I don't know what the tables you have look like but her <span>displacement</span> would be -1m
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h = 0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
Answer:
The spring's maximum compression will be 2.0 cm
Explanation:
There are two energies in this problem, kinetic energy
and elastic potential energy
(with m the mass, v the velocity, x the compression and k the spring constant. ) so the total mechanical energy at every moment is the sum of the two energies:

Here we have a situation where the total mechanical energy of the system is conserved because there are no dissipative forces (there's no friction), so:


Note that at the initial moment where the hockey puck has not compressed the spring all the energy of the system is kinetic energy, but for a momentary stop all the energy of the system is potential elastic energy, so we have:

(1)
Due conservation of energy the equality (1) has to be maintained, so if we let k and m constant x has to increase the same as v to maintain the equality. Therefore, if we increase velocity to 2v we have to increase compression to 2x to conserve the equality. This is 2(1.0) = 2.0 cm
Answer:
160 meters
Relative of speed Vr = 10 -2 = 8 m/s (horizontal speed)
20 sec * 8 m/s = 160 m since ball travels 20 sec