Answer:
Part a)
P = 13.93 kW
Part b)
R = 8357.6 Cents
Explanation:
Part A)
heat required to melt the aluminium is given by

here we have



Since this is the amount of aluminium per hour
so power required to melt is given by



Since the efficiency is 85% so actual power required will be

Part B)
Total energy consumed by the furnace for 30 hours



now the total cost of energy consumption is given as



The energy of a photon is given by

where

is the Planck constant
f is the frequency of the photon
In our problem, the frequency of the light is

therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
Answer:
The value of the centripetal forces are same.
Explanation:
Given:
The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.
The expression for centripetal force is given by,

where,
is the mass of the object,
is the velocity of the object and
is the radius of the path.
The value of the centripetal force depends on the mass of the object, not on its weight.
As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.
Answer:
Final mass=0.89kg
Final pressure=5.6bar
Explanation:
To find mass,m=v/v1
But v1=vf + x(vg-vf)
Vf= 0.001093m^3/kg
Vg= 0.3748m^3/kg
V1= 0.001093+0.5(0.3748-0.001093)
V1= 0.225m^3/kg
M= 0.20/0.225 =0.89kg
Final pressure will be:
V/V1= P/P1
Cross multiply
VP1=V1P
P1= 0.225×5/0.2
P1=:5.6 bar
You are currently converting Distance and Length units from Centimeters to Feet 321 Centimeters (cm) = 10.5315 Feet (ft) This is a hard one but see if this helps if not let me now and i can try again..