Answer:
The compression in the spring is 5.88 meters.
Explanation:
Given that,
Mass of the car, m = 39000 kg
Height of the car, h = 19 m
Spring constant of the spring, 
We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

x is the compression in spring

So, the compression in the spring is 5.88 meters.
A).
It would decrease because the speed of sound and temperature are proportional.
There are three forces acting on the book.
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces
Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).
∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N
Therefore, the normal force is 42 N.
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
Net pulling force on the chain is due to weight of the part of the chain which is over hanging
So we know that mass of overhanging part of chain is given as

now net pulling force on the chain is given as

now acceleration is given as



Part b)
Tension force in the part of the chain is given as





Part c)
velocity of the last link of the chain is given as


now integrate both sides



Answer:2m/s^2
Explanation:
mass=5kg
Force=10N
Acceleration=force ➗ mass
Acceleration=10 ➗ 5
Acceleration=2m/s^2