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| author | Trupti Kini | 2016-07-20 23:30:30 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-07-20 23:30:30 +0600 |
| commit | d60e20de7489eb21a2964bbcfb3e32bd20057691 (patch) | |
| tree | b9389af95c70af62eceb6361969d49abba3f6916 | |
| parent | 10c80b1dd1eb129240f34e72a7404b771a9bbdc2 (diff) | |
| download | Python-Textbook-Companions-d60e20de7489eb21a2964bbcfb3e32bd20057691.tar.gz Python-Textbook-Companions-d60e20de7489eb21a2964bbcfb3e32bd20057691.tar.bz2 Python-Textbook-Companions-d60e20de7489eb21a2964bbcfb3e32bd20057691.zip | |
Added(A)/Deleted(D) following books
A Principles_of_Physics_by_F.J.Bueche/Chapter1.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter10.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter13.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter14.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter15.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter16.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter18.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter19.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter2.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter20.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter21.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter22.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter23.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter24.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter25.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter26.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter27.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter3.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter4.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter5.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter6.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter7.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter8.ipynb
A Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb
A Principles_of_Physics_by_F.J.Bueche/chapter12.ipynb
A Principles_of_Physics_by_F.J.Bueche/screenshots/11.2.png
A Principles_of_Physics_by_F.J.Bueche/screenshots/24.4.png
A Principles_of_Physics_by_F.J.Bueche/screenshots/8.4.png
30 files changed, 9150 insertions, 0 deletions
diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter1.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter1.ipynb new file mode 100644 index 00000000..420e6d18 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter1.ipynb @@ -0,0 +1,197 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:70a6c85604d1bf040682390393f1b3040cc3d95dac3089997f6535e7c7e77e35"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 01: Vectors and their use"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.1:pg-25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 1_1\n",
+ " \n",
+ "import math \n",
+ " #To add the given Displacements Graphically\n",
+ "d1=25 #units in cm\n",
+ "d2=10 #units in cm\n",
+ "d3=30 #units in cm\n",
+ "R=math.sqrt(d1**2+d2**2+d3**2) #units in cm\n",
+ "theta1=30 #units in degrees\n",
+ "theta2=90 #units in degrees\n",
+ "theta3=120 #units in degrees\n",
+ "theta=360-(theta1+theta2+theta3) #units in degrees\n",
+ "print \"The Resultant R=\",round(R,2),\" cm\\n\"\n",
+ "print \"Theta=\",round(theta),\" degrees\"\n",
+ " #In text book the answer is printed wrong as R=49cm and theta=82 degrees but the correct answer is R=40.31cm and theta=120 degrees\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Resultant R= 40.31 cm\n",
+ "\n",
+ "Theta= 120.0 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.2:pg-25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 1_2\n",
+ "\n",
+ "import math\n",
+ " \n",
+ " # To add the given vector displacements\n",
+ "a=1 #units in meters\n",
+ "b=3 #units in meters\n",
+ "c=5 #units in meters\n",
+ "d=6 #units in meters\n",
+ "theta1=90 #units in degrees\n",
+ "Rx_a=a*math.sin(theta1*math.pi/180) #units in meters\n",
+ "Rx_b=round(b*math.cos(theta1*math.pi/180)) #units in meters\n",
+ "theta2=37 #units in degrees\n",
+ "Rx_c=-round(c*math.cos(theta2*math.pi/180)) #units in meters\n",
+ "theta3=53 #units in degrees\n",
+ "Rx_d=-d*math.cos(theta3*math.pi/180)\n",
+ "Ry_a=round(a*math.cos(theta1*math.pi/180)) #units in meters\n",
+ "Ry_b=round(c*math.sin(theta2*math.pi/180)) #units in meters\n",
+ "Ry_c=round(c*math.sin(theta2*math.pi/180)) #units in meters\n",
+ "Ry_d=-(d*math.sin(theta3*math.pi/180)) #units in meters\n",
+ "Rx=Rx_a+Rx_b+Rx_c+Rx_d #units in meters\n",
+ "Ry=Ry_a+Ry_b+Ry_c+Ry_d #units in meters\n",
+ "R=sqrt(Rx**2+Ry**2) #units in meters\n",
+ "phi=round(math.atan(Ry/-(Rx))*180/math.pi) #units in degrees\n",
+ "phi=180-phi #units in degrees\n",
+ "print \"The Resultant R=\",round(R,2),\" Meters\\n\"\n",
+ "print \"The Angle theta=\",round(phi),\" degrees\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Resultant R= 6.72 Meters\n",
+ "\n",
+ "The Angle theta= 170.0 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.3:pg-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 1_3\n",
+ "\n",
+ "import math\n",
+ " #To subtract vector B from Vector A\n",
+ "Ax=8.7 #units in meters\n",
+ "Ay=5 #units in meters\n",
+ "Bx=-6 #units in meters\n",
+ "By=0 #units in meters\n",
+ "Rx=Ax-Bx #units in meters\n",
+ "Ry=Ay-By #units in meters\n",
+ "R=sqrt(Rx**2+Ry**2) #units in meters\n",
+ "theta=round(math.atan(Ry/(Rx))*180/math.pi) #units in degrees\n",
+ "print \"Resultant R=\",round(R,1),\" Meters\\n\"\n",
+ "print \"Angle Theta=\",round(theta),\" Degrees\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant R= 15.5 Meters\n",
+ "\n",
+ "Angle Theta= 19.0 Degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.4:pg-30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 1_4\n",
+ "\n",
+ "\n",
+ " #To calculate the Volume\n",
+ "r=3*10**-5 #units in meters\n",
+ "L=0.20 #units in meters\n",
+ "V=math.pi*r**2*L #Units in meter**3\n",
+ "print \"Volume V=\",round(V,12),\"Meter**3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volume V= 5.65e-10 Meter**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter10.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter10.ipynb new file mode 100644 index 00000000..b0755268 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter10.ipynb @@ -0,0 +1,399 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c6d905c981cd31d9d5ba9137cf0938aeaccbb0b1cd83275a36f68b061bf96083"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10: Gases and the Kinetic Theory"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.1:pg-288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_1\n",
+ " \n",
+ " #To find out the pressure in Lungs\n",
+ "h=6 #units in cm Hg\n",
+ "Pa=76 #Units in cm Hg\n",
+ "Pl=(h+Pa) #units in cm Hg\n",
+ "\n",
+ "Pl=Pl*10**-2 #units in Meters Hg\n",
+ "g=9.8 #Units in Meters/cm**2\n",
+ "H=13600 #Constant \n",
+ "Pl=Pl*H*g #Units in Pa\n",
+ "print \"The pressure in the lungs is Pl=\",round(Pl,1),\" Pa\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure in the lungs is Pl= 109289.6 Pa\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.2:pg-288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_2\n",
+ " \n",
+ " \n",
+ " #To find the mass of copper atom\n",
+ "maa=63.5 #Units in Kgs\n",
+ "n=6.022*10**26 #Units in number of atoms\n",
+ "Mass=maa/n #units in Kg/atom\n",
+ "print \"The Mass per atom is=\",round(Mass,28),\"Kg/Atom\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Mass per atom is= 1.054e-25 Kg/Atom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.3:pg-289"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_3\n",
+ " \n",
+ " \n",
+ " #To find te volume associted with mercury atom in liquid mercury\n",
+ "M=201.0 #Units in Kg/Kmol\n",
+ "n=6.02*10**26 #units in K mol**-2\n",
+ "mo=M/n #units in Kg\n",
+ "n1=13600.0 #units in Kg/Meter**3\n",
+ "noatoms=n1/mo #units in atoms/Meter**3\n",
+ "volume_atom=1/noatoms #units in Meter**3/Atom\n",
+ "print \"The volume associated is \",round(volume_atom,31),\"Meter**3/Atom\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The volume associated is 2.46e-29 Meter**3/Atom\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.4:pg-289"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_4\n",
+ " \n",
+ " \n",
+ " #To find the volume that one kilomole of an ideal gas occupies\n",
+ "p=1.013*10**5 #units in Pa\n",
+ "t=273.15 #units in K\n",
+ "n=1 #units in K mol\n",
+ "R=8314 #units in J/Kmol K\n",
+ "v=(n*R*t)/p #units in Meter**3/Kmol\n",
+ "print \"Volume occupied is V=\",round(v,1),\" Meter**3/Kmol\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volume occupied is V= 22.4 Meter**3/Kmol\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.5:pg-290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_5\n",
+ " \n",
+ " \n",
+ " #To find the gas pressure in the container\n",
+ "v=5*10**-3 #units in meter**3\n",
+ "t=300.0 #units in K\n",
+ "m1=14*10**-6 #Units in Kg\n",
+ "M=28 #Units in Kg/Kmol \n",
+ "n=m1/M #units in K mol\n",
+ "R=8314 #units in J/Kmol K\n",
+ "p=(n*R*t)/v #units in Meter**3/Kmol\n",
+ "print \"The pressure in the container is P=\",round(p),\" Pa\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure in the container is P= 249.0 Pa\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.6:pg-291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_6\n",
+ " \n",
+ " \n",
+ " #To determine the mass of the air in flask\n",
+ "p=1.013*10**5 #Units in Pa\n",
+ "v=50*10**-6 #Units in meter**3\n",
+ "M=28.0 #Units in Kg/Mol\n",
+ "R=8314.0 #units in J/Kmol K\n",
+ "T=293 #units in K\n",
+ "m=(p*v*M)/(R*T) #Units in Kg\n",
+ "print \"The mass of air in flask is=\",round(m,7),\"Kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of air in flask is= 5.82e-05 Kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.7:pg-292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_7\n",
+ " \n",
+ " \n",
+ " #To find out the final pressure in the drum\n",
+ "p1=1 #Units in atm\n",
+ "t2=333.0 #units in K\n",
+ "t1=293.0 #units in K\n",
+ "p2=p1*(t2/t1) #units in atm\n",
+ "print \"The final pressure in the drum is P2=\",round(p2,2),\" atm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The final pressure in the drum is P2= 1.14 atm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.8:pg-292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_8\n",
+ " \n",
+ " \n",
+ " #To find the final volume of gas\n",
+ "\n",
+ "t1=27.0 #Units in Centigrade\n",
+ "t1=t1+273 #Units in Kelvin\n",
+ "t2=547.0 #Units in Centigrade\n",
+ "t2=t2+273 #Units in Kelvin\n",
+ "t1=27.0 #Units in Centigrade\n",
+ "t1=t1+273 #Units in Kelvin\n",
+ "t1=27.0 #Units in Centigrade\n",
+ "t1=t1+273 #Units in Kelvin\n",
+ "p2=3700.0 #units in cm Hg\n",
+ "p1=74.0 #units in cm Hg\n",
+ "v1_v2=1/((t1/t2)*(p2/p1)) #In terms of V1\n",
+ "print \"The final volume of gas in terms of original volume is V2=\",round(v1_v2,5),\"*V1\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The final volume of gas in terms of original volume is V2= 0.05467 *V1\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.9:pg-293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_9\n",
+ " \n",
+ " \n",
+ " #To find the pressure after the car has been driven at high speed\n",
+ "t2=308 #Units in K\n",
+ "t1=273.0 #Units in K\n",
+ "p2_p1=(t2)/t1 #In terms of P1\n",
+ "P1=190.0 #Units in K Pa \n",
+ "P2=101 #Units in K Pa \n",
+ "P2=p2_p1*(P1+P2) #Units in K Pa \n",
+ "print \"The Final pressure is P2=\",round(P2),\" K Pa\" \n",
+ " #In text book the answer is printed wrong as P2=329 K Pa but the correct answer is 328 K Pa \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Final pressure is P2= 328.0 K Pa\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.10:pg-294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 10_10\n",
+ "\n",
+ "import math\n",
+ " \n",
+ " #To findout how fast the nitrogen molecule moving in air\n",
+ "M=28.0 #Units in Kg/Mol\n",
+ "Na=6.02*10**26 #Units in K mol**-1\n",
+ "mo=M/Na #Units in Kg\n",
+ "k=1.38*10**-23 #units in J/K\n",
+ "T=27+273.0 #Units in K\n",
+ "v2=(3*k*T)/mo #unit in Meter**2/Sec**2\n",
+ "v=math.sqrt(v2) #Units in meter/sec\n",
+ "print \"The nitrogen molecule goes at a speed of V=\",round(v),\" meter/sec\"\n",
+ " #In text book the answer is printed wrong as v=517 m/sec the correct answer is v=516 meter/ sec\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The nitrogen molecule goes at a speed of V= 517.0 meter/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb new file mode 100644 index 00000000..0781dd9b --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter11.ipynb @@ -0,0 +1,493 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11:Thermal Properties of Matter" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.1:pg-303" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat required for 400 gm of water is Q= 2000.0 Cal\n", + "\n", + "The heat required for 400 gm of copper is Q= -186.0 Cal\n", + "\n" + ] + } + ], + "source": [ + " #Example 11_1\n", + " \n", + " \n", + " #To find out how much heat is required to change the temperature\n", + " #With 400 Grams of water\n", + "c=1 #units in cal/g Centigrade\n", + "m=400 #Units in gm\n", + "t=5 #Units in centigrade\n", + "q=c*m*t #Units in Cal\n", + "print \"The heat required for 400 gm of water is Q=\",round(q),\" Cal\\n\"\n", + " #With 400 grams of copper\n", + "c=0.093 #units in cal/g Centigrade\n", + "m=400 #Units in gm\n", + "t=-5 #Units in centigrade\n", + "q=c*m*t #Units in Cal\n", + "print \"The heat required for 400 gm of copper is Q=\",round(q),\" Cal\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.2:pg-303" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When it crystallizes heat required is Q= 4000.0 Cal\n", + "\n", + "When it condenses heat required is Q= 26950.0 Cal\n", + "\n" + ] + } + ], + "source": [ + " #Example 11_2\n", + " \n", + " \n", + " #To findout how much water is released\n", + " #When it crystallizes\n", + "m=50 #Units in gm\n", + "h=80 #Units in Cal/gm\n", + "q=m*h #Units in Cal\n", + "print \"When it crystallizes heat required is Q=\",round(q),\" Cal\\n\"\n", + " #When it Condenses\n", + "m=50 #Units in gm\n", + "h=539 #Units in Cal/gm\n", + "q=m*h #Units in Cal\n", + "print \"When it condenses heat required is Q=\",round(q),\" Cal\\n\"\n", + " #In textbook answer is printed wrong as Q=27000 cal but the correct answer is Q=26950 Cal\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.3:pg-304" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amount of ice that has to be added is M= 54.0 gm\n" + ] + } + ], + "source": [ + " #Example 11_3\n", + " \n", + " \n", + " #To findout the amount of Ice that has to be added\n", + "m=200 #Units in gm\n", + "c=1 #Units in Cal/gm Centigrade\n", + "tf=60 #Units in Centigrade\n", + "to=98 #Units in Centigrade\n", + "change=m*c*(tf-to) #units in Cal\n", + "tf=60 #Units in centigrade\n", + "to=0 #Units in centigrade\n", + "Hf=80 #Units in Cal/gm\n", + "change1=Hf+c*(tf-to) #Units in Cal/gm\n", + "M=change/-(change1)\n", + "print \"The amount of ice that has to be added is M=\",round(M,1),\" gm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.4:pg-305" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The specific heat of metal is Cm= 0.216 cal/gm C\n" + ] + } + ], + "source": [ + " #Example 11_4\n", + " \n", + " \n", + " #To findout the specific heat capacity of the metal\n", + "m=400 #Units in gm\n", + "c=0.65 #Units in Cal/gm Centigrade\n", + "tf=23.1 #Units in Centigrade\n", + "to=18 #Units in Centigrade\n", + "oil=m*c*(tf-to) #units in cal\n", + "m1=80 #Units in gm\n", + "tf=23.1 #Units in Centigrade\n", + "to=100 #Units in Centigrade\n", + "cm=m1*(tf-to) #units in in terms of cm and gm Centigrade\n", + "cmm=oil/-cm #Units in Cal/gm Centigrade\n", + "print \"The specific heat of metal is Cm=\",round(cmm,3),\" cal/gm C\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.5:pg-305" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time taken is t= 449.0 sec\n" + ] + } + ], + "source": [ + " #Example 11_5\n", + " \n", + " \n", + " #To findout how long does the heater takes to heat\n", + "m=500 #Units in gm\n", + "c=0.033 #Units in Cal/gm Centigrade\n", + "tf=357 #Units in Centigrade\n", + "to=20.0 #Units in Centigrade\n", + "m1=30 #Units in gm\n", + "hv=65 #Units in cal/gm\n", + "Hg=((m*c*(tf-to))+(m1*hv))*4.1808135 #units in Joules\n", + "delivered=70 #Units in Joule/Sec\n", + "t=Hg/delivered #Units in sec\n", + "print \"The time taken is t=\",round(t),\" sec\"\n", + " #In textbook answer printed wrong as t=450 sec correct answer is t=448 sec\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.6:pg-306" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the rise in temperature is DeltaT= 38.6 C\n" + ] + } + ], + "source": [ + " #Example 11_6\n", + " \n", + " \n", + " #To findout the rise in temperature\n", + "m=0.01 #Units in Kg\n", + "v=100 #Units in meters/sec\n", + "KE=(0.5*m*v**2)/4.1808135 #units in Cal\n", + "m=10 #units in gm\n", + "c=0.031 #units in cal/gm Centigrade\n", + "t=KE/(m*c)\n", + "print \"the rise in temperature is DeltaT=\",round(t,1),\" C\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.8:pg-307" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The slab is longer by= 0.01 meters\n" + ] + } + ], + "source": [ + " #Example 11_8\n", + " \n", + " \n", + " #To findout how much longer is at 35 degrees\n", + "alpha=10*10**-6 #Units in Centigrade\n", + "dist=20.0 #Unis in meters\n", + "t=50 #Units in centigrade\n", + "L=alpha*dist*t #Units in meters\n", + "print \"The slab is longer by=\",round(L,3),\" meters\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.9:pg-308" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The new diameter of the hole is= 2.0076 cm\n" + ] + } + ], + "source": [ + " #Example 11_9\n", + " \n", + " \n", + " #To findout how large a diameter when the sheet is heated\n", + "dist=2 #Units in cm\n", + "delta=19*10**-6 #Units in Centigrade**-1\n", + "t=200 #Units in centigrade\n", + "L=dist*delta*t #Units in cm\n", + "print \"The new diameter of the hole is=\",round(2+L,4),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.10:pg-309" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The change in benzene volume is V30= 101.253 cm**3\n" + ] + } + ], + "source": [ + " #Example 11_10\n", + " \n", + " \n", + " #To findout the change in benzene volume\n", + "delta=1.24*10**-3 #Units in Centigrade**-1\n", + "t=10 #Units in Centigrade\n", + "v10=100.0 #Units in cm**3 \n", + "v20=delta*t+v10 #Units in cm**3\n", + "V=v20*delta*t #Units in cm**3\n", + "v30=V+v20 #Units in cm**3\n", + "print \"The change in benzene volume is V30=\",round(v30,3),\" cm**3\"\n", + " #In textbook the answer is printed wrng as V3=0102.5 cm**3 the correct answer is V3=101.253 cm**3 \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.11:pg-309" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ice melts by 116.0 gm\n" + ] + } + ], + "source": [ + " #Example 11_11\n", + " \n", + " \n", + " #To findout how much ice melts each hour\n", + "s=30 #Units in cm\n", + "a=s*s*10**-4 #units in meter**2\n", + "k=0.032 #Units in W/K meter\n", + "t=25 #Units in K\n", + "l=0.040 #Units in meters\n", + "q_t=(6*k*((a*t)/l))/4.1808135 #Units in cal/sec\n", + "Q=3600*q_t #Units in cal\n", + "qq=80 #Units in cal/gm\n", + "melted=Q/qq #Units in gm\n", + "print \"The ice melts by \",round(melted),\" gm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.12:pg-310" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The radiation defers by 34.0 percent\n" + ] + } + ], + "source": [ + " #Example 11_12\n", + " \n", + " \n", + " #To compare the energy emitted per unit area of our body to with the same emissivity\n", + "t1=37.0 #Units in Centigrade\n", + "t1=273+t1 #Units in K\n", + "t2=15 #Units in Centigrade\n", + "t2=273+t2 #Units in K\n", + "tb_tc=(t1/t2)**4 #Units in terms of (Tb/Tc)**4\n", + "tb_tc=tb_tc*100 #In terms of percentage\n", + "print \"The radiation defers by \",round(tb_tc-100),\" percent\"\n", + "\n", + " #In textbook answer is printed wrong as 40% the correct answer is 34%\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex11.13:pg-310" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amount of heat lost is Q= 736363.6 J\n" + ] + } + ], + "source": [ + " #Example 11_13\n", + " \n", + " \n", + " #To findout how much heat is lost through it\n", + "a=15 #Unis in meter**2\n", + "t=30.0 #Units in K\n", + "R=2.2 #Units in Meter**2 K/W\n", + "q_t=(a*t)/R #Units in W\n", + "T=3600.0 #Units in sec\n", + "Q=q_t*T #Units in J\n", + "print \"The amount of heat lost is Q=\",round(Q,1),\" J\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter13.