)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"Radius = 2.7 //in meter\n", +"\n", +"//Sample Problem 6-7\n", +"printf('**Sample Problem 6-7**\n')\n", +"//The velocity at highest point is equal to sqrt(gR)\n", +"velocity_topmost = sqrt(g * Radius)\n", +"printf('The velocity of ball at the topmost point would be %f m/s', velocity_topmost)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"R = 2.1 //in m\n", +"mu_s = 0.40\n", +"\n", +"//Sample Problem 6-8a\n", +"printf('**Sample Problem 6-8a**\n')\n", +"//N = mv^2/R\n", +"//mg = mu_s * N\n", +"//mg = mu_s * m*v^2/R\n", +"//v = sqrt(g*R/mu_s)\n", +"v_min = sqrt(g*R/mu_s)\n", +"printf('The minimum spped of the cylender should be %f m/s\n', v_min)\n", +"\n", +"//Sample Problem 6-8b\n", +"printf('\n**Sample Problem 6-8b**\n')\n", +"mass = 49 //in kg\n", +"c_force = mass * v_min^2/R\n", +"printf('The Centripetal force on the rider would be %f N', c_force)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"mass = 1600 //in kg\n", +"v_cons = 20 //in m/s\n", +"R_track = 190 //in meter\n", +"\n", +"//Sample Problem 6-9\n", +"printf('**Sample Problem 6-9**\n')\n", +"N = mass * g\n", +"f_s = mass * v_cons^2 /R_track\n", +"mu_s = f_s/N\n", +"printf('The coefficient of static friction for the given surface is %f', mu_s)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/7-Kinetic_Energy_and_Work.ipynb b/Fundamental_Of_Physics_by_D_Haliday/7-Kinetic_Energy_and_Work.ipynb new file mode 100644 index 0000000..81d79a4 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/7-Kinetic_Energy_and_Work.ipynb @@ -0,0 +1,413 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: Kinetic Energy and Work" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.10: Sample_Problem_10.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"//taking right direction as positive direction\n", +"F1 = [-2, 0] //in N\n", +"v = [3, 0] //in m/s\n", +"mag = [4,6]\n", +"\n", +"//Sample Problem 7-10a&b\n", +"Qnum = ['a', 'b']\n", +"count = 1\n", +"for x = mag\n", +" printf('\n**Sample Problem 7-10%s**\n', Qnum(count))\n", +" F2 = [x*cos(dtor(60)), x*sin(dtor(60))] //in N\n", +" //from the definition of the power\n", +" P1 = F1 * v'\n", +" P2 = F2 * v'\n", +" Pnet = P1 + P2\n", +" printf('The power transferred by F1 is %dW\n', P1)\n", +" printf('The power transferred by F2 is %dW\n', P2)\n", +" printf('The net power transferred is %dW\n', Pnet)\n", +" Pnet = round(Pnet)\n", +" if Pnet == 0\n", +" printf('The Kinetic energy is not changing\n')\n", +" else\n", +" printf('The Kinetic Energy is changing\n')\n", +" end\n", +" count = count + 1\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"weight_locomotive = 1.2 * 10^6 //Ng = 9.8 //in m/s^2\n", +"acceleration = 0.26 //m/s^2\n", +"v_final = 0 //m/s\n", +"distance = 3.2 * 10^3 //m\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 7-1\n", +"printf('**Sample Problem 7-1**\n')\n", +"//using newton's second equation of motion\n", +"v_initial = sqrt(v_final^2 + 2 * acceleration * distance)\n", +"total_kinetic_energy = 2 * .5 * weight_locomotive/g * v_initial^2\n", +"printf('The total kinetic energy of two locomotive just before the collision is %e J', total_kinetic_energy)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"mass = 225 //in kg\n", +"displacement = 8.5 //in meter\n", +"F1 = 12 //in N\n", +"Theta1 = dtor(30) //in rad\n", +"F2 = 10 //in N\n", +"Theta2 = dtor(40) //in rad\n", +"\n", +"//Sample Problem 7-2a\n", +"printf('**Sample Problem 7-2a**\n')\n", +"//form the definition of Work done\n", +"W1 = F1 * displacement * cos(Theta1)\n", +"W2 = F2 * displacement * cos(Theta2)\n", +"//The net work done\n", +"W = W1 + W2\n", +"printf('The spies transfer %eJ amount of energy during the given displacement\n', W)\n", +"\n", +"//Sample Problem 7-2b\n", +"printf('\n**Sample Problem 7-2b**\n')\n", +"printf('Work done by the normal in the process is 0J\n')\n", +"printf('Work done by the gravity in the process is 0J\n')\n", +"\n", +"//Sample Problem 7-2c\n", +"printf('\n**Sample Problem 7-2c**\n')\n", +"//Using Work Energy theorem\n", +"//W = Kf - Ki\n", +"//Ki = 0\n", +"v_final = sqrt(2* W/mass)\n", +"printf('The final velocity of the safe is equal to %e', v_final)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"d = [-3, 0] //in meter\n", +"F = [2, -6] //in N\n", +"\n", +"//Sample Problem 7-3a\n", +"printf('**Sample Problem 7-3a**\n')\n", +"W = F * d'\n", +"printf('The work done is equal to %dJ\n', W)\n", +"\n", +"//Sample Problem 7-3b\n", +"printf('\n**Sample Problem 7-3b**\n')\n", +"Ki = 10 //in J\n", +"//Using work energy theorem\n", +"Kf = Ki + W\n", +"printf('The final kinetic energy of the crate is %dJ', Kf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"mass = 260 //in kg\n", +"d = 2 //in meter\n", +"g = 9.8 //on m/^2\n", +"\n", +"//Sample Problem 7-4a\n", +"printf('**Sample Problem 7-4a**\n')\n", +"//Using definition of work done\n", +"Wg = mass* g * (-1) * d\n", +"printf('The work done by the force of gravity is %dJ\n', Wg)\n", +"\n", +"//Sample Problem 7-4b\n", +"printf('\n**Sample Problem 7-4b**\n')\n", +"//Using work energy theorem\n", +"Wc = -1 * Wg\n", +"printf('The work done by Chemerkins force is %dJ\n', Wc)\n", +"\n", +"//Sample Problem 7-4c\n", +"printf('\n**Sample Problem 7-4c**\n')\n", +"printf('The Work done in helding the object stationary is 0, as the dispacement is 0\n')\n", +"\n", +"//Sample Problem 7-4d\n", +"printf('\n**Sample Problem 7-4d**\n')\n", +"Weight = 27900 //in N\n", +"d = 1 * 10^-2 //in meter\n", +"Wg = Weight * d\n", +"printf('The work done by the gravity is %dJ', Wg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"mass = 15 //in kg\n", +"L = 5.7 //in meter\n", +"h = 2.5 //in meter\n", +"g = 9.8\n", +"\n", +"//Sample Problem 7-5a\n", +"printf('**Sample Problem 7-5a**\n')\n", +"//Using the definition of work done\n", +"Wg = - mass * g * h\n", +"printf('The work done by the gravity is during the pull is %eJ\n', Wg)\n", +"\n", +"//Sample Problem 7-5b\n", +"printf('\n**Sample Problem 7-5b**\n')\n", +"//Using Work-Energy theorem\n", +"//as there is no change in kinetic energy\n", +"Wt = - Wg\n", +"printf('The work done by the tension during the pull is %eJ', Wt)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"g = 9.8 //in m/s^2\n", +"m = 500 //in kg\n", +"Vi = 4 //in m/s\n", +"a = g/5 //in m/s\n", +"d = 12 //in meter\n", +"\n", +"//Sample Problem 7-6a\n", +"printf('**Sample Problem 7-6a**\n')\n", +"//Using the definition of the work done\n", +"Wg = m * g * d\n", +"printf('The work done by the gravity during the fall is %eJ\n', Wg)\n", +"\n", +"//Sample Problem 7-6b\n", +"printf('\n**Sample Problem 7-6b**\n')\n", +"//from Example7-6_FBD\n", +"//m*g - T = m*a\n", +"T = m * (g - a)\n", +"Wt = - T * d\n", +"printf('The work done by the tension is %eJ\n', Wt)\n", +"\n", +"//Sample Problem 7-6c\n", +"printf('\n*Sample Problem 7-6c**\n')\n", +"Wnet = Wt + Wg\n", +"printf('The net work done is %eJ\n', Wnet)\n", +"\n", +"//Sample Problem 7-6d\n", +"printf('\n**Sample Problem 7-6d**\n')\n", +"//Using work energy theorem\n", +"//Kf - Ki = Wnet\n", +"Kf = Wnet + .5 * m * Vi^2\n", +"printf('The