Answer
Given,
Energy absorbed, 
Energy expels,
Temperature of cold reservoir, T = 27°C
a) Efficiency of engine



b) Work done by the engine



c) Power output
t = 0.296 s



Answer:
R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi
If you do it in steps
R = 9880 yd * 3 ft/yd = 29640 ft
R = 29640 ft / 5280 ft/mi = 5.61 mi
Time = (distance) / (speed)
Time = (4.12x10^16 m) / (3 x10^8 m/s)
Time = 1.37 x 10^8 seconds
Divide the seconds by 86,400 to get days. Then divide the days by 365 to get years.
Time = about 4.35 years
To find a general equilibrium point for a spring based on the hook law, it is possible to start from the following premise:
Hook's law is given by:

Where,
k= Spring Constant
Change in Length
F = Force
When there is a Mass we have two force acting at the System:
W= mg
Where W is the force product of the weigth. Then the force net can be defined as,

But we have a system in equilibrium, so

We find the equilibrium for any location when
