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sergey [27]
3 years ago
14

1. A composite when tested bounces half as high as the original material. How can

Physics
1 answer:
Mkey [24]3 years ago
7 0

I creating another composite to jump one-quarter as high as the initial composite you most bear in mind factors of weight,hardness ,center of mass will be one-quarter of the prior composite in other for it to bounces only one-quarter .

From the question we are told

A composite when tested bounces half as high as the original material. How can  you create another composite that bounces only one-quarter as high

Generally We know that

A composite Material is a material that is made up of more than be substance i.e is it requires substance of variant properties and make up to come together to form it.

Therefore

I creating another composite to jump one-quarter as high as the initial composite you most bear in mind factors of weight,hardness ,center of mass will be one-quarter of the prior composite in other for it to bounces only one-quarter

For more information on this visit

brainly.com/question/13025901?referrer=searchResults

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Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

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3 years ago
A child sleds down a hill with an acceleration of 2.94 m/s2. If her initial speed is 0.0 m/s and her final speed is 17.5 m/s, ho
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Answer:

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

Explanation:

Given that the child accelerates uniformly and that both initial (v_{o}) and final speeds (v_{f}), measured in meters per second, and acceleration (a), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (t), measured in seconds, is:

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If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 17.5\,\frac{m}{s} and a = 2.94\,\frac{m}{s^{2}}, then the time taken is:

t = \frac{17.5\,\frac{m}{s}-0\,\frac{m}{s} }{2.94\,\frac{m}{s^{2}} }

t = 5.952\,s

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

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