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter13.ipynb new file mode 100644 index 00000000..3c083749 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter13.ipynb @@ -0,0 +1,249 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13:Vibrations and Waves" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex13.1:pg-508## " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum velocity is Vmax= 1.4 Meter/sec\n", + "\n", + "Maximum acceleration is Amax= 4.9 meter/sec**2\n", + "\n", + "Velocity at x=0.1 meters is= 1.36 meters/sec\n", + "\n", + "Acceleration at x=0.1 meters is= -1.23 meters/sec**2\n", + "\n" + ] + } + ], + "source": [ + " #Example 13_1\n", + "import math \n", + " #To find the maximum velocity and acceleration and the same when x=10cm\n", + "xo=0.4 #Units in Meters\n", + "k=24.5 #Units in N/M\n", + "m=2 #Units in Kg\n", + "vmax=xo*(math.sqrt(k/m)) #Units in meters/sec\n", + "print \"Maximum velocity is Vmax=\",round(vmax,1),\" Meter/sec\\n\"\n", + "amax=(k*xo)/m #Units in meter/sec**2\n", + "print \"Maximum acceleration is Amax=\",round(amax,1),\" meter/sec**2\\n\"\n", + "x=0.1 #Units in meters\n", + "v=math.sqrt((k/m)*(xo**2-x**2)) #Units in meters/Sec\n", + "print \"Velocity at x=0.1 meters is= \",round(v,2),\" meters/sec\\n\"\n", + "a=-(k*x)/m #Units in meter/sec**2\n", + "print \"Acceleration at x=0.1 meters is= \",round(a,2),\" meters/sec**2\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex13.2:pg-512## " + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency of vibrations is f= 0.56 Hz\n" + ] + } + ], + "source": [ + " #Example 13_2\n", + " \n", + "import math \n", + " #To find the frequency of the vibrations\n", + "spring=24.5 #Units in N/m\n", + "m=2 #Units in Kg\n", + "f=(1/(2*math.pi))*math.sqrt(spring/m) #Units in Hz\n", + "print \"The frequency of vibrations is f=\",round(f,2),\" Hz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex13.3:pg-513" + ] + }, + { + "cell_type": "code", + "execution_count": 0, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 13_3\n", + " \n", + " \n", + " #To find the tension required in string\n", + "m=0.002 #Units in Kg\n", + "l=0.6 #Units in meters\n", + "v=300 #Units in meters/sec\n", + "T=(m/l)*v**2 #Units in N\n", + "print \"Tension required in the string is T=\",round(T),\" N\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex13.4:pg-514" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The first resonance frequency is F1= 2.0 Hz\n", + "\n", + "The second resonance frequency is F2= 4.0 Hz\n", + "\n", + "The third resonance frequency is F3= 6.0 Hz\n", + "\n" + ] + } + ], + "source": [ + " #Example 13_4\n", + " \n", + " \n", + " #To draw a picture on the first three resonance frequencies\n", + "l=6 #Units in meters\n", + "n=1\n", + "lamda1=(2*l)/n #Units in meters\n", + "n=2\n", + "lamda2=(2*l)/n #Units in meters\n", + "n=3\n", + "lamda3=(2*l)/n #Units in meters\n", + "speed=24 #Units in meters/sec\n", + "f1=speed/lamda1 #Units in Hz\n", + "f2=speed/lamda2 #Units in Hz\n", + "f3=speed/lamda3 #Units in Hz\n", + "print \"The first resonance frequency is F1=\",round(f1),\" Hz\\n\"\n", + "print \"The second resonance frequency is F2=\",round(f2),\" Hz\\n\"\n", + "print \"The third resonance frequency is F3=\",round(f3),\" Hz\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex13.5:pg-515 " + ] + }, + { + "cell_type": "code", + "execution_count": 0, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 13_5\n", + " \n", + " \n", + " #To find the speed of the wave\n", + "l=300*10**-2 #Units in Meters\n", + "lamda3=(l*2)/3 #Units in meters\n", + "f=20 #Units in sec**-1 or Hz\n", + "v=f*lamda3 #Units in meters/sec\n", + "print \"The speed of the wave is v=\",round(v),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex13.6:pg-516" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The youngs modulus is Y=\n", + "1.961071965e+11 N/meters**2\n" + ] + } + ], + "source": [ + " #Example 13_6\n", + " \n", + " \n", + " #To find the youngs modulus\n", + "lamda=1.85 #Units in meters\n", + "f=2700 #units in sec**-1\n", + "v=lamda*f #Units in meters/sec\n", + "density=7.86*10**3 #Units in Kg/meter**3\n", + "y=v**2*density #Units in N/meters**2\n", + "print \"The youngs modulus is Y=\"\n", + "print y,\"N/meters**2\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter14.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter14.ipynb new file mode 100644 index 00000000..c0ff2f9f --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter14.ipynb @@ -0,0 +1,241 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14:Sound" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex14.1:pg-623" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of the sound in neon is v= 432.0 meters/sec\n" + ] + } + ], + "source": [ + " #Example 14_1\n", + "import math \n", + " #To find the speed of sound in neon\n", + "gama=1.66 #units in Constant\n", + "r=8314 #Units in J/Kmol\n", + "t=273 #Units in K\n", + "m=20.18 #Units in Kg/Kmol\n", + "v=math.sqrt((gama*r*t)/m) #Units in meters/sec\n", + "print \"The speed of the sound in neon is v=\",round(v),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex14.2:pg-625" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sound level of the sound wave is= 70.0 dB\n" + ] + } + ], + "source": [ + " #Example 14_2\n", + "import math \n", + " #To find the sound level of a sound wave\n", + "i1=10**-5 #Units in W/meter**2\n", + "i2=10**-12 #Units in W/meter**2\n", + "level=10*math.log10(i1/i2) #units in dB\n", + "print \"The sound level of the sound wave is=\",round(level),\" dB\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex14.3:pg-625" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The intensity of sound is I=\n", + "3.16227766017e-09\n", + "W/meter**2\n" + ] + } + ], + "source": [ + " #Example 14_3\n", + "import math \n", + " #To find the intensity of sound\n", + "level=3.5 #Units in dB\n", + "i2=10**-12 #Units in W/meter**2\n", + "i=10**(level+math.log10(i2)) #Units in W/meter**2\n", + "print \"The intensity of sound is I=\"\n", + "print i\n", + "print \"W/meter**2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex14.4:pg-626" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The distance it has to be moved is= 70.0 cm\n" + ] + } + ], + "source": [ + " #Example 14_4\n", + " \n", + " \n", + " #To find how far it has to be moved before the sound becomes weak\n", + "lamda=70 #units in cm\n", + "lamda1=0.5*lamda #Units in cm\n", + "print \"The distance it has to be moved is=\",round(lamda),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex14.5:pg-627" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The frequency we hear is f= 531.0 Hz\n", + "\n", + "The frequency of the receding is f= 472.0 Hz\n", + "\n" + ] + } + ], + "source": [ + " #Example 14_5\n", + " \n", + " \n", + " #To find the frequency heard and the receding\n", + "f=500 #Units in Hz\n", + "vw=340.0 #Units in meters/sec\n", + "dist=20 #Units in meters/sec\n", + "f1=f*(vw/(vw-dist)) #Units in Hz\n", + "print \"The frequency we hear is f=\",round(f1),\" Hz\\n\"\n", + "f1=f*(vw/(vw+dist)) #Units in Hz\n", + "print \"The frequency of the receding is f=\",round(f1),\" Hz\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex14.6:pg-628" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The difference between the both frequencies is= 1667.0 Hz\n" + ] + } + ], + "source": [ + " #Example 14_6\n", + " \n", + " \n", + " #To find the difference between the frequency of wave reaching the officer and the car\n", + "fo=10**10 #Units in Hz\n", + "vw=3*10.0**8 #Units in meters/sec\n", + "vc=25 #Units in meters/sec\n", + "f1=fo*((vw+vc)/(vw-vc)) #Units in Hz\n", + "f1=f1-10**10 #Units in Hertz\n", + "print \"The difference between the both frequencies is=\",round(f1),\" Hz\"\n", + " #In text book answer printed wrong as 1670 Hz correct answer is 1666 Hz\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter15.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter15.ipynb new file mode 100644 index 00000000..fef24844 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter15.ipynb @@ -0,0 +1,348 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15:Electric Forces and Fields" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.1:pg-719" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of q is= 3.61e-06 C\n", + "\n", + "Number of electrons to be removed is=\n", + "2.25625e+13\n", + "Fraction of atoms lost is=\n", + "7.52083333333e-10\n" + ] + } + ], + "source": [ + " #Example 15_1\n", + "import math \n", + " #To find the value of q and how many electrons must be removed and fraction of atoms lost\n", + "dist=2 #Units in meters\n", + "f=0.0294 #Units in N\n", + "s=9*10**9 #Units in N meter**2/C**2\n", + "q=math.sqrt((dist**2*f)/s) #Units in C\n", + "print \"The value of q is=\",round(q,8),\" C\\n\"\n", + "charge=3.61*10**-6 #Units in C\n", + "c_elec=1.6*10**-19 #Units in C\n", + "n=charge/c_elec #Units in number\n", + "print \"Number of electrons to be removed is=\"\n", + "print n\n", + "f1=3*10.0**22 #Units in number\n", + "fraction=n/f1 #Units of number\n", + "print \"Fraction of atoms lost is=\"\n", + "print fraction\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.2:pg-721" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force on the center charge is= 0.02813 N\n" + ] + } + ], + "source": [ + " #Example 15_2\n", + " \n", + " \n", + " #To find the force on the center charge\n", + "k=9*10**9 #Units in N meter**2/C**2\n", + "q1=4*10.0**-6 #Units in C\n", + "q2=5*10.0**-6 #Units in C\n", + "r1=2 #Units in meters\n", + "r2=4 #Units in meters\n", + "q3=6*10.0**-6 #Units in C\n", + "f1=(k*q1*q2)/r1**2 #Units in N\n", + "f2=(k*q2*q3)/r2**2 #Units in N\n", + "f=f1-f2 #Units in C\n", + "print \"The force on the center charge is=\",round(f,5),\" N\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.3:pg-722" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resultant force is f= 19.0 N \n", + "The resultant angle is theta= 71.6 degrees\n" + ] + } + ], + "source": [ + " #Example 15_3\n", + "import math \n", + " \n", + " #To find the resultant force\n", + "f1=6 #Units in N\n", + "f2=18 #Units in N\n", + "f=math.sqrt(f1**2+f2**2) #Units in N\n", + "theta=math.atan(f2/f1)*180/math.pi #Units in degrees\n", + "print \"The resultant force is f=\",round(f),\" N \\nThe resultant angle is theta=\",round(theta,1),\" degrees\"\n", + " #In text book answer printed wrong as f=19 N correct answer is f=18N \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.4:pg-724" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resultant force is f= 3.4 N \n", + " The resultant angle is theta= 65.0 degrees\n" + ] + } + ], + "source": [ + " #Example 15_4\n", + " \n", + "import math\n", + " #To find the resultant force on 20 micro C\n", + "f1=2 #Units in N\n", + "f2=1.8 #Units in N\n", + "theta=37.0 #Units in degrees\n", + "f2x=f2*math.cos(theta*math.pi/180) #Units in N\n", + "f2y=f2*math.sin(theta*math.pi/180) #Units in N\n", + "fy=f1+f2y #Units in N\n", + "f=math.sqrt(fy**2+f2x**2) #Units in N\n", + "theta=math.atan(fy/f2x)*180/math.pi #Units in degrees\n", + "print \"The resultant force is f=\",round(f,1),\" N \\n The resultant angle is theta=\",round(theta,1),\" degrees\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.6:pg-726 " + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of E is=\n", + "7198876.9\n", + "N/C\n" + ] + } + ], + "source": [ + " #Example 15_6\n", + " \n", + " \n", + " #To find the magnitude of E\n", + "k=9*10**9 #Units in N meter**2/C**2\n", + "q=3.6*10**-6 #Units in C\n", + "theta=37 #Units in degrees\n", + "r=10*math.sin(theta*math.pi/180)*10**-2 #Units in meters \n", + "e1=(k*q)/r**2 #Units in N/C\n", + "q2=5*10**-6 #Units in C\n", + "theta=37 #Units in degrees\n", + "r1=10*10**-2 #Units in meters \n", + "e2=(k*q2)/r1**2 #Units in N/C\n", + "e1y=e1 #Units in N/C\n", + "e2x=e2*math.cos(theta*math.pi/180) #Units in N/C\n", + "e2y=-e2*math.sin(theta*math.pi/180) #Units in N/C\n", + "ex=e2x #Units in N/C\n", + "ey=e1y+e2y #Units in N/C\n", + "e=math.sqrt(ex**2+ey**2) #Units in N/C\n", + "print \"The magnitude of E is=\"\n", + "print round(e,2)\n", + "print \"N/C\"\n", + " #In text book the answer isprinted wrong as E=7.26*10**6 N/C but the correct answer is E=7198876.9 N/C\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.7:pg-726" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The charge occured is q= 8.33e-07 C\n" + ] + } + ], + "source": [ + " #Example 15_7\n", + " \n", + " \n", + " #To find out how much charge occurs\n", + "e=3*10.0**6 #Units in N/C\n", + "r=0.050 #Units in meters\n", + "k=9*10.0**9 #Units in N meter**2/C**2\n", + "q=(e*r**2)/k #Units in C\n", + "print \"The charge occured is q=\",round(q,9),\" C\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.8:pg-727" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lines of force come out of positive charge q suspended in cavity.\n", + "Cavity \n", + "surface must possess a negative charge since lines of force go and terminate on q.\n", + "Therefore a charge +q must exist on outer portions.\n" + ] + } + ], + "source": [ + " #Example 15_8\n", + " \n", + " \n", + " #To show using lines of force that a charge suspended with in cavity induces an equal and opposite charge on surface\n", + "print \"Lines of force come out of positive charge q suspended in cavity.\\nCavity \\nsurface must possess a negative charge since lines of force go and terminate on q.\\nTherefore a charge +q must exist on outer portions.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex15.9:pg-728" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The field goes by a speed of 47952.0 meters/sec\n" + ] + } + ], + "source": [ + " #Example 15_9\n", + "import math \n", + " #To find the speed just before the field strikes\n", + "e=6000 #Units in N/C\n", + "q=1.6*10**-19 #Units in C\n", + "f=e*q #Units in N\n", + "m=1.67*10**-27 #Units in Kg\n", + "a=f/m #Units in meters/sec**2\n", + "vo=0 #Units in meters/sec\n", + "x=2*10**-3 #Units in meters\n", + "v=math.sqrt(vo**2+(2*a*x)) #Units in meters/sec\n", + "print \"The field goes by a speed of \",round(v),\" meters/sec\"\n", + " #In text book answer printed wrong as v=48000 meters/sec the correct answer is v=47952 meters/sec \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter16.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter16.ipynb new file mode 100644 index 00000000..48f13aa4 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter16.ipynb @@ -0,0 +1,398 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16:Electric Potential" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.1:pg-731" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of electric field is E= 2400.0 V/meters\n" + ] + } + ], + "source": [ + " #Example 16_1\n", + " \n", + " \n", + "#To find the magnitude of the electric field\n", + "v=12.0 #Units in V\n", + "d=5.0*10**-3 #units in Meters\n", + "e=v/d #Units in V/meter\n", + "print \"The magnitude of electric field is E=\",round(e),\" V/meters\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.2:pg-731" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of the proton is Vab= 92858.79 meters/sec\n" + ] + } + ], + "source": [ + " #Example 16_2\n", + " \n", + " \n", + "#To calculate the speed of the proton\n", + "q=1.6*10**-19 #Units in C\n", + "vab=45 #Units in V\n", + "m=1.67*10**-27 #Units in Kg\n", + "va=math.sqrt((2*q*vab)/m) #Units in meters/sec\n", + "print \"The speed of the proton is Vab=\",round(va,2),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.3:pg-733" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of the electron is Vab= 3975777.37 meters/sec\n" + ] + } + ], + "source": [ + " #Example 16_3\n", + "import math \n", + "#To find the sped of an electron\n", + "e=1.6*10**-19 #Units in C\n", + "vab=45 #Units in V\n", + "m=9.11*10**-31 #Units in Kg\n", + "va=math.sqrt((2*e*vab)/m) #Units in meters/sec\n", + "print \"The speed of the electron is Vab=\",round(va,2),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.4:pg-735" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Charged metal object have an equi potential volume so its surface is am equi potential volume\n", + "Because lines of force must be perpendiculat ro equipotential lines and surfaces.\n" + ] + } + ], + "source": [ + " #Example 16_4\n", + " \n", + " \n", + " #To sketch the equipotentials and electric field lines near a charged metal object\n", + "print \"Charged metal object have an equi potential volume so its surface is am equi potential volume\\nBecause lines of force must be perpendiculat ro equipotential lines and surfaces.