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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 09: Mechanical properties of Matter"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex9.1:pg-269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Mass Mu= 0.149 Kg\n",
+ "\n",
+ "Side length of ice cube is= 5.46 cm\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example9_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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex9.2:pg-269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "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"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example9_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=math.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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex9.7:pg-273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The flow rate is reduced by a factor of 16.0\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example9_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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex9.9:pg-274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Pressure Difference at A and B is Pa-Pb= 1980.0 Pa therefore Preasure at A is High than at B\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example9_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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex9.10:pg-276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The speed of the rain drop is Vc= 0.049 meters/sec\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example9_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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex9.11:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The required power is= 95.0 hp\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example9_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"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex9.12:pg-278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Sedimentation is vt= 0.0041 cm/sec\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math #Example9_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"
+ ]
+ }
+ ],
+ "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
+}
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