final kinetic energy of the cab is %eJ', Kf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Fa = 4.9 //in N\n", +"x1 = 12 * 10^-3 //in meter\n", +"x2 = 17 * 10^-3 //in meter\n", +"x3 = -12 * 10^-3 //in meter\n", +"\n", +"//Sample Problem 7-7a\n", +"printf('**Sample Problem 7-7a**\n')\n", +"//k * x1 = Fa\n", +"//Spring constant\n", +"k = Fa/x1\n", +"Ws1 = -.5* k * x2^2\n", +"printf('The work done by the spring force is %fJ\n', Ws1)\n", +"\n", +"//Sample Problem 7-7b\n", +"printf('\n**Sample Problem 7-7b**\n')\n", +"Ws2 = .5 * k * (x2^2 - x3^2)\n", +"printf('The work done by the spring is %fJ', Ws2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m=.4 //in kg\n", +"Vi = .5 //in m/s\n", +"k = 750 //in N/m\n", +"\n", +"//Sample Problem 7-8\n", +"printf('**Sample Problem 7-8**\n')\n", +"//Using work energy theorem\n", +"//Wnet = Kf - Ki\n", +"//Kf = 0\n", +"//.5*k*x^2 = Ki\n", +"x = sqrt(m*Vi^2/k)\n", +"printf('The compression in the spring is %em', x)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 7-9\n", +"printf('**Sample Problem 7-9**\n')\n", +"W = integrate('3*x^2', 'x', 2, 3) + integrate('4', 'x', 3, 0)\n", +"printf('The net work done is %dJ', W)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/8-Potential_and_Conservation_of_Energy.ipynb b/Fundamental_Of_Physics_by_D_Haliday/8-Potential_and_Conservation_of_Energy.ipynb new file mode 100644 index 0000000..d27144a --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/8-Potential_and_Conservation_of_Energy.ipynb @@ -0,0 +1,319 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Potential and Conservation of Energy" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"h = 0.8 //in meter\n", +"l = 2.0 //in meter\n", +"m = 2 //in kg\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 8-1\n", +"printf('**Sample Problem 8-1**\n')\n", +"//Using the definition of \n", +"Wg = m* g* h\n", +"printf('The work done by the gravity is %eJ', Wg)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"mass = 2 //in kg\n", +"y1 = 5 //in meter\n", +"g = 9.8 //in m/s^2\n", +"ref = [0, 3, 5, 6]\n", +"\n", +"//Sample Problem 8-2a\n", +"printf('**Sample Problem 8-2a**\n')\n", +"for x = ref\n", +" U = mass* g* (y1 - x)\n", +" printf('The potential energy at y1 is %dJ if reference is assumed to be at y=%d\n', U, x)\n", +"end\n", +"\n", +"//sample Probelm 8-2b\n", +"printf('\n**Sample Problem 8-2b**\n')\n", +"y2 = 0\n", +"//The change in potential energy doesn't depend on choice of reference\n", +"deltaY = y2 - y1\n", +"//because the deltaY doesn't depend upon the choice of references\n", +"deltaU = mass* g* deltaY\n", +"printf('The change in potential energy is %dJ same for all the references', deltaU)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"h = 8.5 //in meter\\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 8-3\n", +"printf('**Sample Problem 8-3**\n')\n", +"//Using conservation of energy\n", +"//Change in Potential energy = Change in Kinetic energy\n", +"//m* g* h = .5* m* v^2\n", +"v = sqrt(2* g * h)\n", +"printf('The velocity of child at the bottom will be %fm/s', v)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 61 //in kg\n", +"Hi = 45 //in meter\n", +"L = 25 //in meter\n", +"k = 160 //in N/m\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 8-4\n", +"printf('**Sample Problem 8-4**\n')\n", +"//Assume that the jumper goes x meter down the bridge\n", +"//Using energy conservation energy\n", +"s=poly(0,'s')\n", +"p = .5*k*(s-25)^2 - m*g*s //equal to 0\n", +"x = roots(p)\n", +"printf('The height fell down is %fm', x(2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"n = 25\n", +"Ms = 9000 //in kg\n", +"Mm = 80 //in kg\n", +"d = [45, 10000] //in meter\n", +"t = 2 //in minute\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 8-5\n", +"count = 0\n", +"for x = d\n", +" //Sample Problem 8-5a\n", +" if count == 0 then\n", +" printf('**Sample Problem 8-5a**\n')\n", +" count = count + 1\n", +" end\n", +" //We can assume that the force applied by each of man is equal to twice his weight\n", +" Wnet = n * (2* Mm * g) * x\n", +" printf('The total work done is equal to %eJ\n', Wnet)\n", +" \n", +" //Sample Problem 8-5b\n", +" if count == 1 then\n", +" printf('\n**Sample Problem 8-5b**\n')\n", +" count = count + 1 \n", +" end\n", +" //All the work done must be converted into thermal energy\n", +" TE = Wnet\n", +" printf('The total thermal energy generated is equal to %eJ\n', TE)\n", +" \n", +" //Sample Problem 8-5c\n", +" if count == 2 then\n", +" printf('\n**Sample Problem 8-5c**\n')\n", +" count = count + 1 \n", +" end\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 14 //in kg\n", +"F = 40 //in N\n", +"d = 0.50 //in meter\n", +"Vo = 0.60 //in m/s\n", +"V = 0.20 //in m/s\n", +"\n", +"//Sample Problem 8-6a\n", +"printf('**Sample Problem 8-6a**\n')\n", +"//Using the definition of Work done\n", +"Wf = F* d\n", +"printf('The work done by the force F is equal to %dJ\n', Wf)\n", +"\n", +"//Sample Problem 8-6b\n", +"printf('\n**Sample Problem 8-6b**\n')\n", +"//Using Work-Energy theorem\n", +"//TE = decrease in Kinetic Energy + Work done by the force F\n", +"TE = .5*m*(Vo^2 - V^2) + Wf\n", +"printf('The incrase in the thermal energy is equal to %fJ', TE)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"mass = 2.0 //in kg\n", +"v1 = 4.0 //in m/s^2\n", +"Ff = 15 //in N\n", +"k = 10^4 //in N/m\n", +"\n", +"//Sample Problem 8-7\n", +"printf('**Sample Problem 8-7**\n')\n", +"//Using energy conservation\n", +"//Ki = Uf + TEf\n", +"Ki = .5* mass* v1^2\n", +"//Uf = .5*k*x^2\n", +"//TEf = Ff* x\n", +"s=poly(0,'s')\n", +"p = .5*k*s^2 + Ff* s - Ki\n", +"x = roots(p)\n", +"printf('The compression in the spring is equal to %fcm', x(1)*100)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 6.0 //in kg\n", +"Vo = 7.8 //in m/s\n", +"Yo = 8.5 //in meter\n", +"Y = 11.1 //in meter\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 8-8\n", +"printf('**Sample Problem 8-8**\n')\n", +"//initial mechanical energy\n", +"Mi = .5* m* Vo^2 + m* g* Yo\n", +"//final mechanical energy\n", +"Mf = 0 + m* g* Y\n", +"Eth = Mi - Mf\n", +"printf('The thermal energy generated is equal to %fJ', Eth)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamental_Of_Physics_by_D_Haliday/9-System_of_Particles.ipynb b/Fundamental_Of_Physics_by_D_Haliday/9-System_of_Particles.ipynb new file mode 100644 index 0000000..8043cbf --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/9-System_of_Particles.ipynb @@ -0,0 +1,318 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: System of Particles" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"m1 = 1.2 //in kg\n", +"m2 = 2.5 //in kg\n", +"m3 = 3.4 //in kg\n", +"a = 140 //in cm\n", +"\n", +"//Sample Problem 9-1\n", +"printf('**Sample Problem 9-1**\n')\n", +"r1 = [0, 0]\n", +"r2 = [a, 0]\n", +"r3 = [a*cos(dtor(60)), a*sin(dtor(60))]\n", +"Rc = (m1*r1 + m2*r2 + m3*r3)/(m1 + m2 + m3)\n", +"printf('The co-ordinate of center of mass are (%f, %f)', Rc(1), Rc(2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Sample Problem 9-2\n", +"printf('**Sample Problem 9-2**\n')\n", +"R = poly(0, 'R')\n", +"//Mass