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.5:pg-735" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The work done in carrying proton is=\n", + "1.44e-18\n", + "Joules\n", + "\n", + "The work done in carrying electron is=\n", + "-1.44e-18\n", + "Joules\n", + "\n" + ] + } + ], + "source": [ + " #Example 16_5\n", + " \n", + " \n", + "#To find the work done in carrying a proton and for an electron\n", + "q=1.6*10**-19 #Units in C\n", + "vab=9.0 #Units in V\n", + "work=q*vab #Units in J\n", + "print \"The work done in carrying proton is=\"\n", + "print work\n", + "print \"Joules\\n\"\n", + "q=-1.6*10**-19 #Units in C\n", + "work=q*vab #Units in J\n", + "print \"The work done in carrying electron is=\"\n", + "print work\n", + "print \"Joules\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.6:pg-736" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of proton before it strikes is Vb= 7756781.9 meters/sec\n" + ] + } + ], + "source": [ + " #Example 16_6\n", + "import math \n", + " #To calculate the speed just befor it strikes it\n", + "va=8*10**6 #Units in meters/sec\n", + "q=1.6*10**-19 #Units in C\n", + "m=1.67*10**-27 #Units in Kg\n", + "vab=20000 #Units in V\n", + "vb=math.sqrt(va**2-((2*q*vab)/m)) #Units in meters/sec\n", + "print \"The speed of proton before it strikes is Vb=\",round(vb,1),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.7:pg-737" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since each proton has a minimum energy of 13.6 eV and a charge of 1.602*10**-19 C\n", + " The required potential difference is=13.6 eV\n" + ] + } + ], + "source": [ + " #Example 16_7\n", + " \n", + " \n", + " #To calculate the minimum value of Vab needed\n", + "print \"Since each proton has a minimum energy of 13.6 eV and a charge of 1.602*10**-19 C\\n The required potential difference is=13.6 eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.8:pg-737" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of electron is v= 4150286.78 meters/sec\n" + ] + } + ], + "source": [ + " #Example 16_8\n", + "import math \n", + " #To find out the speed of the proton\n", + "k=9*10**9 #Units in N meter**2/C**2\n", + "q=5*10.0**-6 #Units in C\n", + "r=0.5 #Units in meters\n", + "v1=(k*q)/r #Units in V\n", + "q=1.6*10**-19 #Units in V\n", + "m=1.672*10**-27 #Units in Kg\n", + "v=math.sqrt((v1*q*2)/m) #Units in V\n", + "print \"The speed of electron is v=\",round(v,2),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.9:pg-739" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Due to 5*10**-8 C V1= 4500.0 V\n", + "Due to 8*10**-8 C V2= 7200.0 V\n", + "Due to 40*10**-8 C V3= -18000.0 V\n", + " Absolute potential at B is Vb= -6300.0 V\n" + ] + } + ], + "source": [ + " #Example 16_9\n", + " \n", + " \n", + " #To compute the absolute potential at B\n", + "k=9*10**9 #Units in N meter**2/C**2\n", + "q=5*10.0**-8 #Units in C\n", + "r=0.1 #Units in meters\n", + "v1=(k*q)/r #Units in V\n", + "q=8*10**-8 #Units in C\n", + "r=0.1 #Units in meters\n", + "v2=(k*q)/r #Units in V\n", + "q=40*10**-8 #Units in C\n", + "r=0.2 #Units in meters\n", + "v3=-(k*q)/r #Units in V\n", + "vb=v1+v2+v3 #Units in V \n", + "print \"Due to 5*10**-8 C V1=\",round(v1),\" V\\nDue to 8*10**-8 C V2=\",round(v2),\" V\\nDue to 40*10**-8 C V3=\",round(v3),\" V\\n Absolute potential at B is Vb=\",round(vb),\" V\"\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex16.10:pg-739" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The absolute potential is V= 27.2 V\n", + "\n", + "The energy that is required is W=\n", + "4.35e-18 J\n" + ] + } + ], + "source": [ + " #Example 16_10\n", + " \n", + " \n", + " #To find the absolute potential and how much energy is needed to pull the electrons from atom\n", + "k=9*10**9 #Units in N meter**2/C**2\n", + "q=1.6*10**-19 #Units in C\n", + "r=5.3*10**-11 #Units in meters\n", + "v=(k*q)/r #Units in V\n", + "print \"The absolute potential is V=\",round(v,1),\" V\\n\"\n", + "Vinfinity=0 #Units in V\n", + "deltaV=Vinfinity-v #Units in V\n", + "work=-q*deltaV #Units in J\n", + "print \"The energy that is required is W=\"\n", + "print round(work,20),\"J\" \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb new file mode 100644 index 00000000..c718557a --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter17.ipynb @@ -0,0 +1,641 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 17:DC Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.1:pg-866" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The number of electrons that pass through bulb is=\n", + "9.375e+17\n", + "electrons\n" + ] + } + ], + "source": [ + "#Example 17_1\n", + " \n", + " \n", + "#To find number of electrons flow through bulb\n", + "current=0.15 #Units in C\n", + "q=1.6*10**-19 #Units in C/electron\n", + "noe=current/q #Units in number of Electrons\n", + "print \"The number of electrons that pass through bulb is=\"\n", + "print noe\n", + "print \"electrons\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.2:pg-867" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance in bulb is= 19.4 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_2\n", + " \n", + " \n", + "#To find the resistance in bulb\n", + "v=1.55 #Units in V\n", + "i=0.08 #Units in A\n", + "r=v/i #Units in Ohms\n", + "print \"The resistance in bulb is=\",round(r,1),\" Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.3:pg-867" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance in wire is= 0.205 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_3\n", + " \n", + " \n", + "#To find the resistance in wire\n", + "row=1.7*10**-8 #Units in Ohm meter\n", + "l=40 #Units in meters\n", + "a=0.0331*10**-4 #Units in meters**2\n", + "r=(row*l)/a #Units in Ohms\n", + "print \"The resistance in wire is=\",round(r,3),\" Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.4:pg-868" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The appropriate resistance in wire is Ro= 26.6 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_4\n", + " \n", + " \n", + "#To find the appropriate resistance of the wire\n", + "alpha=0.0045 #Units in Centigrade**-1\n", + "t=1780 #Units in Centigrade\n", + "deltaR=240 #Units in Ohms\n", + "ro=deltaR/(1+(alpha*t)) #Units in ohms\n", + "print \"The appropriate resistance in wire is Ro=\",round(ro,1),\" Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.5:pg-868" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat generated in bulb is= 48000.0 J\n" + ] + } + ], + "source": [ + "#Example 17_5\n", + " \n", + " \n", + "#To find out the amount of heat developed in bulb\n", + "t=20.0*60 #Units in sec\n", + "pow=40.0 #Units in W\n", + "heat=t*pow #Units in J\n", + "print \"Heat generated in bulb is=\",round(heat),\" J\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.6:pg-869" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cost needed to operate is= 0.035 Dollars\n" + ] + } + ], + "source": [ + "#Example 17_6\n", + " \n", + " \n", + "#To calculate the cost needed to operate\n", + "power=0.7 #Units in KW\n", + "time=0.5 #Units in h\n", + "heat=power*time #Units in K Wh\n", + "cost=0.10 #Units in Dollars\n", + "tcost=cost*heat #Units in Dollars\n", + "print \"Cost needed to operate is=\",round(tcost,4),\" Dollars\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.7:pg-869" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in circuit is I= -0.82 A\n" + ] + } + ], + "source": [ + "#Example 17_7\n", + " \n", + " \n", + "#To find the current in circuit\n", + "v1=3 #Units in V\n", + "v2=12.0#Units in V\n", + "r1=5.0 #Units in Ohms\n", + "r2=6 #Units in Ohms\n", + "i=(v1-v2)/(r1+r2) #Units in A\n", + "print \"The current in circuit is I=\",round(i,2),\" A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.8:pg-870" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 1.75 A\n", + "Current in wire 2 is I2= -0.5 A\n", + "Current in wire 3 is I3= 1.25 A\n", + "\n" + ] + } + ], + "source": [ + "#Example 17_8\n", + " \n", + " \n", + "#To find the current in all wires\n", + "v=9 #Units in V\n", + "r1=18.0 #Units in Ohms\n", + "i2=-v/r1 #Units in A\n", + "v1=6.0 #Units in V\n", + "r2=12.0 #Units in Ohms\n", + "i3=(v+v1)/r2 #Units in A\n", + "i1=i3-i2 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1,2),\" A\\nCurrent in wire 2 is I2=\",round(i2,2),\" A\\nCurrent in wire 3 is I3=\",round(i3,2),\" A\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.9:pg-871" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current I= 2.0 A\n" + ] + } + ], + "source": [ + "#Example 17_9\n", + " \n", + " \n", + "#To find the current I in the battery\n", + "r1=3 #Units in Ohms\n", + "r2=6.0 #Units in Ohms\n", + "rbc=(r1*r2)/(r1+r2)#Units in Ohms\n", + "r3=4#Units in Ohms\n", + "rac=r3+rbc #Units in Ohms\n", + "v=12.0 #Units in V\n", + "i=v/rac #Units in A\n", + "print \"The current I=\",round(i),\" A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.10:pg-872" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "ename": "SyntaxError", + "evalue": "invalid syntax (<ipython-input-10-e97062f38a8b>, line 17)", + "output_type": "error", + "traceback": [ + "\u001b[1;36m File \u001b[1;32m\"<ipython-input-10-e97062f38a8b>\"\u001b[1;36m, line \u001b[1;32m17\u001b[0m\n\u001b[1;33m print \"The current in battery is I=\",round( ,2),\" A\",i)\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n" + ] + } + ], + "source": [ + "#Example 17_10\n", + " \n", + " \n", + "#To find the current in battery\n", + "r1=3 #Units in Ohms\n", + "r2=6 #Units in Ohms\n", + "ra=(r1*r2)/(r1+r2)#Units in Ohms\n", + "r3=2 #Units in Ohms\n", + "r4=4 #Units in Ohms\n", + "rb=r3+r4 #Units in Ohms\n", + "r5=6 #Units in Ohms\n", + "rc=(r5*rb)/(r5+rb) #Units in Ohms\n", + "r6=9 #Units in Ohms\n", + "r=r6+rc #Units in Ohms\n", + "v=6 #Units in V\n", + "i=v/r #Units in Ohms\n", + "print \"The current in battery is I=\",round( ,2),\" A\",i)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.11:pg-872" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 1.2 A\n", + "Current in wire 2 is I2= 0.6 A\n", + "Current in wire 3 is I3= 0.6 A\n", + "\n" + ] + } + ], + "source": [ + "#Example 17_11\n", + " \n", + " \n", + "#To find the current in the wires\n", + "v1=12.0 #Units in V\n", + "r3=20.0 #Units in Ohms\n", + "v2=6 #Units in V\n", + "r2=10.0 #Units in Ohms\n", + "r1=5 #Units in Ohms\n", + "i3=((v1*r3)-(v2*r1))/((r2*r3)+(r1*r3)+(r1*r2)) #Units in A\n", + "i2=((r2*i3)+v2)/r3 #Units in A\n", + "i1=i3+i2 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1,1),\" A\\nCurrent in wire 2 is I2=\",round(i2,1),\" A\\nCurrent in wire 3 is I3=\",round(i3,1),\" A\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.12:pg-873" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 0.667 A\n", + "Current in wire 2 is I2= 0.444 A\n", + "Current in wire 3 is I3= -0.222 A\n", + "\n" + ] + } + ], + "source": [ + "#Example 17_12\n", + " \n", + " \n", + "#To find I1, I2 and I3 in the circuit\n", + "v1=40 #Units in V\n", + "r1=10.0 #Units in Ohms\n", + "r2=30.0 #Units in Ohms\n", + "v2=60.0 #Units in V\n", + "r3=15.0 #Units in Ohms\n", + "v3=50 #Units in V\n", + "i1=((-v1*r2)+(-r3*v1)+(60*r3)+(v3*r2))/((r1*r2)+(r2*r3)+(r3*r1)) #Units in A\n", + "i=2 #Units in A\n", + "i2=(i-i1)/3 #Units in A\n", + "i3=i2-i1 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1,3),\" A\\nCurrent in wire 2 is I2=\",round(i2,3),\" A\\nCurrent in wire 3 is I3=\",round(i3,3),\" A\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.13:pg-874" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The current I= 2.5 A\n", + " Resistance is R= 20.0 Ohms\n", + " The value E is= 41.0 V\n" + ] + } + ], + "source": [ + "#Example 17_13\n", + " \n", + " \n", + "#To find the values of e, R and I\n", + "i1=2 #Units in A\n", + "i2=0.5 #Units in A\n", + "i=i1+i2 #Units in A\n", + "v1=6 #Units in V\n", + "v2=16.0 #Units in V\n", + "r=-(v1-v2)/0.5 #Units in Ohms\n", + "v3=25.0 #Units in V\n", + "e=v2+v3 #Units in V\n", + "print \" The current I=\",round(i,1),\" A\\n Resistance is R=\",round(r),\" Ohms\\n The value E is=\",round(e),\" V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.14:pg-874" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current in wire 1 is I1= 2.0 A\n", + "Current in wire 2 is I2= 4.0 A\n", + "Current in wire 3 is I3= 2.0 A\n", + "\n", + "The charge on the capacitor is q= 1e-05 C\n" + ] + } + ], + "source": [ + "#Example 17_14\n", + " \n", + " \n", + "#To find the I1,I2,I3 values and charge on the capacitor\n", + "v1=12.0 #Units in V\n", + "r1=6.0 #Units in Ohms\n", + "i1=v1/r1 #Units in A\n", + "v2=4.0 #Units in V\n", + "r2=8.0 #Units in Ohms\n", + "i3=(v1+v2)/r2 #Units in A\n", + "i2=i1+i3 #Units in A\n", + "print \"Current in wire 1 is I1=\",round(i1),\" A\\nCurrent in wire 2 is I2=\",round(i2),\" A\\nCurrent in wire 3 is I3=\",round(i3),\" A\\n\"\n", + "v3=10.0 #Units in V\n", + "vfg=-v3+(r1*i1) #Units in V\n", + "c=5*10**-6 #Units in F\n", + "q=c*vfg #Units in C\n", + "print \"The charge on the capacitor is q=\",round(q,5),\" C\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.15:pg-874" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Potential difference between d to c is= 23.8 V\n", + "\n", + "The potential difference between b to a is= 7.8 V\n" + ] + } + ], + "source": [ + "#Example 17_15\n", + " \n", + " \n", + "#To find the terminal potential of each battery\n", + "v=18 #Units in V\n", + "r=9 #Units in Ohms\n", + "i=v/r #Units in A\n", + "r1=0.1 #Units in Ohms\n", + "v1=-i*r1 #Units in V\n", + "v2=24 #Units in V\n", + "v11=v1+v2 #Units in V\n", + "r2=0.9 #Units in Ohms\n", + "v3=i*r2 #Units in V\n", + "v4=6 #Units in V\n", + "v22=v3+v4 #Units in V \n", + "print \"The Potential difference between d to c is=\",round(v11,1),\" V\"\n", + "print \"\\nThe potential difference between b to a is=\",round(v22,1),\" V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex17.16:pg-875" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance of the recording device is= 1000000.0 Ohms\n" + ] + } + ], + "source": [ + "#Example 17_16\n", + " \n", + " \n", + "#To findout how large a a resistance must the recording device must have\n", + "r1=10000.0 #Units in Ohms\n", + "percent=1.0 #Units in Percentage \n", + "vo=1/(r1*(percent*100)) #Units In terms of Ro\n", + "Ro=1/vo #Units in Ohms\n", + "print \"The resistance of the recording device is=\",round(Ro),\" Ohms\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter18.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter18.ipynb new file mode 100644 index 00000000..30da03c6 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter18.ipynb @@ -0,0 +1,211 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18:Magnetism" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.1:pg-901" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force on the wire is F= 0.000958363 N\n" + ] + } + ], + "source": [ + " #Example 18_1\n", + "import math \n", + " \n", + "#To find the force on the wire\n", + "b=2*10**-4 #Units in T\n", + "i=20 #Units in A\n", + "l=0.3 #Units in meters\n", + "theta=53 #Units in degrees\n", + "thetaa=math.sin(theta*math.pi/180) #Units in Radians \n", + "f=b*i*l*thetaa #Units in N\n", + "print \"The force on the wire is F=\",round(f,9),\" N\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.2:pg-902" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of magnetic field is B= 0.2609 T\n" + ] + } + ], + "source": [ + " #Example 18_2\n", + " \n", + " \n", + " #To find the magnitude of the magnetic field\n", + "m=1.67*10**-27 #Units in Kg\n", + "v=10**6 #Units in meters/sec\n", + "r=4*10**-2 #Units in Meters\n", + "q=1.6*10**-19 #Units in C or eV\n", + "b=(m*v)/(r*q) #Units in T\n", + "print \"The magnitude of magnetic field is B=\",round(b,4),\" T\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.3:pg-903" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnetic field exerts a force of q*v*B upwards on the particle.\n", + "The particle doesnot deflect because the two forces are equal\n", + "Hence v=(E/B)\n", + "A particle with this speed will pass through the region of the crossfields and undeflected\n" + ] + } + ], + "source": [ + " #Example 18_3\n", + " \n", + " \n", + "#To show that the particles does not deflect from its straight line path\n", + "print \"The magnetic field exerts a force of q*v*B upwards on the particle.\\nThe particle doesnot deflect because the two forces are equal\\nHence v=(E/B)\\nA particle with this speed will pass through the region of the crossfields and undeflected\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.4:pg-903" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of B is= 1.2 G\n" + ] + } + ], + "source": [ + " #Example 18_4\n", + " \n", + " \n", + "#To calculate the value of B at a radial distance of 5 cm\n", + "u=4*math.pi*10**-7 #Units in T m/A\n", + "i=30 #Units in A\n", + "r=0.05 #Units in Meters\n", + "b=(u*i)/(2*math.pi*r) #Units in T\n", + "b=b*10**4 #Units in G\n", + "print \"The value of B is=\",round(b,2),\" G\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.5:pg-904" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnetic moment of Hydrogen atom is=\n", + "9.319e-24\n", + "A meters**2\n" + ] + } + ], + "source": [ + " #Example 18_5\n", + " \n", + " \n", + "#To find the magnetic moment of hydrogen atom\n", + "r=0.53*10**-10 #Units in meters\n", + "a=math.pi*r**2 #Units in meters**2\n", + "q=1.6*10**-19 #Units in C\n", + "f=6.6*10**15 #Units in sec**-1\n", + "i=q*f #Units in A\n", + "u=i*a #Units in A meter**2\n", + "print \"The magnetic moment of Hydrogen atom is=\"\n", + "print round(u,27)\n", + "print \"A meters**2\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter19.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter19.ipynb new file mode 100644 index 00000000..b47005ea --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter19.ipynb @@ -0,0 +1,237 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 19:Electromagnetic Induction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex19.1:pg-938" + ] + }, + { + "cell_type": "code", + "execution_count": 0, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 19_1\n", + " \n", + " \n", + " #To find the flux in the room\n", + "l=4 #Units in meters\n", + "b=0.8 #Units in meters\n", + "theta=20 #Units in degrees\n", + "a=l*b #Units in meters**2\n", + "b=4*10**-5 #Units in T\n", + "thetaa=math.cos(theta*math.pi/180) #Units in radians\n", + "phi=b*thetaa*a #Units in T meters**2\n", + "print \"The flux in the room is Phi=\",round( ,5),\" T meters**2\",phi)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex19.2:pg-939" + ] + }, + { + "cell_type": "code", + "execution_count": 0, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 19_2\n", + " \n", + " \n", + " #To find how large is the average EMF induced\n", + "b=0.5 #Units in T\n", + "a=4*10**-4 #Units in meters**2\n", + "phi2=b*a #Units in Wb\n", + "phi1=0 #Units in Wb\n", + "deltaPHI=phi2-phi1 #Units in Wb\n", + "n=100 #Units in Constant\n", + "deltaT=2*10**-2 #Units in sec\n", + "emf=(n*deltaPHI)/deltaT #Units in V\n", + "print \"The average emf Induced is emf=\",round( ),\" V\",emf)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex19.3:pg-939" + ] + }, + { + "cell_type": "code", + "execution_count": 0, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 19_3\n", + " \n", + " \n", + " #To findout how large an emf is generated\n", + "m=0.5 #Units in H\n", + "i=1 #Units in A\n", + "t=0.01 #Units in sec\n", + "emf=m*(i/t) #Units in V\n", + "print \"The emf generated is emf=\",round( ),\" V\",emf)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex19.4:pg-939" + ] + }, + { + "cell_type": "code", + "execution_count": 0, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 19_4\n", + " \n", + " \n", + " #To Calculate the value of selfinductance\n", + "print \"The Self Inductance is L=Uo*n**2*D*A\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex19.5:pg-939" + ] + }, + { + "cell_type": "code", + "execution_count": 0, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 19_5\n", + " \n", + " \n", + " #To find the time constant of the circuit and the final energy stored\n", + "l=0.5 #Units in H\n", + "r1=2 #Units in Ohms\n", + "r2=4 #Units in Ohms\n", + "r=r1+r2 #Units in Ohms\n", + "l_r=l/r #Units in sec\n", + "i=2 #Units in A\n", + "ene=0.5*l*i**2\n", + "print \"The time constant is L/R=\",round( ,4),\" Sec\\n The energy stored is=\",round( ),\" J\",l_r,ene)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex19.6:pg-940" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emf induced in the rod is emf= 0.000542 V\n" + ] + } + ], + "source": [ + " #Example 19_6\n", + "import math \n", + "#To find the emf induced in the rod\n", + "b=0.6*10**-4 #Units in T\n", + "v=3 #Units in meters/sec\n", + "d=5 #Units in meters\n", + "theta=53 #Units in degrees\n", + "thetaa=math.cos(theta*math.pi/180) #Units in radians\n", + "emf=b*v*d*thetaa #Units in V\n", + "print \"The emf induced in the rod is emf=\",round(emf,6),\" V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex19.7:pg-940" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The back emf developed is EMF= 104.0 V\n" + ] + } + ], + "source": [ + " #Example 19_7\n", + " \n", + " \n", + "#To calculate the Back emf developed\n", + "i=3 #Units in A\n", + "r=2.0 #Units in Ohms\n", + "v=110.0 #Units in Ohms\n", + "e=v-(i*r) #Units in V\n", + "print \"The back emf developed is EMF=\",round(e),\" V\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter2.