import math
#example 13.1
#calculation of the force exerted by the water on the bottom
#given data
h=20.*10.**-2.#height(in m) of the flask
r=10.*10.**-2.#radius(in m) of the bottom of the flask
P0=1.01*10.**5.#atmospheric pressure(in Pa)
rho=1000.#density of water(in kg/m**3)
g=10.#gravitational acceleration(in m/s**2) of the earth
#calculation
P=P0+(h*rho*g)#pressure at the bottom
A=math.pi*r**2.#area of the bottom
F=P*A#force on the bottom
print '%s %.2f %s' %('the force exerted by the water on the bottom is',F,'N\n')
import math
#example 13.2
#calculation of the volume of the cube outside the water
#given data
m=700.*10.**-3.#mass(in kg) of the cube
l=10.*10.**-2.#length(in m) of the cube
rho=1000.#density of water(in kg/m**3)
#calculation
V=m/rho#weight of displaced water = V*rho*g
Vtotal=l**3.#total volume of the cube
Vout=Vtotal-V#volume of the cube outside the water
print '%s %.2f %s' %('the volume of the cube outside the water is',Vout*10**6,'cm**3')
#example 13.3
#calculation of the speed of the outgoing liquid
#given data
A1=1.*10.**-4.#area(in m**2) of the inlet of the tube
A2=20.*10.**-6.#area(in m**2) of the outlet of the tube
v1=2.#speed(in cm/s) of the ingoing liquid
#calculation
v2=A1*v1/A2#equation of continuity
print '%s %.2f %s' %('the speed of the outgoing liquid is',v2,'cm/s\n')
#example 13.4
#calculation of the difference in the pressures at A and B point
#given data
A1=1.*10.**-4.#area(in m**2) at point A of the tube
A2=20.*10.**-6.#area(in m**2) at point B of the tube
v1=10.*10.**-2.#speed(in m/s) of the ingoing liquid
rho=1200.#density of the liquid(in kg/m**3)
#calculation
v2=A1*v1/A2#equation of continuity
#by Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)
deltaP=(1./2.)*rho*(v2**2.-v1**2.)
print '%s %.2f %s' %('the difference in the pressures at A and B point is',deltaP,'Pa\n')
import math
#example 13.5
#calculation of the speed of the water coming out of the tap
#given data
h=6.#depth(in m) of the tap
g=9.8#gravitational acceleration(in m/s**2) of the earth
#calculation
v=math.sqrt(2.*g*h)#torricellis theorem
print '%s %.2f %s' %('the speed of the water coming out of the tap is',round(v),'m/s\n')
import math
#example 13.1w
#calculation of the force exerted by the mercury on the bottom of the beaker
#given data
h=10.*10.**-2.#height(in m) of the mercury
r=4.*10.**-2.#radius(in m) of the beaker
g=10.#gravitational acceleration(in m/s**2) of the earth
P0=1.*10.**5.#atmospheric pressure(in Pa)
rho=13600.#density of mercury(in kg/m**3)
#calculation
P=P0+(h*rho*g)#pressure at the bottom
A=math.pi*r**2.#area of the bottom
F=P*A#force on the bottom
print '%s %.2f %s' %('the force exerted by the mercury on the bottom of the beaker is',F,'N\n')
import math
#example 13.2w
#calculation of the height of the atmosphere to exert the same pressure as at the surface of the earth
#given data
P0=1.*10.**5.#atmospheric pressure(in Pa)
rho=1.3#density of air(in kg/m**3)
g=9.8#gravitational acceleration(in m/s**2) of the earth
#calculation
h=P0/(g*rho)
print '%s %.2f %s' %("the height of the atmosphere to exert the same pressure as at the surface of the earth is",round(h),"m\n")
#example 13.3w
#calculation of the height of the water coloumn
#given data
h1=2.*10.**-2.#difference in the height(in m)
s=13.6#specific gravity of mercury
#calculation
#P = P0 + (h*rho*g)........using this equation
h=h1*s#height of the water coloumn
print '%s %.2f %s' %('the height of the water coloumn is',h*10**2,'cm\n',)
#example 13.5w
#calculation of the force applied on the water in the thicker arm
