#C = (5/9)*(F-32);
#F = 32+(9*C/5);
#Letting C = F in equation;
#F = (5/9)*(F-32);
#Therefore
F = -160/4; #fahrenheit
# Results
print "Both fahrenheit and celsius temperature scales indicate same temperature at %.2f"%(F);
#Given
Mm = 0.0123 #Unit:lb #Mass of the moon;
Me = 1 #Unit:lb #Mass of the earth;
Dm = 0.273 #Unit:feet #Diameter of the moon;
De = 1. #Unit:feet #Diameter of the earth;
Rm = Dm/2; #Radius of the moon; #Unit:feet
Re = De/2; #Radius of the earth; #Unit:feet
#F = (K*M1*M2)/d**2 #Law of universal gravitation;
#Fe = (K*Me*m)/Re**2; #Fe = Force exerted on the mass;
#Fm = (K*Mm*m)/Rm**2; #Fm = Force exerted on the moon;
F = (Me/Mm)*(Rm/Re)**2; #F = Fe/Fm;
print "Relation of force exerted on earth to mass is"
print "Fe/Fm = %.2f"%F
#Given
M = 5; #Unit:kg #mass of body;
g = 9.81; #Unit:m/s**2 #the local acceleration of gravity
# Calculations
W = M*g; #W = the weight of the body #Unit:Newton # 1 N = 1 kg*m/s**2
# Results
print "The weight of the body is %.2f N"%(W);
print "Solution for a";
#given
M = 10. #Unit:kg #mass of body;
g = 9.5 #Unit:m/s**2 #the local acceleration of gravity
W = M*g; #W = the weight of the body; #Unit:Newton # 1 N = 1 kg*m/s**2
print "The weight of the body is %.2f N"%(W);
print "Solution for b";
#Given
F = 10; #Unit:Newton #Horizontal Force
a = F/M; # #newton's second law of motion
print "The horizontal acceleration of the body is %.2f m/s**2"%(a);
#Conversion Problem
# 1 inch = 0.0254 meter so, 1 = 0.0254 meter/inch #Eq.1
# 1 ft = 12 inch so, 1 = 12 inch/ft......... #Eq.2
#Multiplying Eq.1 & Eq.2 # We get 1 = 0.0254*12 meter/ft
#Taking Square both side
#1**2 = (0.0254*12)**2 meter**2/ft**2
print "1 ft**2 = %.2f meter**2"%((0.0254*12)**2);
#The Specific gravity of mercury is 13.6 #Given
#Converting the unit of weight of grams per cubic centimeter to pounds per cubic foot
# 1 lbf = 454 gram #1 inch = 2.54 cm
#So 1 gram = 1/454 lbf and 1 ft = 12*2.54 cm
#Gamma = (gram/cm**3)*(lb/gram)*(cm**3/ft**3) = lb/ft**3
#Gamma = (1 gram/cm**3)*(1 lbf/454 gram)*(2.54*12)**3 *cm**3/ft**3
Gamma = (1./454)*(2.54*12)**3; #lbf/ft**3 #conversion factor
print "Conversion Factor = %.2f"%Gamma
p = (1./12)*(Gamma*13.6); #lbf/ft**2 #gage pressure
p = (1./12)*Gamma*13.6*(1./144) #ft**2/inch**2 #gage pressure
print "Guage Pressure is %.2f psi"%(p);
print "Local atmospheric pressure is 14.7 psia";
P = p+14.7; #Pressure on the base of the column #Unit:psia
print " So Pressure on the base of the column is %.2f psia"%(P);
#Given
Rho = 13.595; #Unit: kg/m**3 #The density of mercury
h = 25.4; #Unit: mm #Height of column of mercury
g = 9.806; #Unit:m/s**2 #the local acceleration of gravity
#Solution
p = Rho*g*h; #P = Pressure at the base of a column of mercury #Unit:Pa
# Results
print "Pressure at the base of a column of mercury is %.2f Pa"%(p);
#Given
Patm = 30.0; #in. #pressure of mercury at smath.radians(numpy.arcmath.tan(ard temperature
Vacuum = 26.5; #in. #vaccum pressure
Pabs = Patm-Vacuum; #Absolute pressure of mercury #in.
# 1 inch mercury exerts a pressure of 0.491 psi
p = Pabs*0.491; #Absolute pressure in psia
# Results
print "Absolute pressure of mercury in is %.2f psia"%(p);
#Given
Rho = 2000; #Unit: kg/m**3 #The density of fluid
h = -10; #Unit: mm #Height of column of fluid #the height is negative because it is measured up from the base
g = 9.6 #Unit:m/s**2 #the local acceleration of gravity
#Solution
p = -Rho*g*h; #P = Pressure at the base of a column of fluid #Unit:Pa
# Results
print "Pressure at the base of a column of fluid is %.2f Pa"%(p);
#Given
Patm = 30.0 #in. #pressure of mercury at smath.radians(numpy.arcmath.tan(ard temperature
Vacuum = 26.5 #in. #vaccum pressure
Pabs = Patm-Vacuum; #Absolute pressure of mercury #in.
# (3.5 inch* (ft/12 inch) * (13.6*62.4)LBf/ft**3 * kg/2.2 LBf * 9.806 N/kg)/((12 inch**2/ft**2) * (0.0254 m/inch)**2)
p = (3.5*(1./12)*13.6*62.4*(1/2.2)*9.806)/(12**2*0.0254**2*1000); #kPa #Absolute pressure in psia
# Results
print "Absolute pressure of mercury is %.2f kPa"%(p)