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter2.ipynb new file mode 100644 index 00000000..45af3540 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter2.ipynb @@ -0,0 +1,371 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 02:Static Equilibrium" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.1:pg-41" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tension in String 1 is F1= 48.0 N\n", + "\n", + "Tension in String 2 is F2= 64.0 N\n" + ] + } + ], + "source": [ + " #Example 2_1\n", + "\n", + "\n", + " #To find the tension in the other two Strings\n", + " #As Sigma(Fx)=0\n", + "F3=80 #units in Newtons\n", + "Fx1=F3*math.sin(37*math.pi/180) #units in Newtons\n", + "Fy1=F3*math.cos(37*math.pi/180) #units in Newtons\n", + "F2=round(Fy1+0) #units in Newtons\n", + "F1=round(Fx1+0) #units in Newtons\n", + "print \"Tension in String 1 is F1=\",round(F1),\" N\\n\"\n", + "print \"Tension in String 2 is F2=\",round(F2),\" N\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.2:pg-41" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tension in string 1 is F1= 240.0 N\n", + "\n", + "Tension in string 2 is F2= 319.0 N\n", + "\n" + ] + } + ], + "source": [ + " #Example 2_2\n", + "\n", + "\n", + " #To find the tension in the three cords that hold the object\n", + " #As Sigma(Fx)=0\n", + "theta1=37 #units in degrees\n", + "theta2=53 #units in degrees\n", + "F1_F2=math.cos(theta2*math.pi/180)/math.cos(theta1*math.pi/180)\n", + " #As Sigma(Fy)=0\n", + "F3=400 #units in Newtons\n", + "F2=round((F3*math.cos(theta1*math.pi/180))/(math.cos(theta1*math.pi/180)**2+math.cos(theta2*math.pi/180)**2)) #units in Newtons\n", + "F1=(math.cos(theta2*math.pi/180)/math.cos(theta1*math.pi/180))*F2 #units in Newtons\n", + "print \"Tension in string 1 is F1=\",round(F1),\" N\\n\"\n", + "print \"Tension in string 2 is F2=\",round(F2),\" N\\n\"\n", + " #In textbook the Answer for F2 is printed wrong as 320 N But the correct answer is 319 N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.3:pg-42" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Weight W= 133.0 N\n", + "\n", + "Tension in the chord is F= 166.0 N\n" + ] + } + ], + "source": [ + " #Example 2_3\n", + "\n", + "\n", + " #To find the weight and the Tension in the cords\n", + " #As Sigma(Fx)=0\n", + "theta1=53 #units in degrees\n", + "theta2=37 #units in degrees\n", + "F1=100 #units in Newtons\n", + "F=F1/math.cos(theta1*math.pi/180) #units in Newtons\n", + "W=math.cos(theta2*math.pi/180)*F #units in Newtons\n", + "print \"The Weight W=\",round(W),\" N\\n\"\n", + "print \"Tension in the chord is F=\",round(F),\" N\"\n", + " #In text book the answers are printed wrong as F=167N and W=133N but the correct answers are W=132N and F=166N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.5:pg-48" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tension in the Supporting Cable T= 2309.0 N\n" + ] + } + ], + "source": [ + " #Example 2_5\n", + "\n", + "\n", + " #To find the Tension T in the Supporting Cable\n", + " #As Sigma(Fx)=0\n", + "theta1=30 #units in degrees\n", + "theta2=90-theta1 #units in degrees\n", + "H_T=math.sin(theta1*math.pi/180)\n", + "W=2000 #Units in Newtons\n", + "T=W/math.sin(theta2*math.pi/180) #units in Newtons\n", + "H=T*H_T #units in Newtons\n", + "print \"Tension in the Supporting Cable T=\",round(T),\" N\"\n", + " #In textbook The answer is printed wrong as T=2310N but the correct answer is T=2309N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.6:pg-52" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The First Force F1= -1800.0 N\n", + "\n", + "The Second Force F2= 2700.0 N\n", + "\n" + ] + } + ], + "source": [ + " #Example 2_6\n", + "\n", + "\n", + " #To find the forces exerted bythe pedestals on the board\n", + "tou=900 #units in Newtons\n", + "d1=3 #units in Meters\n", + "d2=1.5 #Units in Meters\n", + "F1=-(tou*d1)/d2 #Units in Newtons\n", + "F2=tou-F1 #units in Newtons\n", + "print \"The First Force F1=\",round(F1),\" N\\n\"\n", + "print \"The Second Force F2=\",round(F2),\" N\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.7:pg-53" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tension T= 291.0 N\n", + "\n", + "H= 232.0 N\n", + "\n", + "V= -25.13 N\n" + ] + } + ], + "source": [ + " #Example 2_7\n", + "\n", + "\n", + " #To find tension in the supporting cable and Components of the force exerted by the hinge\n", + "F1=50 #units in Newtons\n", + "d1=0.7 #units in meters\n", + "F2=100 #units in Newtons\n", + "d2=1.4 #units in meters\n", + "d3=1 #units in meters\n", + "theta2=53 #units in degrees\n", + "T=round(((F1*d1)+(F2*d2))/(d3*math.cos(theta2*math.pi/180))) #units in Newtons\n", + "theta1=37 #units in degrees\n", + "H=math.cos(theta1*math.pi/180)*T #units in Newtons\n", + "\n", + "V=F1+F2-(math.cos(theta2*math.pi/180)*T) #units in Newtons\n", + "print \"Tension T=\",round(T),\" N\\n\"\n", + "print \"H=\",round(H),\" N\\n\"\n", + "print \"V=\",round(V,2),\" N\"\n", + " #In text book the answer is printed wrong as H=234N but the correct answer is H=232N\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.8:pg-55" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tension T= 410.0 N\n", + "\n", + "H= 140.0 N\n", + "\n", + "V= -301.0 N\n" + ] + } + ], + "source": [ + " #Example 2_8\n", + "\n", + "\n", + " #To find the tension in the Muscle and the Component Forces at elbow\n", + "F1=65 #units in Newtons\n", + "d1=0.1 #units in Meters\n", + "F2=20 #Units in Newtons\n", + "d2=0.35 #units in meters\n", + "theta1=20 #units in degrees\n", + "d3=0.035 #units in Meters\n", + "Tm=((F1*d1)+(F2*d2))/(math.cos(theta1*math.pi/180)*d3) #units in Newtons\n", + "V=F1+F2-(Tm*math.cos(theta1*math.pi/180))\n", + "H=Tm*math.sin(theta1*math.pi/180)\n", + "print \"Tension T=\",round(Tm),\" N\\n\"\n", + "print \"H=\",round(H),\" N\\n\"\n", + "print \"V=\",round(V),\" N\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex2.9:pg-55" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Force P= 276.0 N\n", + "\n", + "Force V= 600.0 N\n", + "\n", + "Force H= 276.0 N\n" + ] + } + ], + "source": [ + " #Example 2_9\n", + "\n", + "\n", + "#To find the forces at the wall and the ground\n", + "theta1=53 #units in degrees\n", + "d1=3 #units in meters\n", + "F1=200 #units in Newtons\n", + "d2=4 #units in Meters\n", + "F2=400 #units in Newtons\n", + "theta2=37 #units in degrees\n", + "d3=6 #units in meters\n", + "P=((math.cos(theta1*math.pi/180)*d1*F1)+(math.cos(theta1*math.pi/180)*d2*F2))/(math.cos(theta2*math.pi/180)*d3) #units in Newtons\n", + "H=P #units in Newtons\n", + "V=F1+F2 #units in Newtons\n", + "print \"Force P=\",round(P),\" N\\n\"\n", + "print \"Force V=\",round(V),\" N\\n\"\n", + "print \"Force H=\",round(H),\" N\"\n", + " #In text book the answer is printed wrong as P=H=275N but the correct answer is P=H=276N\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter20.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter20.ipynb new file mode 100644 index 00000000..57a9e2b3 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter20.ipynb @@ -0,0 +1,235 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 20: Alternating Currents and Electronics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex20.1:pg-982" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We have to wait for a time of t= 10.0 sec\n" + ] + } + ], + "source": [ + " #Example 20_1\n", + " \n", + " \n", + "#To findout the time that it has to wait after turning off the set before it is safe to touch capacitor\n", + "r=10.0**6 #Units in Ohms\n", + "c=10.0**-5 #Units in F\n", + "ti=r*c #Units in Sec\n", + "print \"We have to wait for a time of t=\",round(ti),\" sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex20.2:pg-983" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The RMS current when f=20 Hz is= 0.00402 Ohms\n", + "The RMS current when f=2*10**6 Hz is= 402.12 Ohms\n" + ] + } + ], + "source": [ + " #Example 20_2\n", + "\n", + "import math\n", + " \n", + "#To find the rms current in the circuit \n", + "f=20 #Units in Hz\n", + "c=4*10.0**-7 #Units in F\n", + "xc=1/(2.0*math.pi*f*c) #Units in Ohms/sec\n", + "f=2*10**6 #Units in Hz\n", + "xc1=1/(2*math.pi*f*c) #Units in Ohms/sec\n", + "v=80.0 #Units in V\n", + "i=v/xc #Units in A\n", + "i1=v/xc1 #Units in A\n", + "print \"The RMS current when f=20 Hz is=\",round(i,5),\" Ohms\\nThe RMS current when f=2*10**6 Hz is=\",round(i1,2),\" Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex20.3:pg-984" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in the inductor when frequency=60 Hz is I= 7.07 A\n", + "\n", + "The current in the inductor when frequency=6*10**2 Hz is I= 0.000707 A\n" + ] + } + ], + "source": [ + " #Example 20_3\n", + " \n", + " \n", + "#To find the current through the inductor\n", + "f=60 #Units in Hz\n", + "l=15.0*10**-3 #Units in H\n", + "xl=2*math.pi*f*l #Units in Ohms\n", + "v=40 #Units in V\n", + "i=v/xl #Units in A\n", + "print \"The current in the inductor when frequency=60 Hz is I=\",round(i,2),\" A\"\n", + "f=6.0*10**5 #Units in Hz\n", + "l=15.0*10**-3 #Units in H\n", + "xl=2*math.pi*f*l #Units in Ohms\n", + "v=40 #Units in V\n", + "i=v/xl #Units in A\n", + "print \"\\nThe current in the inductor when frequency=6*10**2 Hz is I=\",round(i,6),\" A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex20.4:pg-984" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "ename": "SyntaxError", + "evalue": "invalid syntax (<ipython-input-4-42e5d01775ba>, line 15)", + "output_type": "error", + "traceback": [ + "\u001b[1;36m File \u001b[1;32m\"<ipython-input-4-42e5d01775ba>\"\u001b[1;36m, line \u001b[1;32m15\u001b[0m\n\u001b[1;33m print \"The current in circuit is I=\",round( ,4),\" A\\nVolt meter readings across resistor Vr=\",round( ,1),\" V\\nReadings across capacitor is Vc=\",round( ,1),\" V\\nPower loss in circuit is=\",round( ,1),\" W\",i,vr,vc,p)\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n" + ] + } + ], + "source": [ + " #Example 20_4\n", + " \n", + " \n", + "#To find current in circuit, Voltmeter reading, reading across capacitor and power loss\n", + "f=2000 #Units in Hz\n", + "c=0.6*10**-6 #Units in F\n", + "xc=1/(2*math.pi*f*c) #Units in Ohms\n", + "r=300 #Units in Ohms\n", + "z=sqrt(r**2+xc**2) #Units in Ohms\n", + "v=80 #Units in V\n", + "i=v/z #Units in A\n", + "vr=i*r #Units in V\n", + "vc=i*xc #Units in V\n", + "p=i**2*r #Units in W\n", + "print \"The current in circuit is I=\",round( ,4),\" A\\nVolt meter readings across resistor Vr=\",round( ,1),\" V\\nReadings across capacitor is Vc=\",round( ,1),\" V\\nPower loss in circuit is=\",round( ,1),\" W\",i,vr,vc,p)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex20.5:pg-984" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current in circuit is I= 2.17 A\n", + "Volt meter reading across R Vr= 43.4 V\n", + "Volt meter reading across L Vl= 32.7 V\n", + "Volt meter reading across c Vc= 57.6 V\n", + "\n" + ] + } + ], + "source": [ + " #Example 20_5\n", + "\n", + "import math \n", + "#To find the current in circuit and voltmeters reading across R C and L\n", + "f=600 #Units in Hz\n", + "l=4.0*10**-3 #Units in H\n", + "xl=2*math.pi*f*l #Units in Ohms\n", + "c=10.0*10**-6 #Units in F\n", + "xc=1/(2*math.pi*f*c) #Units in Ohms\n", + "r=20.0 #Units in Ohms\n", + "z=math.sqrt(r**2+(xl-xc)**2) #Units in Ohms\n", + "v=50.0 #Units in V\n", + "i=v/z #Units in A\n", + "vr=i*r #Units in V\n", + "vl=i*xl #Units in V\n", + "vc=i*xc #Units in V\n", + "print \"The current in circuit is I=\",round(i,2),\" A\\nVolt meter reading across R Vr=\",round(vr,1),\" V\\nVolt meter reading across L Vl=\",round(vl,1),\" V\\nVolt meter reading across c Vc=\",round(vc,1),\" V\\n\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter21.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter21.ipynb new file mode 100644 index 00000000..19fc26f8 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter21.ipynb @@ -0,0 +1,143 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 21: Elecromagnetics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex21.1:pg-1037" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Wavelength of the Electromagnetic wave is lamda= 294.0 meters\n" + ] + } + ], + "source": [ + " #Example 21_1\n", + " \n", + " \n", + " #To find the wavelength of the electromagnetic wave\n", + "v=3*10.0**8 #Units in meters/sec\n", + "f=1.02*10**6 #Units in Hz\n", + "lamda=v/f #Units in Meters\n", + "print \"The Wavelength of the Electromagnetic wave is lamda=\",round(lamda),\" meters\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex21.2:pg-1038" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of the magnetic field is Bo=\n", + "1.4e-11\n", + "T\n" + ] + } + ], + "source": [ + " #Example 21_2\n", + " \n", + " \n", + " #To find the value of magnetic field\n", + "eo=4.2*10**-3 #units in V/m\n", + "c=3*10**8 #Units in meters/sec\n", + "bo=eo/c #Units in T\n", + "print \"The value of the magnetic field is Bo=\"\n", + "print bo\n", + "print \"T\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex21.3:pg-1039" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Eo= 0.02448 N/C\n", + " The value of Bo= 8.16124749653e-11\n", + "T\n" + ] + } + ], + "source": [ + " #Example 21_3\n", + "import math \n", + " #To find the values of Eo and Bo in the wave\n", + "power=1000 #Units in W\n", + "r=10000 #units in meters\n", + "area=4*math.pi*r**2 #units in meter**2\n", + "P_a=power/area #unts in W/meter**2\n", + "c=3*10**8 #units in meters/sec\n", + "eeo=8.85*10**-12 #units in C**2/N*meter**2\",round( ,5),\" \n", + "eo=math.sqrt((2*P_a)/(c*eeo)) #units in N/C\n", + "bo=eo/c #Units in T\n", + "print \"The value of Eo=\",round(eo,5),\" N/C\\n The value of Bo=\",\n", + "print bo\n", + "print \"T\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter22.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter22.ipynb new file mode 100644 index 00000000..83a6cca0 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter22.ipynb @@ -0,0 +1,314 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 22:The properties of Light" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.1:pg-1067" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of the image is i= 6.0 cm\n", + "The Size of the image is I= 0.0 cm High\n" + ] + } + ], + "source": [ + " #Example 22_1\n", + " \n", + " \n", + "#To find the position and size of the image\n", + "d1=5 #units in cm\n", + "d2=30 #units in cm\n", + "i=(d1*d2)/(d2-d1) #Units in cm\n", + "d3=2 #units in cm\n", + "I=(i/d2)*d3 #units in cm\n", + "print \"The position of the image is i=\",round(i),\" cm\\nThe Size of the image is I=\",round(I,2),\" cm High\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.2:pg-1068" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of the image is i= -10.0 cm\n" + ] + } + ], + "source": [ + " #Example 22_2\n", + " \n", + " \n", + "#To find the location of the image\n", + "d1=10 #units in cm\n", + "d2=5 #units in cm\n", + "i=(d1*d2)/(d2-d1) #Units in cm\n", + "print \"The position of the image is i=\",round(i),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.3:pg-1069" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The location of the image is i= -30.0 cm\n", + " The relative size of the image is I_O= 0.4 cm\n" + ] + } + ], + "source": [ + " #Example 22_3\n", + " \n", + " \n", + "#To find the location of the image and its relative size\n", + "r=100.0 #Unts in cm\n", + "d1=-r/2 #units in cm\n", + "d2=75.0 #units in cm\n", + "i=(d1*d2)/(d2-d1) #Units in cm\n", + "p=75 #units in cm\n", + "sizee=-i/p #units in cm\n", + "print \"The location of the image is i=\",round(i),\" cm\\n The relative size of the image is I_O=\",round(sizee,2),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.4:pg-1069" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angle at which the light emerges in air is theta= 53.0 degrees\n" + ] + } + ], + "source": [ + "#Example 22_4\n", + " \n", + "import math \n", + "#To find the angle at which the light emerge in to the air\n", + "theta=37 #Units in degrees\n", + "n1=1.33 #Units in constant\n", + "n2=1 #Units in constant\n", + "thetaa=math.asin((n1*math.sin(theta*math.pi/180))/n2)*180/math.pi #units in degrees\n", + "print \"The angle at which the light emerges in air is theta=\",round(thetaa),\" degrees\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.5:pg-1068" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "We have Theta1=Theta4 \n", + "Which shows that A unform layer of transparent material does not change the direction of the beam of light\n" + ] + } + ], + "source": [ + " #Example 22_5\n", + " \n", + " \n", + "#At what angle does the light emerges from the bottom of the dish\n", + "print \"We have Theta1=Theta4 \\nWhich shows that A unform layer of transparent material does not change the direction of the beam of light\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.6:pg-1068" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the diagram we notice that eyes will assume that the three rays come from image position indicated and as we see the image is virtual, erect and enlarged\n", + "\n", + "The image is located at i= -10.0 cm\n" + ] + } + ], + "source": [ + " #Example 22_6\n", + " \n", + " \n", + "#To draw a ray diagram to locate the image\n", + "print \"From the diagram we notice that eyes will assume that the three rays come from image position indicated and as we see the image is virtual, erect and enlarged\"\n", + "d1=10 #units in cm\n", + "d2=5 #units in cm\n", + "i=(d1*d2)/(d2-d1) #Units in cm\n", + "print \"\\nThe image is located at i=\",round(i,2),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.7:pg-1068" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the ray diagram we have noticed that the image is virtual, erect and dimnished in size\n", + "\n", + "The image is located at i= 3.0 cm\n" + ] + } + ], + "source": [ + " #Example 22_7\n", + " \n", + " \n", + "#To find the image position by means of the ray diagram\n", + "print \"From the ray diagram we have noticed that the image is virtual, erect and dimnished in size\"\n", + "d1=5 #units in cm\n", + "d2=-10 #units in cm\n", + "i=(d1*d2)/(d2-d1) #Units in cm\n", + "print \"\\nThe image is located at i=\",round(i,2),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex22.8:pg-1069" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + "The image is located at i= -13.33 cm\n", + "\n", + "The Size of the image is I= 1.0 cm\n" + ] + } + ], + "source": [ + " #Example 22_8\n", + " \n", + " \n", + "#To find the image positon and size\n", + "d1=-20 #units in cm\n", + "d2=40.0 #units in cm\n", + "i=(d1*d2)/(d2-d1) #Units in cm\n", + "print \"\\nThe image is located at i=\",round(i,2),\" cm\"\n", + "d3=3.0 #units in cm\n", + "I=(-i*d3)/d2 #units in cm\n", + "print \"\\nThe Size of the image is I=\",round(I),\" cm\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter23.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter23.ipynb new file mode 100644 index 00000000..59609590 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter23.ipynb @@ -0,0 +1,134 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 23:Optical Devices" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex23.1:pg-1094" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The focal length of the reading glasses is f= 37.5 cm\n" + ] + } + ], + "source": [ + " #Example 23_1\n", + " \n", + " \n", + "#To find the focal length of the reading glasses\n", + "d1=25.0 #units in cm\n", + "d2=-75 #units in cm\n", + "f=(d1*d2)/(d2+d1) #Units in cm\n", + "print \"The focal length of the reading glasses is f=\",round(f,2),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex23.2:pg-1094" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The focal length of the corrective lens is f= 50.0 cm\n" + ] + } + ], + "source": [ + " #Example 23_2\n", + " \n", + " \n", + " #To find the focal length of the corrective lens\n", + "d1=-50.0 #units in cm\n", + "f=-(d1) #Units in cm\n", + "print \"The focal length of the corrective lens is f=\",round(f),\" cm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex23.3:pg-1095" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The combined focal length is= 0.3 cm\n" + ] + } + ], + "source": [ + " #Example 23_3\n", + " \n", + " \n", + "#To find the focal length of the combination\n", + "d1=20.0 #units in cm\n", + "d2=-30.0 #units in cm\n", + "d3=60.0 #units in cm\n", + "p1=100/d1 #units in dipoters\n", + "p2=100/d2 #units in dipoters\n", + "p3=100/d3 #units in dipoters\n", + "p=1/(p1+p2+p3) #Units in diopters\n", + "print \"The combined focal length is=\",round(p,1),\" cm\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter24.