Answer:
Using the log combination rules to reduce the famous Sakur-Tetrode equation, The change in entropy is given as:
∆S = NK*ln(V"V$/V").
Where V"V$ is final Volume (Vf) after constraint's removal,
V" is Initial Volume (Vi) before constraint's removal.
Temperature (T) is constant, Internal Energy, U is constant, N and K have their usual notations
Explanation:
Given in the question, the container is an adiabatic container.
For an adiabatic contain, it does not permit heat to the environment due to its stiff walls. This implies that the Internal Energy, U is kept constant(Q = U). The temperature is also constant (Isothermal). Thus, the famous Sakur-Tetrode equation will reduce to ∆S = NK* In(Vf/Vi).
Vf is the volume after the constraint is removed(Vf = V"V$). Vi is the volume occupied before the constraint is removed (Vi = V")
It is B since they are traveling at the same speed one is positive so the other had to be negative hope this helps;)))))
Answer:
The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent. The closer to 100 percent a machine's efficiency is, the better it is at reducing friction
Explanation:
A heat pump? it might be furnace but i think its heat pump
