Commercials for a toy bouncy ball advertise that it will bounce to a height that is greater than the

Physics

1 answer:

storchak [24]3 years ago

7 0

Answer:

because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!

You might be interested in

The two types of vitamins are fat and water soluble?

lianna [129]

Its true hopefully this helps you.

4 0

3 years ago

12. Which of the following statements about

aliya0001 [1]

A seems to be correct

8 0

2 years ago

1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the

Anettt [7]

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = $\frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}$

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= $\frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}$

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

4 0

3 years ago

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

erica [24]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$