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter24.ipynb new file mode 100644 index 00000000..493417e2 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter24.ipynb @@ -0,0 +1,172 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 24:Interference and Diffraction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex24.1:pg-1130" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angle at which the reinforcement line occurs is theta2= 44.0 degrees\n" + ] + } + ], + "source": [ + " #Example 24_1\n", + "import math \n", + " \n", + " #To find the angle at which the reinforcement line occurs\n", + "n=2 #units in constant\n", + "lamda=0.7 #units in cm\n", + "d=2 #units in cm\n", + "theta2=math.asin((n*lamda)/d)*180/math.pi #Units in degrees\n", + "print \"The angle at which the reinforcement line occurs is theta2=\",round(theta2),\" degrees\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex24.2:pg-1131" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The thickness of air gap increases by= 294.0 nm\n" + ] + } + ], + "source": [ + " #Example 24_2\n", + " \n", + " \n", + " #To find by how much does thickness of air gap increases\n", + "lamda=589 #units in nm\n", + "gap=round(lamda/2) #units in nm\n", + "print \"The thickness of air gap increases by=\",round(gap),\" nm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex24.3:pg-1132" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The thickness that should be coated for minimum reflection is L= 99.6 nm\n" + ] + } + ], + "source": [ + " #Example 24_3\n", + " \n", + " \n", + " #To find the thickness that should be coated for minimum reflection\n", + "lamda=550 #units in nm\n", + "n=1.38 #units in constant\n", + "L=(lamda/2)/(2*n) #units in nm\n", + "print \"The thickness that should be coated for minimum reflection is L=\",round(L,1),\" nm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex24.4:pg-1133" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For the first order theta1= 32.0 degrees\n", + "For the second order theta2= 53.0 degrees\n", + "\n", + " As it is impossible for the sine of angle that is= 1.18 radians to be greater that unity this second order and higher order images doesnot exist\n" + ] + } + ], + "source": [ + " #Example 24_4\n", + "\n", + "import math \n", + " \n", + "#To find out the angle at which the line appears\n", + "line=5.89*10**-7 #Units in meters\n", + "noline=1/10.0**6 #units in Lines per meter\n", + "theta1=math.sin(line/noline)*180/math.pi #units in degrees\n", + "#For second order\n", + "theta2=math.sin(2*line/noline)*180/math.pi #units in degrees\n", + "print \"For the first order theta1=\",round(theta1),\" degrees\\nFor the second order theta2=\",round(theta2),\" degrees\"\n", + "sinevalue=2*line/noline #units in radians\n", + "print \"\\n As it is impossible for the sine of angle that is=\",round(sinevalue,2),\" radians to be greater that unity this second order and higher order images doesnot exist\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter25.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter25.ipynb new file mode 100644 index 00000000..31ea6b76 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter25.ipynb @@ -0,0 +1,447 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 25:Three Revolutionary Concepts" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.1:pg-1164" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The particle lves by a time of=\n", + "8.33e-08 Sec\n" + ] + } + ], + "source": [ + " #Example 25_1\n", + "import math \n", + " #To find out how long does a particle lives when shooted\n", + "l=2.6*10**-8 #units in sec\n", + "t=0.95 #units in c\n", + "life=l/math.sqrt(1-t**2) #units in sec\n", + "print \"The particle lves by a time of=\"\n", + "print round(life,10),\"Sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.2:pg-1164" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The original time that is required to complete a round trip is= 0.4 Years or 5.0 Months\n" + ] + } + ], + "source": [ + " #Example 25_2\n", + "import math \n", + " \n", + "#How log it would take according to earth clockfor a space ship to make a round trip\n", + "fac=0.9990 #Units in c\n", + "relfactor=math.sqrt(1-fac**2) #units in constant\n", + "time1=4.5 #Units in Years\n", + "time=2*time1 #Units in Years\n", + "oritime=relfactor*time #Units in years\n", + "print \"The original time that is required to complete a round trip is=\",round(oritime,1),\" Years or \",round(12*oritime),\" Months\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.3:pg-1164" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In every day life our clocks never come any where close to such high speeds. The electrons in a beam such as that in television tube are easily accelerated to relativistic speeds\n" + ] + } + ], + "source": [ + " #Example 25_3\n", + " \n", + " \n", + " #To graph the relativistic factor and explain why we do not observe relativistic time delaton n everyfay phenomena\n", + "print \"In every day life our clocks never come any where close to such high speeds. The electrons in a beam such as that in television tube are easily accelerated to relativistic speeds\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.4:pg-1165" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "She notices there is no change in stick. The length contraction effect concerns objects moving at high speed relative to observer. The meter stick is at rest relative to observer.\n" + ] + } + ], + "source": [ + " #Example 25_4\n", + " \n", + " \n", + " #To find out what does the women notice about the length of the stick as she starts rotating\n", + "print \"She notices there is no change in stick. The length contraction effect concerns objects moving at high speed relative to observer. The meter stick is at rest relative to observer.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.5:pg-1166" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy that is obtained by changing all mass to energy is E=\n", + "9e+15\n", + "J\n" + ] + } + ], + "source": [ + " #Example 25_5\n", + " \n", + " \n", + " #To compare the energy that obtained by changing all mass to energy\n", + "m=0.1 #units in Kg\n", + "c=3*10**8 #Units in meters/sec\n", + "e=m*c**2 #units in J \n", + "print \"The energy that is obtained by changing all mass to energy is E=\"\n", + "print e\n", + "print \"J\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.6:pg-1167" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The apparent mass of High speed electron is mo=\n", + "1.01823376491e-30\n", + "Kg\n" + ] + } + ], + "source": [ + " #Example 25_6\n", + "import math \n", + "#To find the apparent mass of a high speed electron\n", + "rati=1/3.0 #units in constant\n", + "mo=9.6*10**-31 #units in Kg\n", + "m=mo/(math.sqrt(1-rati**2)) #Units in Kg\n", + "print \"The apparent mass of High speed electron is mo=\"\n", + "print m\n", + "print \"Kg\"\n", + " #In textbook answer printed wrong as m=9.*10**-31 Kg the correct answer is m=1.018*10**-30 \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.7:pg-1168" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy of photon is E= 1.0 eV\n" + ] + } + ], + "source": [ + " #Example 25_7\n", + " \n", + " \n", + "#To find the energy of the photon in a beam\n", + "h=6.626*10**-34 #units in J\n", + "c=3*10.0**8 #units in meters/sec\n", + "lamda=1240*10**-9 #units in meters\n", + "e=(h*c)/lamda #units in J\n", + "e=e/(1.6*10**-19) #Units in eV\n", + "print \"The energy of photon is E=\",round(e),\" eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.8:pg-1168" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy with radio waves is E1=\n", + "1.24e-08\n", + "eV\n", + "\n", + "The energy with green light is E2=\n", + "2.25454545455\n", + "eV\n", + "\n", + "The energy with photon is E3=\n", + "6200.0\n", + "eV\n", + "\n" + ] + } + ], + "source": [ + " #Example 25_8\n", + " \n", + " \n", + " #To find the energy of photonn each case\n", + "dist1=1240.0*10**-9 #units in meters\n", + "lamda1=100.0 #units in meters\n", + "e1=dist1/lamda1 #Units in eV\n", + "dist2=1240.0 #units in nano meters\n", + "lamda2=550.0 #units in meters\n", + "e2=dist2/lamda2 #Units in eV\n", + "dist3=1240.0 #units in nano meters\n", + "lamda3=0.2 #units in meters\n", + "e3=dist3/lamda3 #Units in eV\n", + "print \"The energy with radio waves is E1=\"\n", + "print e1\n", + "print \"eV\\n\"\n", + "print \"The energy with green light is E2=\"\n", + "print e2\n", + "print \"eV\\n\"\n", + "print \"The energy with photon is E3=\"\n", + "print e3\n", + "print \"eV\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.9:pg-1165" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of work function for material is Phi= 1.89 eV\n" + ] + } + ], + "source": [ + " #Example 25_9\n", + " \n", + " \n", + "#To find the value of work function for material\n", + "h=6.63*10**-34 #units in J\n", + "c=3*10.0**8 #units in meters/sec\n", + "lamda=5*10.0**-7 #units in meters\n", + "vo=0.6 #units in V\n", + "e=1.6*10**-19 #units in eV\n", + "phi=((h*c)/lamda)-(vo*e) #Units in J\n", + "\n", + "phi=phi/(1.6*10**-19) #units in eV\n", + "\n", + "print \"The value of work function for material is Phi=\",round(phi,2),\" eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.10:pg-1167" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The be-broglies wavelength is lamda=\n", + "1.45714285714e-11\n", + "Meters\n" + ] + } + ], + "source": [ + " #Example 25_10\n", + " \n", + " \n", + "#To calculate the be-broglies wavelength\n", + "h=6.63*10**-34 #units in J\n", + "c=5*10**7 #units in meters/sec\n", + "m=9.1*10**-31 #Units in Kg\n", + "lamda=h/(m*c) #units in meters\n", + "print \"The be-broglies wavelength is lamda=\"\n", + "print lamda\n", + "print \"Meters\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex25.11:pg-1167" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The diffraction pattern that would be obtained by shooting bullet is math.sin(theta)=\n", + "1.6575e-29\n", + "Radians\n", + " The diffraction angles are so small that the particles will travel essentially straight through the slit\n" + ] + } + ], + "source": [ + " #Example 25_11\n", + " \n", + " \n", + " #To describe the diffraction pattern that would be obtained by shooting bullet\n", + "h=6.63*10**-34 #units in J\n", + "m=10**-4 #Units in Kg\n", + "c=200 #units in meters/sec\n", + "p=m*c #units in Kg meter/sec\n", + "lamda=h/p #units in meters\n", + "width=0.2*10**-2 #units in meters\n", + "sintheta=lamda/width #units in radians\n", + "print \"The diffraction pattern that would be obtained by shooting bullet is math.sin(theta)=\"\n", + "print sintheta\n", + "print \"Radians\\n The diffraction angles are so small that the particles will travel essentially straight through the slit\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter26.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter26.ipynb new file mode 100644 index 00000000..ab561997 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter26.ipynb @@ -0,0 +1,216 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 26: Energy levels and Spectra" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.1:pg-1208" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ionization energy of the hydrogen atom is E= 13.6 eV\n" + ] + } + ], + "source": [ + " #Example 26_1\n", + " \n", + " \n", + "#To find the ionization energy of the hydrogen atom\n", + "e=13.6 #units in eV\n", + "print \"The ionization energy of the hydrogen atom is E=\",round(e,1),\" eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.2:pg-1209" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The wavelength of fourth line in Paschen series is= 1002.0 nm\n" + ] + } + ], + "source": [ + " #Example 26_2\n", + " \n", + " \n", + " #To find the wavelength of fourth line in Paschen series\n", + "n1=3 #Units in constant\n", + "n2=7 #Units in constant\n", + "r=1.0974*10**7 #units in meter**-1\n", + "lamda=round((1/r)*((n1**2*n2**2)/(n2**2-n1**2))*10**9) #Units in nm\n", + "print \"The wavelength of fourth line in Paschen series is=\",round(lamda),\" nm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.3:pg-1209" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy associated with line 1 is E1= -54.4 eV\n", + "The energy associated with line 2 is E2= -13.6 eV\n", + "The energy associated with line 3 is E3= -6.04 eV\n", + "\n", + "The first line of balmer series is lamda= 163.0 nm and belongs to the ultraviolet region\n" + ] + } + ], + "source": [ + " #Example 26_3\n", + " \n", + " \n", + " #To draw the energy level diagram and the find the first line of balmer type series\n", + "n=1\n", + "e1=-54.4/n**2 #units in ev\n", + "n=2\n", + "e2=-54.4/n**2 #units in ev\n", + "n=3\n", + "e3=-54.4/n**2 #units in ev\n", + "print \"The energy associated with line 1 is E1=\",round(e1,1),\" eV\\nThe energy associated with line 2 is E2=\",round(e2,1),\" eV\\nThe energy associated with line 3 is E3=\",round(e3,2),\" eV\\n\"\n", + "e1=1 #units in eV\n", + "e2=7.6 #Units in eV\n", + "lamda1=1240 #units in nm\n", + "lamda=(e1/e2)*lamda1 #Units in nm\n", + "print \"The first line of balmer series is lamda=\",round(lamda),\" nm and belongs to the ultraviolet region\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.4:pg-1209" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The longest wavelength of light capable of ionizing hydrogen atom is lamda= 91.2 nm\n" + ] + } + ], + "source": [ + " #Example 26_4\n", + " \n", + " \n", + " #To find the longest wavelength of light capable of ionizing hydrogen atom\n", + " #First method\n", + "R=1.097*10**7 #Units in meter**-1\n", + "lamda=(1/R)*10**9 #Units in meters\n", + " #Second method\n", + "E=13.6 #units in eV\n", + "e1=1 #units in eV\n", + "lamda3=1240.0 #Units in eV\n", + "lamda2=(e1/E)*(lamda3) #Units in nm\n", + "print \"The longest wavelength of light capable of ionizing hydrogen atom is lamda=\",round(lamda2,1),\" nm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.5:pg-1210" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy difference between n=1 and n=2 level is E= 17713.0 eV\n" + ] + } + ], + "source": [ + " #Example 26_5\n", + " \n", + " \n", + " #To find the energy difference between the n is 1 and n is 2 level\n", + "e1=1 #Units in eV\n", + "lamda2=1240 #Units in eV\n", + "lamda3=0.07 #Units in eV\n", + "e2=lamda2/lamda3 #Units in eV\n", + "e=e2-e1 #Units in eV\n", + "print \"The energy difference between n=1 and n=2 level is E=\",round(e),\" eV\"\n", + " #In textbook answer is prinred wrong as E=18000 eV the correct answer is E=17713 eV \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter27.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter27.ipynb new file mode 100644 index 00000000..350d825b --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter27.ipynb @@ -0,0 +1,512 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 27:The Atomic Nucleus" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.1:pg-1242" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The fraction of atomic mass of Uranium is due to its electrons is= 0.000215\n" + ] + } + ], + "source": [ + " #Example 27_1\n", + " \n", + " \n", + "#What fraction of atomic mass of Uranium is due to its electrons\n", + "n=92 #Units in constant\n", + "mass=0.000549 #Units in u\n", + "tmass=235 #units in u\n", + "per=(n*mass)/tmass #Units in fractions\n", + "print \"The fraction of atomic mass of Uranium is due to its electrons is=\",round(per,6)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.2:pg-1243" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The density of gold nucleus is p=\n", + "2.30561808801e+17 Kg/meter**3\n" + ] + } + ], + "source": [ + " #Example 27_2\n", + "import math \n", + " \n", + "#To find the density of gold nucleus\n", + "r=6.97*10**-15 #Units in meters\n", + "a=197 #Units in u\n", + "v=(4/3.0)*math.pi*r**3 #Units in meter**3\n", + "m1=1.66*10**-27 #Units in Kg/u\n", + "mass=a*m1 #Units in Kg\n", + "p=mass/v #Units in Kg/meter**3\n", + "print \"The density of gold nucleus is p=\"\n", + "print p,\"Kg/meter**3\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.3:pg-1243" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy required to change the mass of a system is= 933.7 MeV\n" + ] + } + ], + "source": [ + " #Example 27_3\n", + " \n", + " \n", + "#To calculate the energy required to change the mass of a system\n", + "c=3*10**8 #units in meters/sec\n", + "m=1.66*10**-27 #Units in g\n", + "e=m*c**2 #Units in J\n", + "e=e/(1.6*10**-19)*10**-6 #Units in MeV\n", + "print \"The energy required to change the mass of a system is=\",round(e,1),\" MeV\"\n", + "#In text book answer is printed wrong as e=931.5Mev the correct answer is933.7 MeV\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.4:pg-1243" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The binding energy of deuterium is E= 2.22 MeV \n" + ] + } + ], + "source": [ + " #Example 27_4\n", + " \n", + " \n", + "#To compute the binding energy of deuterium\n", + "m1=2.014102 #Units in u\n", + "m2=0.000549 #Units in u\n", + "total=m1-m2 #Unts in u\n", + "m3=1.007276 #Units in u\n", + "m4=1.008665 #Units in u\n", + "suma=m3+m4 #Units in u\n", + "massdefect=suma-total #units in u\n", + "e1=931.5 #Units in MeV\n", + "m5=1 #Units iin eV\n", + "e=massdefect*e1/m5 #Units in MeV\n", + "print \"The binding energy of deuterium is E=\",round(e,2),\" MeV \"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.5:pg-1245" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "After 48 days only 0.313 mg will remain\n" + ] + } + ], + "source": [ + " #Example 27_5\n", + " \n", + " \n", + "#To find how much of the orignal I will still present\n", + "d1=20.0 #Units in mg\n", + "d2=d1/2 #Units in mg\n", + "d3=d2/2 #Units in mg\n", + "d4=d3/2 #Units in mg\n", + "d5=d4/2 #Units in mg\n", + "d6=d5/2 #Units in mg\n", + "d7=d6/2 #Units in mg\n", + "print \"After 48 days only \",round(d7,3),\" mg will remain\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.6:pg-1246" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The number of dis integrations per sec=\n", + "-36195210827.7\n" + ] + } + ], + "source": [ + " #Example 27_6\n", + " \n", + " \n", + "#To find how many radium atoms in the vial undergo decay\n", + "t1=5.1*10**10 #Units in sec\n", + "lamda=0.693/t1 #Units in sec**-1\n", + "n1=6.02*10**26 #Units in atoms/Kmol\n", + "n2=226 #Units in Kg/Kmol\n", + "m1=0.001 #Units in Kg\n", + "N=n1*m1/n2 #Units in number of atoms\n", + "deltat=1 #Units in sec\n", + "deltan=-lamda*N*deltat #Units in Number\n", + "print \"The number of dis integrations per sec=\"\n", + "print deltan\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.7:pg-1247" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The fraction of uranium remains undecayed