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

d) the number of neutrons plus the number of protons

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop