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")
The 2 sides of magnet are called the north and south pole
Explanation:
Momentum = mass × velocity
p = (65 kg) (5 m/s)
p = 325 kg m/s
Answer:
B.The force of friction between the block and surface will decrease.
Explanation:
The force of friction is given by
where 
is the coefficient of friction and 
is the normal force.
When the student pulls on the block with force 
at an angle 
, the normal force on the block becomes
and hence the frictional force becomes
.
Now, as we increase , 
increases which as a result decreases the normal force 
, which also means the frictional force decreases; Hence choice B stands true.
<em>P.S: Choice D is tempting but incorrect since the weight </em>
<em> is independent of the external forces on the block. </em>