#given data
A1=1.*10.**-4.#area(in m**2) of arm 1
A2=10.*10.**-4.#area(in m**2) of arm 2
f=5.#force(in N) applied on the water in the thinner arm
#calculation
#P = P0 + (h*rho*g)........using this equation
F=f*A2/A1#force applied on the water in the thicker arm
print '%s %.2f %s' %('the force applied on the water in the thicker arm is',F,'N\n')
#example 13.6w
#calculation of the elongation of the spring
#given data
m=10.*10.**-3.#mass(in kg) of the copper piece
l=1.*10.**-2.#elongation(in m) in the spring
g=10.#gravitational acceleration(in m/s**2) of the earth
rho=9000.#density of copper(in kg/m**3)
rho0=1000.#density of water(in kg/m**3)
#calculation
k=m*g/l#spring constant
V=m/rho#volume of copper
Fb=V*rho0*g#force of buoyancy
x=((k*l)-Fb)/k#elongation of the spring
print '%s %.2f %s' %('the elongation of the spring is',x*10**2,'cm\n')
#example 13.7w
#calculation of the maximum weight that can be put on the block without wetting it
#given data
l=3.*10.**-2.#length(in m) of the edge of the cubical block
rho=800.#density of wood(in kg/m**3)
k=50.#spring constant(in N/m)
g=10.#gravitational acceleration(in m/s**2) of the earth
rho0=1000.#density of water(in kg/m**3)
#calculation
s=rho/rho0#specific gravity
hin=l*s#height inside water
hout=l-hin#height outside water
V=l**3.#volume of the block
Fb=V*rho0*g#force of buoyancy
Fs=k*hout#force exerted by the spring
Wdash=V*rho*g#weight of the block
W=Fb+Fs-Wdash#maximum weight
print '%s %.2f %s' %('the maximum weight that can be put on the block without wetting it is',W,'N\n')
import math
from math import acos,sqrt
#example 13.8w
#calculation of the angle that the plank makes with the vertical in equilibrium
#given data
l=1.#length(in m) of the planck
h=0.5#height(in m) of the water level in the tank
s=0.5#specific gravity of the planck
#calculation
#A = OC/2 = l/(2*cosd(theta)
# mg = 2*l*rho*g
#buoyant force Fb=(2*l*rho*g)/cosd(theta)
#m*g*(OB)*sind(theta) = F(OA)*sind(theta)
theta=acos(math.sqrt(1./2.))*57.3
print '%s %.2f %s' %('the angle that the plank makes with the vertical in equilibrium is',theta,'degree\n')
import math
#example 13.10w
#calculation of the rate of water flow through the tube
#given data
A1=30.#area(in cm**2) of the tube at point A
A2=15.#area(in cm**2) of the tube at point B
deltaP=600.#change in pressure(in N/m**2)
rho0=1000.#density of the water(in kg/m**3)
#calculation
r=A1/A2#ratio of area
#from equation of continuity vB/vA = A1/A2 = r = 2
#by Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)
#take vB = vA*2
vA=math.sqrt(deltaP*(r/(r+1.))*(1./rho0))
Rflow=vA*A1#rate of water flow
print '%s %.2f %s' %('the rate of water flow through the tube is',Rflow*10**2,'cm**3/s\n')
import math
#example 13.11w
#calculation of the velocity of the water coming out of the opening
#given data
AA=.5#area(in m**2) of the tank
AB=1.*10.**-4.#area(in m**2) of the cross section at the bottom
m=20.#mass(in kg) of the load
h=50.*10.**-2.#height(in m)of the water level
g=10.#gravitational acceleration(in m/s**2) of the earth
rho=1000.#density of the water(in kg/m**3)
#calculation
#from the equation............P = P0 + (h*rho*g)#pressure at the bottom
r=m*g/AA#in above equation it is the value of (h*rho*g)
#on solving,we get............PA = P0 + (400 N/m**2)
#from Bernoulli equtation.....P1 + (rho*g*h1) + (rho*v1**2/2) = P2 + (rho*g*h2) + (rho*v2**2/2)
#we get
vB=math.sqrt((2.*(r+(rho*g*h)))/rho)
print '%s %.2f %s' %('the velocity of the water coming out of the opening is',vB,'m/s\n')