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