{ "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": {} } ] }