today is= 0.54\n" + ] + } + ], + "source": [ + " #Example 27_7\n", + " \n", + " \n", + "#To find what fraction of uranium remains undecayed today\n", + "t1=4.5*10**9 #Units in Years\n", + "lamda=0.693/t1 #Units in years**-1\n", + "t=4*10.0**9 #Units in Years\n", + "n_no=math.e**(-lamda*t) #Units in Fractions\n", + "print \"The fraction of uranium remains undecayed today is=\",round(n_no,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.8:pg-1248" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The decay constant is lamda= 0.00878 h**-1\n", + " The Half life is t0.5= 79.0 h\n" + ] + } + ], + "source": [ + " #Example 27_8\n", + " \n", + " \n", + "#To calculate the decay constant and half life of substance\n", + "n_no=0.9 #Units in constant\n", + "t=12 #Units in h\n", + "lamda=math.log(1/n_no)/t #Units in h**-1\n", + "t1=round(0.693/lamda) #Units in h\n", + "print \"The decay constant is lamda=\",round(lamda,7),\" h**-1\\n The Half life is t0.5=\",round(t1),\" h\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.9:pg-1249" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The approximate energy of the emitted alpha particle is E= 1.56 MeV\n" + ] + } + ], + "source": [ + " #Example 27_9\n", + " \n", + " \n", + "#To fnd the approximate energy of the emitted alpha particle\n", + "m1=222.01753 #Units in u\n", + "m3=4.00263 #Units in u\n", + "m2=218.00893 #Units in u\n", + "massloss=m1-(m2+m3) #Units in u\n", + "e1=931.5 #Units in MeV\n", + "e=e1/massloss*10**-5 #Units in MeV\n", + "print \"The approximate energy of the emitted alpha particle is E=\",round(e,2),\" MeV\"\n", + " #In textbook answer s printed wrong as E=5.56eV the correct answer is E=1.56 eV\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.10:pg-1250" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage of original amount still remainng is N/No= 78.212 Percent\n" + ] + } + ], + "source": [ + " #Example 27_10\n", + "import math \n", + "#To find the fraction of original amount still existence in earth\n", + "t1=1.41*10**10 #Units in Years\n", + "lamda=0.693/t1 #Units in year**-1\n", + "t=5*10**9 #Units in years\n", + "n_no=math.e**-(lamda*t) #Units in constant\n", + "n_no=n_no*100 #Units in percentage\n", + "print \"The percentage of original amount still remainng is N/No=\",round(n_no,3),\" Percent\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.11:pg-1251" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The activity of sr=\n", + "5.25130010147e+12\n", + "Bq\n" + ] + } + ], + "source": [ + " #Example 27_11\n", + " \n", + " \n", + "#To find the activity of sr\n", + "t1=28 #units in Years\n", + "t1=t1*86400*365 #Units in sec\n", + "acti=6.022*10**26 #Units of Bq\n", + "m1=90 #Units in Kg\n", + "m2=0.001 #Units in Kg\n", + "N=(m2/m1)*acti #Units in constant\n", + "activity=0.693*N/t1 #Units in Bq\n", + "print \"The activity of sr=\"\n", + "print activity\n", + "print \"Bq\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.12:pg-1252" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The age of the axe handle is t= 27959.0 years\n" + ] + } + ], + "source": [ + " #Example 27_12\n", + "import math \n", + " \n", + "#To estimate the age of the axe handle\n", + "n_no=0.034\n", + "t1=5730 #Units in Years\n", + "t=-(math.log(n_no)*t1)/0.693 #Units in Years\n", + "print \"The age of the axe handle is t=\",round(t),\" years\"\n", + " #In textbook answer is printed wrong as t=28000 years correct answer is t=27958 years \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex27.13:pg-1253" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy released in the reaction E= 163.0 MeV\n" + ] + } + ], + "source": [ + " #Example 27_13\n", + " \n", + " \n", + "#To find the energy released in the reaction\n", + "m1=141.91635 #Units in u\n", + "m2=89.91972 #Units in u\n", + "m3=4.03466 #Units in u\n", + "n2=36 #Units in Constant\n", + "n1=56 #Units in Constant\n", + "n4=92 #units in constant\n", + "m5=236.04564 #Units in u\n", + "loss=m5-(m1+m2+m3)+n4-(n1+n2) #Units in u\n", + "e1=931.5 #units n MeV\n", + "energy=round(e1*loss) #units in MeV\n", + "print \"The energy released in the reaction E=\",round(energy),\" MeV\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter3.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter3.ipynb new file mode 100644 index 00000000..731a39e5 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter3.ipynb @@ -0,0 +1,540 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0f7242c47093d1a671ed93f821c40acaeb14b795b5ea8c74623c63dfc9c85a4c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 03:Uniform Accelerated Motion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.1:pg-97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_1\n",
+ "\n",
+ "\n",
+ " #To find the balls instantaneous velocity and Average Velocity\n",
+ "d1=8.6 #units in meters\n",
+ "t1=0.86 #units in sec\n",
+ "vp=d1/t1 #units in meters/sec\n",
+ "print \"The Instantaneous Velocity at P Vp=\",round(vp),\" meters/sec\\n\"\n",
+ " #The ball stops at position Q Hence vp=0 met/sec\n",
+ "vq=0 #units in meters/sec\n",
+ "print \"The Instantaneous Velocity at Q Vq=\",round(vq,10),\" meters/sec\\n\"\n",
+ "d2=-10.2 #units in meters\n",
+ "t2=1.02 #units in sec\n",
+ "vn=d2/t2 #units in meters/sec\n",
+ "print \"The Instantaneous Velocity at N Vn=\",round(vn),\" meters/sec\\n\"\n",
+ "d3=20 #units in meters\n",
+ "t3=2.0 #units in sec\n",
+ "vAQ=d3/t3 #units in meters/sec\n",
+ "print \"The Average Velocity between A and Q is VAQ=\",round(vAQ),\" meters/sec\\n\"\n",
+ "d4=0 #units in meters\n",
+ "t4=4.0 #units in sec\n",
+ "vAM=d4/t4 #units in meters/sec\n",
+ "print \"The Average Velocity between A and M is VAM=\",round(vAM,10),\" meters/sec\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Instantaneous Velocity at P Vp= 10.0 meters/sec\n",
+ "\n",
+ "The Instantaneous Velocity at Q Vq= 0.0 meters/sec\n",
+ "\n",
+ "The Instantaneous Velocity at N Vn= -10.0 meters/sec\n",
+ "\n",
+ "The Average Velocity between A and Q is VAQ= 10.0 meters/sec\n",
+ "\n",
+ "The Average Velocity between A and M is VAM= 0.0 meters/sec\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.2:pg-98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_2\n",
+ "\n",
+ "\n",
+ " #To calculate the Acceleration\n",
+ "v1=20.0 #units in meters/sec\n",
+ "v2=15.0 #units in meters/sec\n",
+ "t1=0 #units in sec\n",
+ "t2=0.5 #units in sec\n",
+ "c_v=v2-v1 #units in meters/sec\n",
+ "c_t=t2-t1 #units in sec\n",
+ "acceleration=c_v/c_t #units in meters/sec**2\n",
+ "print \"Acceleration a=\",round(acceleration,2),\" meters/sec**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceleration a= -10.0 meters/sec**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.3:pg-98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_3\n",
+ "\n",
+ "\n",
+ " #To find acceleration and the distance it travels in time\n",
+ "vf=5.0 #units in meters/sec\n",
+ "v0=0 #units in meters/sec\n",
+ "t=10.0 #units in sec\n",
+ "a=(vf-v0)/t #units in meters/sec**2\n",
+ "v_1=(vf+v0)/2 #unis in meters/sec\n",
+ "x=v_1*t #units in meters\n",
+ "print \"Acceleration is a=\",round(a,1),\" meters/sec\\n\"\n",
+ "print \"Distance travelled is x=\",round(x),\" meters\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceleration is a= 0.5 meters/sec\n",
+ "\n",
+ "Distance travelled is x= 25.0 meters\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.4:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_4\n",
+ "\n",
+ "\n",
+ " #To find acceleration and time taken to stop\n",
+ "v0=5.0 #units in meters/sec\n",
+ "vf=0 #units in meters/sec\n",
+ "v_1=(v0+vf)/2 #units in meters/sec\n",
+ "x=20.0 #units in meters\n",
+ "t=x/v_1 #units in sec\n",
+ "a=(vf-v0)/t #units in meters/sec**2\n",
+ "print \"Acceleration is a=\",round(a,3),\" meters/sec**2\\n\"\n",
+ "print \"Time taken to stop t=\",round(t),\" sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceleration is a= -0.625 meters/sec**2\n",
+ "\n",
+ "Time taken to stop t= 8.0 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.5:pg-100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_5\n",
+ "\n",
+ "import math\n",
+ " \n",
+ " #To calculate the speed and time to cover\n",
+ "a=4.0 #units in meters/sec**2\n",
+ "x=20.0 #units in meters\n",
+ "vf=math.sqrt(a*x*2) #units in meters/sec\n",
+ "t=vf/a #units in sec\n",
+ "print \"Speed vf=\",round(vf,2),\" meters/sec\\n\"\n",
+ "print \"Time taken T=\",round(t,2),\" sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed vf= 12.65 meters/sec\n",
+ "\n",
+ "Time taken T= 3.16 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.6:pg-112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_6\n",
+ " \n",
+ "import math \n",
+ " #To find the time taken by a car to travel\n",
+ "x=98.0 #uniys in meters\n",
+ "a=4.0 #units in meters/sec**2\n",
+ "t=math.sqrt((2*x)/a) #units in sec\n",
+ "print \"Time taken by a car to travel is T=\",round(t),\" sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time taken by a car to travel is T= 7.0 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.7:pg-112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_7\n",
+ " \n",
+ "import math \n",
+ " #To calculate the time taken to travel\n",
+ "v0=16.7 #units in meters/sec\n",
+ "a=1.5 #units in meters/sec**2\n",
+ "x=70 #units in meters\n",
+ "t=-((-v0)+math.sqrt(v0**2-(4*(a/2)*x)))/(2*(a/2)) #units in sec\n",
+ "print \"Time taken to travel T=\",round(t,1),\" sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time taken to travel T= 5.6 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.8:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_8\n",
+ " \n",
+ " \n",
+ " #To calculate the acceleration\n",
+ "vf=30.0 #units in meters/sec\n",
+ "v0=0 #units in meters/sec\n",
+ "t=9.0 #units in sec\n",
+ "a=(vf-v0)/t #units in meters/sec**2\n",
+ "a=a*(1/1000.0)*(3600.0/1)*(3600.0/1) #units in km/h**2\n",
+ "print \"Acceleration a=\",round(a),\" km/h**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceleration a= 43200.0 km/h**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.9:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_9\n",
+ " \n",
+ " \n",
+ " #To find how above the water is the bridge\n",
+ "v0=0 #units in meters/sec\n",
+ "t=3.0 #units in sec\n",
+ "a=-9.8 #units in meters/sec**2\n",
+ "y=(v0*t)+(0.5*a*t**2) #units in meters\n",
+ "print \"The bridge is y=\",round(y),\" meters above the water\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The bridge is y= -44.0 meters above the water\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.10:pg-115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_10\n",
+ " \n",
+ "import math \n",
+ " #To find out how high does it goes and its speed and how long will it be in air \n",
+ "vf=0 #units in meters/sec\n",
+ "v0=15 #units in meters/sec\n",
+ "a=-9.8 #units in meters/sec**2\n",
+ "y=(vf**2-v0**2)/(2*a) #units in meters\n",
+ "print \"Distance it travels is y=\",round(y,1),\" meters\\n\"\n",
+ "vf=-math.sqrt(2*a*-y) #units in meters/sec\n",
+ "print \"The speed is vf=\",round(vf),\" meters/sec\\n\"\n",
+ "t=vf/(0.5*a) #units in sec\n",
+ "print \"Time taken is T=\",round(t,2),\" sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance it travels is y= 11.5 meters\n",
+ "\n",
+ "The speed is vf= -15.0 meters/sec\n",
+ "\n",
+ "Time taken is T= 3.06 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.11:pg-116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_11\n",
+ " \n",
+ " \n",
+ " #To find out how fast a ball must be thrown\n",
+ "a=9.8 #unita in meters/sec**2\n",
+ "t=3 #units in sec\n",
+ "v=(0.5*a*t**2)/t\n",
+ "print \"The speed by which the ball has to be thrown is v=\",round(v,1),\" meters/sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The speed by which the ball has to be thrown is v= 14.7 meters/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.12:pg-117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_12\n",
+ " \n",
+ "import math \n",
+ "#To find out where the ball will hit the ground\n",
+ "#Horizontal\n",
+ "y=2 #units in meters\n",
+ "a=9.8 #units in meters/sec**2\n",
+ "t=math.sqrt(y/(0.5*a)) #units in sec\n",
+ "v=15 #units in meters/sec\n",
+ "x=v*t #units in sec\n",
+ "print \"The ball hits the ground at x=\",round(x,2),\" meters\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ball hits the ground at x= 9.58 meters\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.13:pg-118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 3_13\n",
+ " \n",
+ "import math \n",
+ " #To find out at what height above ground does it hit wall and is it still going up befor it hits or down\n",
+ "v_1=24.0 #units in meters/sec\n",
+ "x=15.0 #units in meters\n",
+ "t=x/v_1 #units in sec\n",
+ "v0=18 #units in meters/sec\n",
+ "a=-9.8 #units in meters/sec**2\n",
+ "y=(v0*t)+(0.5*a*t**2) #units in meters\n",
+ "print \"The arrow hits y=\",round(y,1),\" meters above the straight point\\n\"\n",
+ "v=v0+(a*t) #units in meters/sec\n",
+ "print \"The Vertical componet of velocity is v=\",round(v,1),\" meters/sec\\n\"\n",
+ "print \"As V is Positive the arrow is in its way up\\n\"\n",
+ "vtotal=math.sqrt(v**2+v_1**2) #units in meters/sec\n",
+ "print \"The magnitude of velocity is vtotal=\",round(vtotal,1),\" meters/sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The arrow hits y= 9.3 meters above the straight point\n",
+ "\n",
+ "The Vertical componet of velocity is v= 11.9 meters/sec\n",
+ "\n",
+ "As V is Positive the arrow is in its way up\n",
+ "\n",
+ "The magnitude of velocity is vtotal= 26.8 meters/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter4.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter4.ipynb new file mode 100644 index 00000000..750b65e7 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter4.ipynb @@ -0,0 +1,396 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5482cd080863a0bf132cc69662813c918cd153651feccebbb960c53549f6ef87"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter04: Newtons Law"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.1:pg-147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_1\n",
+ " \n",
+ " \n",
+ " #To calculate the force required\n",
+ "vf=12 #units in meters/sec\n",
+ "v0=0 #units in meters/sec\n",
+ "t=8 #units in sec\n",
+ "a=(vf-v0)/t #units in meters/sec**2\n",
+ "m=900 #units in Kg\n",
+ "F=m*a #units in Newtons\n",
+ "print \"The force required is F=\",round(F),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The force required is F= 900.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.2:pg-147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_2\n",
+ " \n",
+ " \n",
+ " #To find the friction force that opposes the motion\n",
+ "F1=500 #units in Newtons\n",
+ "F2=800 #units in Newtons\n",
+ "theta=30 #units in degrees\n",
+ "Fn=F1+(F2*math.sin(theta*math.pi/180)) #units in Newtons\n",
+ "u=0.6\n",
+ "f=u*Fn #units in Newtons\n",
+ "print \"The Frictional force that is required is f=\",round(f),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Frictional force that is required is f= 540.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.3:pg-153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_3\n",
+ " \n",
+ " \n",
+ " #To find out at what rate the wagon accelerate and how large a force the ground pushing up on wagon\n",
+ "F1=90 #units in Newtons\n",
+ "F2=60 #units in Newtons\n",
+ "P=F1-F2 #units in Newtons\n",
+ "F3=100 #units in Newtons\n",
+ "F4=sqrt(F3**2-F2**2) #units in Newtons\n",
+ "a=9.8 #units in meters/sec**2\n",
+ "ax=(F4*a)/F1 #units in Meters/sec**2\n",
+ "print \"The wagon accelerates at ax=\",round(ax,1),\" meters/sec**2\\n\"\n",
+ "print \"Force by which the ground pushing is P=\",round(P),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wagon accelerates at ax= 8.7 meters/sec**2\n",
+ "\n",
+ "Force by which the ground pushing is P= 30.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.4:pg-153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_4\n",
+ " \n",
+ " \n",
+ " # To calculate How far does the car goes\n",
+ "w1=3300 #units in lb\n",
+ "F1=4.45 #units in Newtons\n",
+ "w2=1 #units in lb\n",
+ "weight=w1*(F1/w2) #units in Newtons\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "Mass=weight/g #units in Kg\n",
+ "speed=38 #units in mi/h\n",
+ "speed=speed*(1.61)*(1/3600) #units in Km/sec\n",
+ "stoppingforce=0.7*(weight) #units in Newtons\n",
+ "a=stoppingforce/-(Mass) #units in meters/sec**2\n",
+ "vf=0\n",
+ "v0=17 #units in meters/sec\n",
+ "x=(vf**2-v0**2)/(2*a)\n",
+ "print \"The car goes by x=\",round(x,1),\" meters\"\n",
+ " #In text book the answer is printed wrong as x=20.9 meters the correct answer is x=21.1 meters\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The car goes by x= 21.1 meters\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.5:pg-155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_5\n",
+ " \n",
+ " \n",
+ " #To find the acceleration of the masses\n",
+ "w1=10 #units in Kg\n",
+ "w2=5 #units in Kg\n",
+ "f1=98 #units in Newtons\n",
+ "f2=49 #units in Newtons\n",
+ "w=w1/w2\n",
+ "T=round((f1+(w*f2))/(w+1)) #units in Newtons\n",
+ "a=(f1-T)/w1 #units in meters/sec**2\n",
+ "print \"Acceleration is a=\",round(a,1),\" meters/sec**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceleration is a= 3.3 meters/sec**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.6:pg-156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_6\n",
+ " \n",
+ "\n",
+ " #To find the acceleration of the objects\n",
+ "w1=0.4 #units in Kg\n",
+ "w2=0.2 #units in Kg\n",
+ "w=w1/w2\n",
+ "a=9.8 #units in meters/sec**2\n",
+ "f=0.098 #units in Newtons\n",
+ "c=w2*a #units in Newtons\n",
+ "T=((w*c)+f)/(1+w) #units in Newtons\n",
+ "a=(T-f)/w1 #units in meters/sec**2\n",
+ "print \"Acceleration a=\",round(a,1),\" meters/sec**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceleration a= 3.1 meters/sec**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.7:pg-157"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_7\n",
+ " \n",
+ " \n",
+ " #To estimate the lower limit for the speed\n",
+ " #In a practical situation u should be atleast 0.5\n",
+ "u=0.5\n",
+ "g=9.8 #units in meter/sec**2\n",
+ "x=7 #units in meters\n",
+ "v0=math.sqrt(2*u*g*x) #units in meters/sec\n",
+ "print \"The lower limit of the speed v0=\",round(v0,1),\" meter/sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lower limit of the speed v0= 8.3 meter/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.9:pg-158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_9\n",
+ " \n",
+ " \n",
+ " #To calculate how large a force must push on car to accelerate\n",
+ "m=1200 #units in Kg\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "d1=4 #units in meters\n",
+ "d2=40 #units in meters\n",
+ "a=0.5 #units in meters/sec**2\n",
+ "P=((m*g)*(d1/d2))+(m*a) #units in Newtons\n",
+ "print \"The force required is P=\",round(P),\" N\"\n",
+ " #In text book the answer is printed wrong as P=1780 N but the correct answer is P=1776 N\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.10:pg-159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_10\n",
+ " \n",
+ " \n",
+ " #To calculate the tension in the rope\n",
+ "u=0.7\n",
+ "sintheta=(6.0/10)\n",
+ "w1=50 #units in Kg\n",
+ "g=9.8 #units in meter/sec**2\n",
+ "costheta=(8.0/10)\n",
+ "Fn=w1*g*costheta #units in Newtons\n",
+ "f=u*Fn #units in Newtons\n",
+ "T=f+(w1*g*sintheta)\n",
+ "print \"The tension in the rope is T=\",round(T),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.11:pg-159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 4_11\n",
+ " \n",
+ " \n",
