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"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
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"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"
]
}
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