Answer:
B: False
Explanation:
The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.
Thus, it means that the entropy change will always be positive.
Therefore, the given statement in the question is false.
a) 0.94 m
The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

where:
is the force applied by the snow
d is the displacement of the man in the snow, so it is the depth of the snow that stopped him
m = 68 kg is the man's mass
v = 0 is the final speed of the man
u = 55 m/s is the initial speed of the man (when it touches the ground)
and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)
Solving the equation for d, we find:

b) -3740 kg m/s
The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

where
m = 68 kg is the mass of the man
is the change in velocity of the man
Substituting,

Answer:
they r schooling fish
Explanation:
they need to be kept in groups of 5.
To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.
From the theory we could consider that the energy change is subject to

Where
Initial Energy
Energy loses
Replacing we have that


Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV