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cupoosta [38]
3 years ago
9

A rubber ball with mass 0.20 kg is dropped vertically from a height of 1.5 m above a floor. The ball bounces off of the floor, a

nd during the bounce 0.60 J of energy is dissipated. What is the maximum height of the ball after the bounce
Physics
2 answers:
Zina [86]3 years ago
8 0

Answer:

The maximum height of the ball after the bounce is 1.2 m

Explanation:

Potential Energy = mass * Height * acceleration of gravity

PE=mgh

= 0.2 x 9.8 x 1.5

P.E = 2.94 J

During bounce of ball 0.60 J of energy is lost. So

  2.94 - 0.6 = 2.34 J

now  new energy is 2.34

New P.E = mgh

2.34 = 0.2 x 9.8 x h

h = 2.34 / 0.2 x 9.8

h =  1.2 m

The maximum height of the ball after the bounce is 1.2 m

zlopas [31]3 years ago
7 0

Explanation:

Given:

Mass = 0.2 kg

Height = 1.5 m

Potential energy of the drop, PE = m × g × h

= 0.2 x 9.81 x 1.5

= 2.94 J

After the drop, 0.6 J of energy is dissipated, so amount of energy left

= 2.94 - 0.6

= 2.34 J

This energy is eqal to thenew potential energy which is:

m × g × h = 2.34

0.2 × 9.81 × h = 2.34

= 2.34/1.962

= 1.19 m.

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The kinetic energy of the bullet is 20.4 kJ.

<u>Explanation:</u>

Kinetic energy of a bullet will be equal to the product of mass of the bullet with the square of velocity or speed of the bullet and then the half of that product value.

But here the mass of the bullet is not given, instead the weight of the bullet is given in terms of force. So from this, we have to first find the mass of the bullet.

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