Answer:
a) 6498.84 kW
b) 0.51
c) 0.379
Explanation:
See the attached picture below for the solution
The ideal mechanical advantage (IMA) can be determined by the following equation:
IMA= Input distance/Output distance
The Input distance and Output distance are:
Input distance=220 meters
Output distance=110 meters
When you substitute in the equation of the ideal mechanical advantage (IMA), you obtain:
IMA= Input distance/Output distance
IMA= 220 meters/110 meters
IMA=2
Answer:
change in internal energy 3.62*10^5 J kg^{-1}
change in enthalapy 5.07*10^5 J kg^{-1}
change in entropy 382.79 J kg^{-1} K^{-1}
Explanation:
adiabatic constant 
specific heat is given as 
gas constant =287 J⋅kg−1⋅K−1

specific heat at constant volume

change in internal energy 

change in enthalapy 

change in entropy


