The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
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
<span>Work: W = Fd. 50(distance) multiplied by 90(force) would equal 4500 J or, answer D</span>