u = 0.077
Explanation:
Work done by friction is
Wf = ∆KE + ∆PE
-umgx = ∆KE,. ∆PE =0 (level ice surface)
-umgx = KEf - KEi = -(1/2)mv^2
Solving for u,
u = v^2/2gx
= (12 m/s)^2/2(9.8 m/s^2)(95 m)
= 0.077
The compression curve would be theoretically given for a system of bodies in which the spring applies the force (Although in the same way the following process can be extrapolated to any system, depending on the type of Force to consider) For a spring mass system, the strength is given by Hooke's law as

Where,
K = Spring constant
x = Displacement
If we integrate based on distance we would have

This integral represents the area under the Force Curve based on each distance segment traveled.



This is the same formula that represents the elastic potential energy of a body. Therefore the correct answer is D.
Answer:
(A) Work done will be 87.992 KJ
(B) Work done will be 167.4 KJ
Explanation:
We have given mass of methane m = 4.5 gram = 0.0045 kg
Volume occupies 
And volume is increased by
so 
Temperature T = 310 K
Pressure is given as 200 Torr = 26664.5 Pa
(a) At constant pressure work done is given by

(b) At reversible process work done is given by 
We have given mass = 4.5 gram
Molar mass of methane = 16
So number of moles 
So work done 
Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,
Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.
v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s
Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2
Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds
Then,
x = 8*0.553 = 4.424 m