The answer is D, because the collision's between molecules are elastic, not inelastic.
Answer:
<h2>4.6 m/s²</h2>
Explanation:
The acceleration of an object given it's velocity and time taken can be found by using the formula
<h3>

</h3>
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
Since the body is from rest u = 0
From the question we have

We have the final answer as
<h3>4.6 m/s²</h3>
Hope this helps you
Answer:
It changes at a rate of 4/3 meter per second
Explanation:
In the given figure below we have
Solving for Y given
we get

Answer:
Part a)

Part b)

Explanation:
Part a)
As we know that orbital velocity at certain height from the surface of Earth is given as

here we know that



now we have


Part b)
When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

The layers form from sand dunes