Answer:
a)
The analytic expression for p(V) is 
b)
The expression of the work in terms of p0 ,V0,V1 is
![W= \frac{5}{2} P_{0}[V_{0}-V_{0}^{7/5}V_{1}^{-2/5}]](https://tex.z-dn.net/?f=W%3D%20%5Cfrac%7B5%7D%7B2%7D%20P_%7B0%7D%5BV_%7B0%7D-V_%7B0%7D%5E%7B7%2F5%7DV_%7B1%7D%5E%7B-2%2F5%7D%5D)
c)
The change in internal energy of the gas
![\delta U = \frac{5}{2}P_{0}[V_{0}^{7/5}V_{1}^{-2/5}-V_{0}]](https://tex.z-dn.net/?f=%5Cdelta%20U%20%3D%20%5Cfrac%7B5%7D%7B2%7DP_%7B0%7D%5BV_%7B0%7D%5E%7B7%2F5%7DV_%7B1%7D%5E%7B-2%2F5%7D-V_%7B0%7D%5D)
Explanation:
The explanation is shown on the first, second and third
Given:
Work done, W = 5 J
Initial energy = 8J
Final energy = 30J
Let's determine if the work done have a positive or nrgative value.
Appy the equation for the first lae of thermodynamics:

Where:
U is the change in internal energy
Q is the added heat
W is the work done
To find the work done here, we have:
Rewrite the formula for W

Where:
ΔU = 30J - 8J = 22J
Q = 5J
Thus, we have:

Therefore, the work done here is -17J.
This means the work done in this scenario has a negative value.
ANSWER:
The work done in this scenario has a negative value
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
Answer:
d) precipitation
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