# Chapter 09: Mechanical properties of Matter¶

## Ex9.1:pg-269¶

In [2]:
  import math   #Example9_1

#To find its mass and how large a cube of ice has the same mass
pu=18680.0     #units in Kg/meter**3
s=2*10**-2     #units in meters
vu=s**3.0     #units in meter**3
mu=pu*vu      #units in Kg
print "Mass Mu=",round(mu,3)," Kg\n"
pi=920     #units in Kg/meter**3
vi=mu/pi     #units in meter**3
ss=vi**(1/3.0)*10**2     #units in cm
print "Side length of ice cube is=",round(ss,2),"cm"

Mass Mu= 0.149  Kg

Side length of ice cube is= 5.46 cm


## Ex9.2:pg-269¶

In [4]:
  import math   #Example9_2

#To calculate the cross sectional area and how far the ball will stretch the wire
m=40.0     #units in Kg
g=9.8     #units in meter/sec**2
F=m*g     #units in Kg meter/sec**2
stress=0.48*10**8     #units in Newton/meter**2
A=F/stress     #units in meter**2
r=math.sqrt(A/math.pi)*10.0**3     #units in mm
print "The radius of the wire should be r=",round(r,1)," mm and the cross sectional area is A=",round(A),"meter**2"
y=200.0*10**9      #units in Newton/meter**2
strain=stress/y
L0=15     #units in meters
deltaL=strain*L0     #units in meters
deltaL=deltaL*10**3      #units in mm
print "\nThe ball stretches the wire a distance of deltaL=",round(deltaL,2),"mm"

The radius of the wire should be r= 1.6  mm and the cross sectional area is A= 0.0 meter**2

The ball stretches the wire a distance of deltaL= 3.6 mm


## Ex9.7:pg-273¶

In [5]:
  import math   #Example9_7

#To find out by what factor the blood flow in an artery is reduced
r1_r2=1/2.0    #The ratio by which the radius is altered in arterys
R1_R2=1/r1_r2**4           #Ratio by which flow is altered
print "The flow rate is reduced by a factor of ",round(R1_R2)

The flow rate is reduced by a factor of  16.0


## Ex9.9:pg-274¶

In [6]:
  import math   #Example9_9

#To compare the pressures at A and at B
p=1000      #Units in Kg/Meter**3
va=0.2     #units in meters/sec
vb=2     #units in meters/sec
Pa_Pb=-0.5*p*(va**2-vb**2)      #units in Pa
print "Pressure Difference at A and B is Pa-Pb=",round(Pa_Pb)," Pa therefore Preasure at A is High than at B"

Pressure Difference at A and B is Pa-Pb= 1980.0  Pa therefore Preasure at A is High than at B


## Ex9.10:pg-276¶

In [7]:
  import math   #Example9_10

#To  find out how fast a raindrop becomes turbulent
Nr=10     #Number of molecules
n=1.9*10**-5     #Units in PI
p=1.29     #Units in Kg/Meter**3
d=3*10**-3     #Units in meters
vc=(Nr*n)/(p*d)       #units in meters/sec
print "The speed of the rain drop is Vc=",round(vc,3)," meters/sec"

The speed of the rain drop is Vc= 0.049  meters/sec


## Ex9.11:pg-277¶

In [8]:
  import math   #Example9_11

#To find out what horsepower is required
p=1.29    #Units in Kg/Meter**3
Cd=0.45
af=2     #Units in Meter**2
v=20     #Units in meters/sec
M=1000     #units in Kg
F=(0.5*p*Cd*af*v**2)+((M/1000)*((110+(1.1*v))))     #Units in Newtons
Power=F*v     #Units in Watts
Power=Power/747.3061       #units in Horse Power
reqHPower=Power**2     #unis in Horse power
print "The required power is=",round(reqHPower)," hp"
#In text book the answer is printed wrong as 80 Hp the correct answer is 95Hp

The required power is= 95.0  hp


## Ex9.12:pg-278¶

In [9]:
  import math   #Example9_12

#To find out the sedimentation rate of sphrical particles
b=2*10**-3     #units in cm
g=9.8      #Units in meters/sec**2
n=1     #units in m PI
Pp_Pt=1050     #units in Kg/Meter**3
vt=(((2*b**2*g)/(9*n))/(2*Pp_Pt))*10**6        #units in cm/sec
print "Sedimentation is vt=",round(vt,4),"cm/sec"
#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

Sedimentation is vt= 0.0041 cm/sec