+ " #To find the acceleration of the system\n",
+ "w1=7.0 #units in Kg\n",
+ "a=9.8 #units in meters/sec**2\n",
+ "w2=5 #units in Kg\n",
+ "w=w1/w2\n",
+ "F1=29.4 #units in Newtons\n",
+ "F2=20 #units in Newtons\n",
+ "f=(F1+F2) #units in Newtons\n",
+ "T1=w1*a #units in Newtons\n",
+ "T=(T1+(w*f))/(1+w) #units in Newtons\n",
+ "a=((w1*a)-T)/w1 #units in meters/sec**2\n",
+ "print \"Acceleration a=\",round(a,2),\" meters/sec**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceleration a= 1.6 meters/sec**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter5.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter5.ipynb new file mode 100644 index 00000000..05c6aa63 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter5.ipynb @@ -0,0 +1,616 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8f34af9d7ab4f4986ce62fe277146eddcd673716e93f679d03460223f2973812"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 05:Work and Energy"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.1:pg-159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_1\n",
+ " \n",
+ "import math \n",
+ " #To calculate the work done\n",
+ "Fs=8.0 #units in meters\n",
+ "W=Fs*round(math.cos(math.pi/2.0)) #units in Joules\n",
+ "print \"The work done is W=\",round(W),\" Joules\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The work done is W= 0.0 Joules\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.2:pg-159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_2\n",
+ " \n",
+ " \n",
+ " #To calculate the work done when lifting object as well as lowering the object\n",
+ "Fs=1.0 #units in terms of Fs\n",
+ "theta=0 #units in degrees\n",
+ "W=Fs*math.cos(theta*math.pi/180) #units in terms of m, g and h\n",
+ "print \"Work done when lifting is W=mgh*\",round(W),\"\\n\"\n",
+ "theta=180.0 #units in degrees\n",
+ "W=Fs*math.cos(theta*math.pi/180) #units in terms of m, g and h\n",
+ "print \"Work done when downing is W=mgh*\",round(W),\"\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done when lifting is W=mgh* 1.0 \n",
+ "\n",
+ "Work done when downing is W=mgh* -1.0 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.3:pg-160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_3\n",
+ " \n",
+ " \n",
+ " #To find the work done by the pulling force\n",
+ "F=20.0 #units in Newtons\n",
+ "d=5 #units in meters\n",
+ "W=F*d #units in joules\n",
+ "print \"Work done is W=\",round(W),\" Joules\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Work done is W= 100.0 Joules\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.4:pg-160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_4\n",
+ " \n",
+ " \n",
+ " #To find out the power being developed in motor\n",
+ "m=200 #units on Kg\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "Fy=m*g #units in Newtons\n",
+ "vy=0.03 #units in meter/sec\n",
+ "P=Fy*vy #units in Watts\n",
+ "P=P*(1.0/746) #units in hp\n",
+ "print \"Power developed P=\",round(P,5),\" hp\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power developed P= 0.07882 hp\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5:pg-161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_5\n",
+ " \n",
+ " \n",
+ " #To calculate the average frictional force developed\n",
+ "m=2000 #units in Kg\n",
+ "vf=20 #units in meters/sec\n",
+ "d=100 #units in meters\n",
+ "f=(0.5*m*vf**2)/d #units in Newtons\n",
+ "print \"Average frictional force f=\",round(f),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average frictional force f= 4000.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.6:pg-162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_6\n",
+ " \n",
+ "import math\n",
+ " #To find out how fast the car is going\n",
+ "f=4000.0 #units in Newtons\n",
+ "s=50.0 #units in meters\n",
+ "theta=180.0 #units in degrees\n",
+ "m=2000.0 #units in Kg\n",
+ "v0=20 #units in meter/sec\n",
+ "vf=math.sqrt((2*((f*s*math.cos(theta*math.pi/180))+(0.5*m*v0**2)))/m) #units in meter/sec\n",
+ "print \"The speed of the car is vf=\",round( ,1),\" meters/sec\",vf)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.7:pg-163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_7\n",
+ "\n",
+ "import math\n",
+ " \n",
+ " #To find the required tension in the rope\n",
+ "m=40 #units in Kg\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "theta=0 #units in degrees\n",
+ "vf=0.3 #units in meters/sec\n",
+ "s=0.5 #units in meters\n",
+ "T=round((m*g)+((0.5*m*vf**2)/(s*math.cos(theta*math.pi/180)))) #units in Newtons\n",
+ "print \"Tension in the rope is T=\",round(T),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Tension in the rope is T= 396.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.8:pg-163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_8\n",
+ "import math \n",
+ " \n",
+ " #To calculate the frictional force\n",
+ "m=900.0 #units in Kg\n",
+ "v0=20.0 #units in meters/sec\n",
+ "s=30 #units in meters\n",
+ "f=(0.5*m*v0**2)/s #units in Newtons\n",
+ "print \"Frictional force required is f=\",round(f),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frictional force required is f= 6000.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9:pg-164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_9\n",
+ " \n",
+ " \n",
+ " #To find out how fast a a ball is going\n",
+ "m=3 #units in Kg\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "hf=0 #units in meters\n",
+ "h0=4 #units in meters\n",
+ "vf=2*sqrt(((m*g*-(hf-h0))*0.5)/m) #units in meters/sec\n",
+ "print \"The ball is moving with a speed of vf=\",round(vf,2),\" meters/sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ball is moving with a speed of vf= 8.85 meters/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.10:pg-164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_10\n",
+ " \n",
+ " \n",
+ " #To calculate how large the average frictional force\n",
+ "a=9.8 #units in meters/sec**2\n",
+ "s=4 #units in meters\n",
+ "v=6 #units in meters/sec\n",
+ "m=3 #units on Kg\n",
+ "f=m*((a*s)-(0.5*v**2))/s #units in Newtons\n",
+ "print \"The average frictional force f=\",round(f,1),\" N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average frictional force f= 15.9 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.11:pg-165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_11\n",
+ " \n",
+ "import math \n",
+ " #To find out how fast a car is going at points B and C\n",
+ "m=300 #units in Kg\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "hb_ha=10 #units in meters\n",
+ "f=20 #units in Newtons\n",
+ "s=60 #units in meters\n",
+ "vf=2*math.sqrt((0.5*((m*g*(hb_ha))-(f*s)))/m) #units in meters/sec\n",
+ "print \"The car is going at a speed of vf=\",round(vf,1),\" meters/sec at point B\\n\"\n",
+ "hc_ha=2 #units in meters\n",
+ "vf=2*math.sqrt((0.5*((m*g*(hc_ha))-(f*s)))/m) #units in meters/sec\n",
+ "print \"The car is going at a speed of vf=\",round(vf,2),\" meters/sec at point C\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The car is going at a speed of vf= 13.7 meters/sec at point B\n",
+ "\n",
+ "The car is going at a speed of vf= 5.59 meters/sec at point C\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.12:pg-166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_12\n",
+ " \n",
+ " \n",
+ " #How far the average velocity and how far beyond B does the car goes\n",
+ "m=2000 #units in Kg\n",
+ "vb=5 #units in meters/sec\n",
+ "va=20 #units in meters/sec\n",
+ "hb_ha=8 #units in meters\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "sab=100 #units in meters\n",
+ "f=-((0.5*m*(vb**2-va**2))+(m*g*(hb_ha)))/sab #units in Newtons\n",
+ "print \"Average frictional force is f=\",round(f),\" N\\n\"\n",
+ "Sbe=(0.5*m*vb**2)/f #units in meters\n",
+ "print \"The distance by which the car goes beyond is Sbe=\",round(Sbe,1),\" meters\"\n",
+ " #In text book answer is printed wrong as f=2180 N but correct answer is f=2182N\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average frictional force is f= 2182.0 N\n",
+ "\n",
+ "The distance by which the car goes beyond is Sbe= 11.5 meters\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.13:pg-167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_13\n",
+ " \n",
+ " \n",
+ " #To find out how large the force is required\n",
+ "m=2 #units in Kg\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "hc_ha=10.03 #units in meters\n",
+ "sbc=0.030 #units in meters\n",
+ "f=(m*g*(hc_ha))/sbc #units in Newtons\n",
+ "print \"The average force required is f=\",round(f),\" N\"\n",
+ " #In text book answer is printed wrong as f=6550 N correct answer is f=6552N\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average force required is f= 6553.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.14:pg-168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_14\n",
+ " \n",
+ " \n",
+ " #To find out how fast the pendulum is moving \n",
+ " #At point A\n",
+ "hb_ha=0.35 #units in Meters\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "vb=sqrt((g*hb_ha)/0.5) #units in meters/sec\n",
+ "print \"The velocity of pendulum at point B is vb=\",round(vb,2),\" meters/sec\\n\"\n",
+ "print \"From A to C hc=ha and Vc=Va=0 so Frictional force is Negligible at point C\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of pendulum at point B is vb= 2.62 meters/sec\n",
+ "\n",
+ "From A to C hc=ha and Vc=Va=0 so Frictional force is Negligible at point C\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.15:pg-169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_15\n",
+ " \n",
+ " \n",
+ " #To find out how large a force is required\n",
+ "m=2000 #units in Kg\n",
+ "vf=15 #units in meters/sec\n",
+ "f1=500 #units in Newtons\n",
+ "F=((0.5*m*(vf**2))/80)+f1 #units in Newtons\n",
+ "print \"Force required is F=\",round(F),\" N\"\n",
+ " #In text book the answer is printed wrong as F=3300 N but the correct answer is 3312 N\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force required is F= 3313.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.16:pg-169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 5_16\n",
+ " \n",
+ " \n",
+ " #To find IMA AMA and Efficiency of the system\n",
+ "si=3.0\n",
+ "so=1\n",
+ "IMA=si/so\n",
+ "Fo=2000.0 #units in Newtons\n",
+ "Fi=800.0 #units in Newtons\n",
+ "AMA=Fo/Fi\n",
+ "effi=AMA/IMA*100\n",
+ "print \"IMA=\",round(IMA,2),\"\\n\"\n",
+ "print \"AMA=\",round(AMA,2),\"\\n\"\n",
+ "print \"Percentage of efficiency is \",round(effi),\" percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "IMA= 3.0 \n",
+ "\n",
+ "AMA= 2.5 \n",
+ "\n",
+ "Percentage of efficiency is 83.0 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter6.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter6.ipynb new file mode 100644 index 00000000..09d307d8 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter6.ipynb @@ -0,0 +1,372 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 06:Linear Momentum" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.1:pg-189" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average retarding force is F= -2500.0 Newtons\n" + ] + } + ], + "source": [ + " #Example 6_1\n", + " \n", + " \n", + " #To calculate how large is the average force retarding its motion\n", + "m=1500 #units in Kg\n", + "vf=15.0 #units in meters/sec\n", + "v0=20 #units in meters/sec\n", + "t=3 #units in sec\n", + "f=((m*vf)-(m*v0))/t #Units in Newtons\n", + "print \"The average retarding force is F=\",round(f),\" Newtons\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.2:pg-190" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average stopping force the tree exerts on the car is F=\n", + "-160000.0 Newtons\n" + ] + } + ], + "source": [ + " #Example 6_2\n", + " \n", + " \n", + " #To estimate the average stopping force the tree exerts on the car\n", + "m=1200 #units in Kg\n", + "vf=0 #units in meters/sec\n", + "v0=20 #units in meters/sec\n", + "v=0.5*(vf+v0) #units in meters/sec\n", + "s=1.5 #units in meters\n", + "t=s/v #units in sec \n", + "f=((m*vf)-(m*v0))/t #Units in Newtons\n", + "print \"The average stopping force the tree exerts on the car is F=\"\n", + "print f,\"Newtons\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.3:pg-191" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The car is moving at vf= 8.0 Meters/sec\n", + "\n", + "The positive sign of vf Indicate the car is moving in the direction the truck was moving\n" + ] + } + ], + "source": [ + " #Example 6_3\n", + " \n", + " \n", + " #To find out how fast and the direction car moving\n", + "m1=30000 #units in Kg\n", + "m2=1200 #units in Kg\n", + "v10=10 #units in meters/sec\n", + "v20=-25 #units in meters/sec\n", + "vf=((m1*v10)+(m2*v20))/(m1+m2) #unis in meters/sec\n", + "print \"The car is moving at vf=\",round(vf,2),\" Meters/sec\\n\"\n", + "print \"The positive sign of vf Indicate the car is moving in the direction the truck was moving\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.5:pg-193" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity V2f= 0.2 meters/sec or 20.0 cm/sec\n", + "\n", + "The velocity V1f= -0.1 meters/sec or 10.0 cm/sec\n", + "\n" + ] + } + ], + "source": [ + " #Example 6_5\n", + " \n", + " \n", + " #To find the velocity of each ball after collision\n", + "m1=0.04 #units in kg\n", + "m2=0.08 #units in kg\n", + "v1=0.3 #units in meters/sec\n", + "v2f=(2*m1*v1)/(m1+m2) #units in meters/sec\n", + "v2f1=v2f*100 #units in cm/sec\n", + "print \"The velocity V2f=\",round(v2f,1),\" meters/sec or \",round(v2f1),\" cm/sec\\n\"\n", + "v1f=((m1*v1)-(m2*v2f))/m1 #units in meters/sec\n", + "v1f1=-v1f*100 #units in cm/sec\n", + "print \"The velocity V1f=\",round(v1f,1),\" meters/sec or \",round(v1f1),\" cm/sec\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.6:pg-196" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of the pelet before collision is V10= 487.0 meters/sec\n" + ] + } + ], + "source": [ + " #Example 6_6\n", + " \n", + "import math \n", + " #To calculate the speed of the pellet before collision\n", + "h=0.30 #units in meters\n", + "g=9.8 #units in meters/sec**2\n", + "v=math.sqrt(2*g*h) #units in meters/sec\n", + "m1=2 #units in Kgs\n", + "m2=0.010 #units in kgs\n", + "v10=((m1+m2)*v)/m2 #units in meters/sec\n", + "print \"The speed of the pelet before collision is V10=\",round(v10),\" meters/sec\"\n", + " #In textbook the answer is printed wrong as V10=486 meters/sec the correct answer is V10=487 meters/sec\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.7:pg-196" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Thrust is F= 65000000.0 Newtons\n" + ] + } + ], + "source": [ + " #Example 6_7\n", + " \n", + " \n", + " #To calculate how large a forward push given to the rocket\n", + "m=1300 #units in Kgs\n", + "vf=50000 #units in meters/sec\n", + "v0=0 #units in meters/sec\n", + "F=((m*vf)-(m*v0)) #units in Newtons\n", + "print \"The Thrust is F=\",round(F),\" Newtons\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.8:pg-197" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Z component of velocity is Vz= 0.0 meters/sec\n", + "\n", + "The Y component of velocity is Vy= -0.6 *V0\n", + "\n", + "The X component of velocity is Vx= -1.8 *V0\n" + ] + } + ], + "source": [ + " #Example 6_8\n", + " \n", + " \n", + " #To determine the velocity of the third peice\n", + "momentumbefore=0 #units in kg meter/s\n", + "m=0.33 #units in Kgs\n", + "vz=momentumbefore/m\n", + "print \"The Z component of velocity is Vz=\",round(vz),\" meters/sec\\n\"\n", + "m=0.33 #units in Kgs\n", + "v0=0.6 #units in meters/sec\n", + "vy=-(m*v0)/m #interms of v0 and meters/sec\n", + "print \"The Y component of velocity is Vy=\",round(vy,1),\"*V0\\n\"\n", + "v01=1 #units in meters/sec\n", + "v02=0.8 #units in meters/sec\n", + "vx=-((v01+v02)*m)/m #interms of v0 and units in meters/sec\n", + "print \"The X component of velocity is Vx=\",round(vx,1),\"*V0\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.9:pg-198" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "After the collision the second ball moves at a speed of v= 4.1 Meters/sec\n" + ] + } + ], + "source": [ + " #Example 6_9\n", + " \n", + "import math \n", + " #To find out the velocity of second ball after collision\n", + "v1=5 #units in meters/sec\n", + "theta=50.0 #units in degrees\n", + "v2=2 #units in meters/sec\n", + "vx=v1/(v2*math.cos(theta*math.pi/180)) #units in meters/sec\n", + "vy=-(v2*math.cos(theta*math.pi/180)) #units in meters/sec\n", + "v=math.sqrt(vx**2+vy**2) #units in meters/sec\n", + "print \"After the collision the second ball moves at a speed of v=\",round(v,2),\" Meters/sec\"\n", + " #in textbook the answer is printed wrong as 4.01 meters/sec the correct answer is 4.1 meters/sec\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex6.10:pg-199" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average speed of the nitrogen molecule in air is V= 492.0 meters/sec\n" + ] + } + ], + "source": [ + " #Example 6_10\n", + " \n", + " \n", + " #To find the average speed of the nitrogen molecule in air\n", + "ap=1.01*10**5 #units in Newton/meter**2\n", + "nofmol=2.69*10**25 #Number of molecules\n", + "nitmass=4.65*10**-26 #units in Kg\n", + "v=sqrt((ap*3)/(nofmol*nitmass)) #units in meters/sec\n", + "print \"The average speed of the nitrogen molecule in air is V=\",round(v),\" meters/sec\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter7.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter7.ipynb new file mode 100644 index 00000000..0925e948 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter7.ipynb @@ -0,0 +1,433 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:83169fd272ce864306940c4a49b0d63f096c19015d841f59df75135834323538"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter07: Motion in a Cirlce"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.1:pg-208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_1\n",
+ " \n",
+ " \n",
+ " #To convert angles to radians and revolutions\n",
+ "theta=70.0 #units in degrees\n",
+ "deg=360.0 #units in degrees\n",
+ "rad=theta*2*math.pi/deg #units in radians\n",
+ "rev=1 #units in revolution\n",
+ "rev=theta*rev/deg #units in revolution\n",
+ "print \" 70 degrees in radians is \",round(rad,2),\"radians \\n 70 degrees in revolutions it is \",round(rev,3),\" revolutions\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " 70 degrees in radians is 1.22 radians \n",
+ " 70 degrees in revolutions it is 0.194 revolutions\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.2:pg-209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_2\n",
+ " \n",
+ " \n",
+ "#To find average angular velocity\n",
+ "theta=1800.0 #units in rev\n",
+ "t=60.0 #units in sec\n",
+ "w=(theta/t) #units in rev/sec\n",
+ "w=w*(2*math.pi) #units in rad/sec\n",
+ "print \"Average angular velocity is w=\",round(w),\" rad/sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average angular velocity is w= 188.0 rad/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.3:pg-210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_3\n",
+ " \n",
+ " \n",
+ " #To find average angular acceleration\n",
+ "wf=240.0 #units in rev/sec\n",
+ "w0=0 #units in rev/sec\n",
+ "t=2.0 #units in minutes\n",
+ "t=t*60 #units in sec\n",
+ "alpha=(wf-w0)/t #units in rev/sec**2\n",
+ "print \"Average angular acceleration is alpha=\",round(alpha),\" rev/sec**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average angular acceleration is alpha= 2.0 rev/sec**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.4:pg-212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_4\n",
+ " \n",
+ " \n",
+ "#To find out how many revolutions does it turn before rest\n",
+ "wf=0 #units in rev/sec\n",
+ "w0=3 #units in rev/sec\n",
