Steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m.
(a) calculate its velocity just before it strikes the floor.
(b) calculate its velocity just after it leaves the floor on its way back up.
(c) calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00×10−5s).
(d) how much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Vb= 5.42 m/s

Vc= 5.33 m/s

Δ L= 0.43 mm

Explanation:

a)  A to B

Initial velocity Va = 0 m/s

As we know that

Vb= 5.42 m/s

b)

C to D

Vd= 0 m/s

Vc= 5.33 m/s

c)

Here

Vb= 5.42 m/s

Vc= 5.33 m/s

We know that

Acceleration is the rate of change of velocity

d)

Now length compressed given as

  Δ L= 0.43 mm

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.