is proportinal to area\n", +"Aw = %pi* (2*R)^2\n", +"Ac = %pi* R^2\n", +"//& the x-co-ordinate of the masses\n", +"CMw = 0\n", +"CMc = -R\n", +"CMf = pdiv((Aw*CMw - Ac*CMc), (Aw - Ac))\n", +"printf('The CM is located at a distace of %fR to the right of the center', horner(CMf, 1))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"m1 = 4 //in kg\n", +"m2 = 8 //in kg\n", +"m3 = 4 //in kg\n", +"F1 = [-6, 0]\n", +"F2 = [12*cos(dtor(45)), 12*sin(dtor(45))]\n", +"F3 = [14, 0]\n", +"\n", +"//Sample Problem 9-3\n", +"printf('**Sample Problem 9-3**\n')\n", +"aC = (F1 + F2 + F3)/(m1 + m2+ m3)\n", +"printf('The acceleration of center of mass is %fm/s^2 at \nan angle of %f degrees to the positive x-axis', norm(aC), rtod(atan(aC(2)/aC(1))))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"m = 2.0 //in kg\n", +"v1 = [0, -0.50]\n", +"v2 = [0.40, 0]\n", +"\n", +"//Sample Problem 9-4\n", +"printf('**Sample Problem 9-4**\n')\n", +"deltaP = m* (v2 - v1)\n", +"printf('The change in mometum vactor in kg.m/sec is %1.1fi + %1.1fj', deltaP(1), deltaP(2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"//Before collision\n", +"m = 6 //in kg\n", +"v = 4 //in m/sec\n", +"//After collision\n", +"m1 = 2 //in kg\n", +"\n", +"//Sample Problem 9-5\n", +"printf('**Sample Problem 9-5**\n')\n", +"m2 = m - m1\n", +"v1 = 8.0 //in m/s\n", +"v2 = (m*v - m1*v1)/m2\n", +"printf('The velocity of peiece having mass m2 is %dm/s', v2)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"M = 1 //(say)to get the answer directly\n", +"Vi = 2100 //in km/h\n", +"//initial momentum\n", +"Pi = M* Vi\n", +"Vrel = 500 //in km/h\n", +"\n", +"//Sample Problem 9-6\n", +"printf('**Sample Problem 9-6**\n')\n", +"//Assuming Vf as the final velocity of the hauler\n", +"Vf = poly(0, 'Vf')\n", +"Pf = 0.20*M*(Vf - Vrel) + (M - 0.20*M)*Vf\n", +"p = Pi - Pf\n", +"Vf = roots(p)\n", +"printf('The final velocity of the hauler is %dkm/h', Vf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('degree_rad.sci',-1)\n", +"\n", +"//Given that\n", +"M = 1 //(say) to directly get the answer\n", +"Mc = 0.30*M\n", +"Vc = [5*cos(dtor(40)), 5*sin(dtor(40))]\n", +"Mb = 0.20*M\n", +"Ma = 0.50*M\n", +"\n", +"//Sample Problem 9-7\n", +"printf('**Sample Problem 9-7**\n')\n", +"deff('[f] = eq_maker(V)', 'f = Ma*V(1)*[cos(dtor(140)),sin(dtor(140))] + Mb*V(2)*[0,-1] + Mc*Vc')\n", +"V= fsolve([0,0], eq_maker)\n", +"printf('The velocity of A is %dm/s & velocity of B is %fm/s after the collision in the given directions', V(1), V(2))" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"Mi = 850 //in kg\n", +"R = 2.3 //kg/s\n", +"Vrel = 2800 //in kg\n", +"\n", +"//Sample Problem 9-8a\n", +"printf('**Sample Problem 9-8a**\n')\n", +"T = R*Vrel\n", +"printf('The Thrust force rocket engine provide is equal to %dN\n', T)\n", +"\n", +"//Sample Problem 9-8b\n", +"printf('\n**Sample Problem 9-8b**\n')\n", +"a = T/Mi\n", +"printf('The initial acceleration of rocket is %fm/s^2\n', a)\n", +"\n", +"//Sample Problem 9-8c\n", +"printf('\n**Sample Problem 9-8c**\n')\n", +"Mf = 180 //in kg\n", +"Vf = Vrel * log(Mi/Mf)\n", +"printf('The final velocity of the rocket is %fm/s', Vf)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"m = 4.0* 10^-6 //in kg\n", +"l = 0.77 * 10^-3 //in meter\n", +"h = 0.30 //in m\n", +"g = 9.8 //in m/s^2\n", +"\n", +"//Sample Problem 9-9\n", +"printf('**Sample Problem 9-9**\n')\n", +"//Using Work-Energy theorem\n", +"//F * l = mgh\n", +"F = m* g* h/l\n", +"printf('The external force on the betal is %fN', F)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |