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| author | kinitrupti | 2017-05-12 18:40:35 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
| commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
| tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb | |
| parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
| download | Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.gz Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.bz2 Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.zip | |
Revised list of TBCs
Diffstat (limited to 'Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb')
| -rwxr-xr-x | Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb | 291 |
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diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb deleted file mode 100755 index 2cc88216..00000000 --- a/Principles_of_Physics_by_F.J.Bueche/Chapter9.ipynb +++ /dev/null @@ -1,291 +0,0 @@ -{
- "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": {}
- }
- ]
-}
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