Answer:
d = 11.1 m
Explanation:
Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

Plugging this into the energy conservation equation and cancelling m, we get

Solving for d,

Answer:
<h2>
m/s ^2</h2><h2 />
Explanation:
Solution,
When a certain object comes in motion from rest, in the case, initial velocity = 0 m/s
Initial velocity ( u ) = 0 m/s
Final velocity ( v ) = 72 km/h ( Given)
We have to convert 72 km /h in m/s


m/s
Final velocity ( v ) = 20 m/s
Time taken ( t ) = 2 seconds
Acceleration (a) = ?
Now,
we have,



m/s ^2
Hope this helps...
Good luck on your assignment..
Answer:
Explanation:
To stop a ball with high momentum in a small-time imparts a high amount of impact on hands. This is the reason for the stinging of hands.
The momentum of the ball is due to the mass and velocity. To prevent stinging in the hand one needs to lower his hands to increase the time of contact. In this way, the momentum transfer to the hands will be lesser.