+ "t=18 #units in sec\n",
+ "alpha=(wf-w0)/t #units in rev/sec**2\n",
+ "theta=(w0*t)+0.5*(alpha*t**2) #units in rev\n",
+ "print \"Number of revolutions does it turn before rest is theta=\",round(theta),\" rev\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of revolutions does it turn before rest is theta= -108.0 rev\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.5:pg-212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_5\n",
+ " \n",
+ " \n",
+ " #To find the angular acceleration and angular velocity of one wheel\n",
+ "vtf=20.0 #units in meters/sec\n",
+ "r=0.4 #units in meters\n",
+ "wf=vtf/r #units in rad/sec\n",
+ "vf=20.0 #units in meters/sec\n",
+ "v0=0 #units in meters/sec**2\n",
+ "t=9.0 #units in sec\n",
+ "a=(vf-v0)/t #units in meters/sec**2\n",
+ "alpha=a/r #units in rad/sec**2\n",
+ "print \"Angular accelertion is a=\",round(a,2),\" meters/sec**2\\n\"\n",
+ "print \"Angular velocity is alpha=\",round(alpha,2),\" rad/sec**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angular accelertion is a= 2.22 meters/sec**2\n",
+ "\n",
+ "Angular velocity is alpha= 5.56 rad/sec**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.6:pg-213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_6\n",
+ " \n",
+ " \n",
+ " #To find out the rotation rate\n",
+ "at=8.6 #units in meters/sec**2\n",
+ "r=0.2 #units in meters\n",
+ "alpha=at/r #units in rad/sec**2\n",
+ "t=3 #units in sec\n",
+ "wf=alpha*t #units in rad/sec\n",
+ "print \"The rotation rate is wf=\",round(wf),\" rad/sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rotation rate is wf= 129.0 rad/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.7:pg-215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_7\n",
+ " \n",
+ " \n",
+ " #To calculate how large a horizontal force must the pavement exert\n",
+ "m=1200.0 #units in Kg\n",
+ "v=8.0 #units in meters/sec\n",
+ "r=9 #units in meters\n",
+ "F=(m*v**2)/r #units in Newtons\n",
+ "print \"The horizontal force must the pavement exerts is F=\",round(F),\" Newtons\"\n",
+ " #In text book the answer is printed wrong as F=8530 N but the correct answer is 8533 N\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The horizontal force must the pavement exerts is F= 8533.0 Newtons\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.9:pg-220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_9\n",
+ " \n",
+ " \n",
+ " #To find out the angle where it should be banked\n",
+ "v=25 #units in meters/sec\n",
+ "r=60 #units in meters\n",
+ "g=9.8 #units in meters/sec**2\n",
+ "tantheta=v**2/(r*g) #units in radians\n",
+ "theta=math.atan(tantheta)*180/math.pi\n",
+ "print \"The angle where it should be banked is theta=\",round(theta),\" degrees\",\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle where it should be banked is theta= 47.0 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.10:pg-220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_10\n",
+ " \n",
+ " \n",
+ " #To find out the ratio of F/W\n",
+ "G=6.67*10**-11 #units in Newton meter**2/Kg**2\n",
+ "m1=0.0080 #units in Kgs\n",
+ "m2=0.0080 #units in Kgs\n",
+ "r=2 #units in Meters\n",
+ "F=(G*m1*m2)/r**2 #units in Newtons\n",
+ "m=m1 #units in Kgs\n",
+ "g=9.8 #units in meter/sec**2\n",
+ "W=m*g #units in Newtons\n",
+ "F_W=F/W\n",
+ "print \"The F/W Ratio is=\",round(F_W,16)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The F/W Ratio is= 1.36e-14\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.11:pg-221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_11\n",
+ " \n",
+ " \n",
+ " #To find the mass of the sun\n",
+ "t=3.15*10**7 #units in sec\n",
+ "r=1.5*10**11 #units in meters\n",
+ "v=(2*math.pi*r)/t #units in meters/sec\n",
+ "G=6.67*10**-11 #units in Newtons\n",
+ "ms=(v**2*r)/G #Units in Kg\n",
+ "print \"The mass of the sun is Ms=\",round(ms,-28),\"Kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of the sun is Ms= 2.01e+30 Kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.12:pg-222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 7_12\n",
+ " \n",
+ " \n",
+ " #To findout the orbital radius and its speed\n",
+ "G=6.67*10**-11 #units in Newtons\n",
+ "me=5.98*10**24 #units in Kg\n",
+ "t=86400.0 #units in sec\n",
+ "r=((G*me*t**2)/(4*math.pi**2))**(1/3.0)\n",
+ "print \"The orbital radius is r= \",round(r),\" meters\\n\"\n",
+ "v=(2*math.pi*r)/t #units in meters/sec\n",
+ "print \"The orbital speed is v=\",round(v),\" meters/sec\"\n",
+ " #in textbook the answer is printed wrong as v=3070 m/sec but the correct answer is v=3073 m/sec\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The orbital radius is r= 42250474.0 meters\n",
+ "\n",
+ "The orbital speed is v= 3073.0 meters/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter8.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter8.ipynb new file mode 100644 index 00000000..b6b6ca79 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter8.ipynb @@ -0,0 +1,299 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 08: Rotational work energy and momentum" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex8.1:pg-240" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [ + " #Example 8_1\n", + " \n", + " \n", + " #To find the rotational kinetic energy\n", + "m=5.98*10**24 #units in Kg\n", + "r=6.37*10**6 #units in meters\n", + "I=(2/5)*m*r**2 #units in Kg meter**2\n", + "t=86400 #units in sec\n", + "w=(2*math.pi)/(t) #units in rad/sec\n", + "KE=0.5*(I*w**2) #units in joules\n", + "print \"The rotational kinetic energy is KE=\")\n", + "print KE)\n", + "print \"Joules\")\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex8.2:pg-242" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angular acceleration is alpha= 0.384 rad/sec**2\n" + ] + } + ], + "source": [ + " #Example 8_2\n", + " \n", + " \n", + " #To find the angular acceleration of the wheel\n", + "m=30 #units in Kg\n", + "k=0.25 #units in meters\n", + "I=m*k**2 #units in Kg meter**2\n", + "force=1.8 #units in Newtons\n", + "levelarm=0.40 #nits in meters\n", + "tou=force*levelarm #units in Newton meter\n", + "alpha=tou/I #units in rad/sec**2\n", + "print \"Angular acceleration is alpha=\",round(alpha,3),\" rad/sec**2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex8.3:pg-242" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time taken is t= 15.7 sec\n", + "\n", + "The wheel goes a distance of theta= 98.7 rad\n", + "\n", + "The rotational kinetic energy is KE= 197.0 Joules\n" + ] + } + ], + "source": [ + " #Example 8_3\n", + " \n", + " \n", + " #To find out how long does it take to accelerate and how far does wheel turn in this time and the rotational kinetic energy\n", + "force=8 #units in Newtons\n", + "arm=0.25 #units in meters\n", + "tou=force*arm #units in Newton meter\n", + "m=80 #units in Kg\n", + "b=arm #units in meters\n", + "I=0.5*m*b**2 #units in Kg meter**2\n", + "alpha=tou/I #units in rad/sec**2\n", + "wf=4*math.pi #units in rad/sec\n", + "w0=0 #units in rad/sec\n", + "t=(wf-w0)/alpha #units in sec\n", + "print \"The time taken is t=\",round(t,1),\" sec\\n\"\n", + "theta=0.5*(wf+w0)*t #units in radians\n", + "print \"The wheel goes a distance of theta=\",round(theta,1),\" rad\\n\"\n", + "KE=0.5*I*wf**2 #units in Joules\n", + "print \"The rotational kinetic energy is KE=\",round(KE),\" Joules\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex8.4:pg-243" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angular acceleration is alpha= 1.37 rad/sec**2\n", + "\n", + "The objects goes a distance of y= 51.4 meters\n" + ] + } + ], + "source": [ + " #Example 8_4\n", + " \n", + " \n", + " #To find out the angular acceleration and the distance the object falls\n", + "f1=29.4 #units in Newtons\n", + "r1=0.75 #units in meters\n", + "m1=40 #units in Kgs\n", + "r2=0.6 #units in meters\n", + "m2=3 #units in Kgs\n", + "alpha=(f1*r1)/((m1*r2**2)+(m2*r1**2)) #units in rad/sec**2\n", + "print \"The angular acceleration is alpha=\",round(alpha,2),\" rad/sec**2\\n\"\n", + "a=r1*alpha #units in meters/sec**2\n", + "t=10 #units in sec\n", + "y=0.5*a*t**2 #units in meters\n", + "print \"The objects goes a distance of y=\",round(y,1),\" meters\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex8.5:pg-244" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The object is moving at v= 1.28 meters/sec\n" + ] + } + ], + "source": [ + " #Example 8_5\n", + " \n", + " \n", + " #To find the speed of the object\n", + "m=3 #units in Kg\n", + "g=9.8 #units in meters/sec**2\n", + "h=0.80 #units in meters\n", + "m1=3 #units in Kg\n", + "m2=14.4 #units in Kg\n", + "r=0.75 #units in meters\n", + "v=sqrt((m*g*h)/((0.5*m1)+((0.5*m2)/r**2)))\n", + "print \"The object is moving at v=\",round(v,2),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex8.8:pg-247" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sun would take for one revolution in time=\n", + "0.000216 sec\n" + ] + } + ], + "source": [ + " #Example 8_8\n", + " \n", + " \n", + " #To find out how long does the sun take to complete one revolution\n", + "ra_rb=10.0**5\n", + "noofrev=1.0/25 #units in rev/day\n", + "wafter=(ra_rb)**2*(noofrev)\n", + "t=86400 #units in sec\n", + "time=t/wafter #units in sec\n", + "print \"The sun would take for one revolution in time=\"\n", + "print time,\"sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex8.9:pg-248" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The rotational speed is Wf= 1.63 rev/sec\n" + ] + } + ], + "source": [ + " #Example 8_9\n", + " \n", + " \n", + " #To find out the rotational speed \n", + "m=0.3 #units in Kg\n", + "r=0.035 #units in meters\n", + "Iw=0.5*m*r**2 #units in Kg meter**2\n", + "Ibt=8*10**-4 #units in Kg meter**2\n", + "w0=2 #units in rev/sec\n", + "wf=(Ibt*w0)/(Ibt+Iw) #units in rev/sec\n", + "print \"The rotational speed is Wf=\",round(wf,2),\" rev/sec\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb new file mode 100644 index 00000000..2cc88216 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb @@ -0,0 +1,291 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:233302014269b24da4024327815295a69860454a17ca20816030528c9e7cb7d4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 09: Mechanical properties of Matter"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.1:pg-269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 9_1\n",
+ " \n",
+ " \n",
+ " #To find its mass and how large a cube of ice has the same mass\n",
+ "pu=18680.0 #units in Kg/meter**3\n",
+ "s=2*10**-2 #units in meters\n",
+ "vu=s**3.0 #units in meter**3\n",
+ "mu=pu*vu #units in Kg\n",
+ "print \"Mass Mu=\",round(mu,3),\" Kg\\n\"\n",
+ "pi=920 #units in Kg/meter**3\n",
+ "vi=mu/pi #units in meter**3\n",
+ "ss=vi**(1/3.0)*10**2 #units in cm\n",
+ "print \"Side length of ice cube is=\",round(ss,2),\"cm\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass Mu= 0.149 Kg\n",
+ "\n",
+ "Side length of ice cube is= 5.46 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.2:pg-269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 9_2\n",
+ " \n",
+ " \n",
+ " #To calculate the cross sectional area and how far the ball will stretch the wire\n",
+ "m=40.0 #units in Kg\n",
+ "g=9.8 #units in meter/sec**2\n",
+ "F=m*g #units in Kg meter/sec**2\n",
+ "stress=0.48*10**8 #units in Newton/meter**2\n",
+ "A=F/stress #units in meter**2\n",
+ "r=sqrt(A/math.pi)*10.0**3 #units in mm\n",
+ "print \"The radius of the wire should be r=\",round(r,1),\" mm and the cross sectional area is A=\",round(A),\"meter**2\"\n",
+ "y=200.0*10**9 #units in Newton/meter**2\n",
+ "strain=stress/y\n",
+ "L0=15 #units in meters\n",
+ "deltaL=strain*L0 #units in meters\n",
+ "deltaL=deltaL*10**3 #units in mm\n",
+ "print \"\\nThe ball stretches the wire a distance of deltaL=\",round(deltaL,2),\"mm\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of the wire should be r= 1.6 mm and the cross sectional area is A= 0.0 meter**2\n",
+ "\n",
+ "The ball stretches the wire a distance of deltaL= 3.6 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.7:pg-273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 9_7\n",
+ " \n",
+ " \n",
+ " #To find out by what factor the blood flow in an artery is reduced\n",
+ "r1_r2=1/2.0 #The ratio by which the radius is altered in arterys\n",
+ "R1_R2=1/r1_r2**4 #Ratio by which flow is altered\n",
+ "print \"The flow rate is reduced by a factor of \",round(R1_R2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The flow rate is reduced by a factor of 16.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.9:pg-274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 9_9\n",
+ " \n",
+ " \n",
+ " #To compare the pressures at A and at B\n",
+ "p=1000 #Units in Kg/Meter**3\n",
+ "va=0.2 #units in meters/sec\n",
+ "vb=2 #units in meters/sec\n",
+ "Pa_Pb=-0.5*p*(va**2-vb**2) #units in Pa\n",
+ "print \"Pressure Difference at A and B is Pa-Pb=\",round(Pa_Pb),\" Pa therefore Preasure at A is High than at B\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure Difference at A and B is Pa-Pb= 1980.0 Pa therefore Preasure at A is High than at B\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.10:pg-276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 9_10\n",
+ " \n",
+ " \n",
+ " #To find out how fast a raindrop becomes turbulent\n",
+ "Nr=10 #Number of molecules\n",
+ "n=1.9*10**-5 #Units in PI\n",
+ "p=1.29 #Units in Kg/Meter**3\n",
+ "d=3*10**-3 #Units in meters\n",
+ "vc=(Nr*n)/(p*d) #units in meters/sec\n",
+ "print \"The speed of the rain drop is Vc=\",round(vc,3),\" meters/sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The speed of the rain drop is Vc= 0.049 meters/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.11:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 9_11\n",
+ " \n",
+ " \n",
+ " #To find out what horsepower is required\n",
+ "p=1.29 #Units in Kg/Meter**3\n",
+ "Cd=0.45\n",
+ "af=2 #Units in Meter**2\n",
+ "v=20 #Units in meters/sec\n",
+ "M=1000 #units in Kg\n",
+ "F=(0.5*p*Cd*af*v**2)+((M/1000)*((110+(1.1*v)))) #Units in Newtons\n",
+ "Power=F*v #Units in Watts\n",
+ "Power=Power/747.3061 #units in Horse Power\n",
+ "reqHPower=Power**2 #unis in Horse power\n",
+ "print \"The required power is=\",round(reqHPower),\" hp\"\n",
+ " #In text book the answer is printed wrong as 80 Hp the correct answer is 95Hp\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required power is= 95.0 hp\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.12:pg-278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " #Example 9_12\n",
+ " \n",
+ " \n",
+ " #To find out the sedimentation rate of sphrical particles\n",
+ "b=2*10**-3 #units in cm\n",
+ "g=9.8 #Units in meters/sec**2\n",
+ "n=1 #units in m PI\n",
+ "Pp_Pt=1050 #units in Kg/Meter**3\n",
+ "vt=(((2*b**2*g)/(9*n))/(2*Pp_Pt))*10**6 #units in cm/sec\n",
+ "print \"Sedimentation is vt=\",round(vt,4),\"cm/sec\"\n",
+ " #in text book answer is printed wrong as vt=4.36*10**-3 cm/sec but the correct answer is vt=4.14*10**-3 cm/sec\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Sedimentation is vt= 0.0041 cm/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Principles_of_Physics_by_F.J.Bueche/chapter12.ipynb b/Principles_of_Physics_by_F.J.Bueche/chapter12.ipynb new file mode 100644 index 00000000..deed5b2f --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/chapter12.ipynb @@ -0,0 +1,245 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12:Thermodynamics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex12.1:pg-413" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The work done by the gas is= -220.0 J\n" + ] + } + ], + "source": [ + " #Example 12_1\n", + " \n", + " \n", + " #To find the work done by the gas\n", + "d1=800 #Units in meter**3\n", + "d2=500 #Units in meter**3\n", + "p1=5*10**5 #Units in Pa\n", + "w1=p1*(d1-d2)*10**-6 #Units in J\n", + "p2=2*10**5 #Units in Pa\n", + "d3=200*10**-6 #Units in meter**3\n", + "p3=3*10**5 #Units in Pa\n", + "w2=(p2*d3)+(0.5*p3*d3) #Units in J\n", + "print \"The work done by the gas is=\",round(-(w1+w2)),\" J\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex12.2:pg-415" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The estimated Cv value of nitric acid is Cv= 693.0 J/Kg K\n" + ] + } + ], + "source": [ + " #Example 12_2\n", + " \n", + " \n", + " #To estimate the Cv of nitric acid\n", + "r=8314 #Units in J/Kmol K\n", + "m=30 #Units in Kg/Kmol\n", + "Cv=2.5*(r/m) #Units in J/Kg K\n", + "print \"The estimated Cv value of nitric acid is Cv=\",round(Cv),\" J/Kg K\"\n", + " #in textbook the answer is printed wrong as Cv=690 J/Kg K correct answer is 692 J/Kg K\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex12.3:pg-415" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The final temperature is T2= 886.0 K\n" + ] + } + ], + "source": [ + " #Example 12_3\n", + " \n", + "import math \n", + " #To find the final temperature\n", + "t1=27 #units in Centigrade\n", + "t1=t1+273 #Units in K\n", + "gama=1.4 #Units in Constant\n", + "p1=1 #units in Pa\n", + "v1_v2=15 #Units of in ratio\n", + "logT2=math.log10(t1)-((gama-1)*(math.log10(p1)-math.log10(v1_v2)))\n", + "T2=10**logT2 #Uniys in K\n", + "print \"The final temperature is T2=\",round(T2),\" K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex12.4:pg-417## " + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "This type of process is termed as throttling process and described by the equation Delta U=- Delta W\n", + "\n", + "Where Delta W is the work done\n" + ] + } + ], + "source": [ + " #Example 12_4\n", + " \n", + " \n", + " #To describe the Temperature changes of the gas\n", + "print \"This type of process is termed as throttling process and described by the equation Delta U=- Delta W\\n\"\n", + "print \"Where Delta W is the work done\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex12.5:pg-418## " + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The entropy is Delta S= 24.5 J/K\n" + ] + } + ], + "source": [ + " #Example 12_5\n", + " \n", + " \n", + " #To findout by how much the entropy of the system changes\n", + "m=20 #Units in gm\n", + "alpha=80 #Units in cal/gm\n", + "t=4.184 #Units in J/Cal\n", + "Q=m*alpha*t #Units in J\n", + "T=273 #Units in K\n", + "S=Q/T #Units in J/K\n", + "print \"The entropy is Delta S=\",round(S,1),\" J/K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex12.6:pg-419## " + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amount of Electricity is required is Delta W= 11666.0 J\n" + ] + } + ], + "source": [ + " #Example 12_6\n", + " \n", + " \n", + " #To findout how much electricity is needed\n", + "Tc=278 #Units in K\n", + "Th=293 #Units in K\n", + "COP=Tc/(Th-Tc) #Units in ratio\n", + "Qc=210000 #Units in J\n", + "W=Qc/COP #Units in J\n", + "print \"The amount of Electricity is required is Delta W=\",round(W),\" J\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Physics_by_F.J.Bueche/screenshots/11.2.png b/Principles_of_Physics_by_F.J.Bueche/screenshots/11.2.png Binary files differnew file mode 100644 index 00000000..cd276d33 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/screenshots/11.2.png diff --git a/Principles_of_Physics_by_F.J.Bueche/screenshots/24.4.png b/Principles_of_Physics_by_F.J.Bueche/screenshots/24.4.png Binary files differnew file mode 100644 index 00000000..e0ed1a69 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/screenshots/24.4.png diff --git a/Principles_of_Physics_by_F.J.Bueche/screenshots/8.4.png b/Principles_of_Physics_by_F.J.Bueche/screenshots/8.4.png Binary files differnew file mode 100644 index 00000000..29ad7ee3 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/screenshots/8.4.png 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