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 false it has straight sirface
Answer:
the time taken for the object to fall is 6 s.
Explanation:
Given;
final velocity of the object, v = 58.8 m/s
initial velocity of the object, u = 0
The height of fall of the object is calculated as;
v² = u² + 2gh
v² = 2gh
The time to fall through the height is calculated as;
Therefore, the time taken for the object to fall is 6 s.
