Answer:

Work done = = 5 kJ
Explanation:
Given data:
volume of nitrogen 



Polytropic exponent n = 1.4
![\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7BT_1%7D%20%3D%20%5B%5Cfrac%7BP_2%7D%7BP_1%7D%5D%5E%7B%5Cfrac%7Bn-1%7D%7Bn%7D)
putting all value
![\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7B473%7D%20%3D%20%5B%5Cfrac%7B80%7D%7B150%7D%5D%5E%7B%5Cfrac%7B1.4-1%7D%7B1.4%7D)

polytropic process is given as



work done 

= 5 kJ
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Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
Explanation:
Given that,
The mass of rock, m = 2.35-kg
It was released from rest at a height of 21.4 m.
(a) The kinetic energy is given by : 
As the rock was at rest initially, it means, its kinetic energy is equal to 0.
(b) The gravitational potential energy is given by : 
It can be calculated as :

(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,
M = 0 J + 492.84 J
M = 492.84 J
Hence, this is the required solution.