Answer:
v = 2.21 m/s
Explanation:
The foreman had released the box from rest at a height of 0.25 m above the ground.
We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :
So, its velocity at the bottom of the ramp is 2.21 m/s.
Fm=Fe and am>ae
Hopefully this helps
Explanation:
When Michelson-Morley apparatus is turned through 
then position of two mirrors will be changed. The resultant path difference will be as follows.
Formula for change in fringe shift is as follows.
n = 
v = 
According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.
l = 11 m
c = 
m/s
Hence, putting the given values into the above formula as follows.
v =
= 
= 
Thus, we can conclude that velocity deduced is .
Answer:
Friction acts in the opposite direction to the motion of the truck and box.
Explanation:
Let's first review the problem.
A moving truck applies the brakes, and a box on it does not slip.
Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.
The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.
