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Masteriza [31]
3 years ago
5

A tennis ball with a velocity of +10.0 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball r

ebounds in the opposite direction with a velocity of â8.0 m/s to the left. If the ball is in contact with the wall for 0.012 s, what is the average acceleration of the ball while it is in contact with the wall?
Physics
1 answer:
netineya [11]3 years ago
6 0

Answer:

-1500 m/s2

Explanation:

So the ball velocity changes from 10m/s into the wall to -8m/s in a totally opposite direction within a time span of 0.012s. Then we can calculate the average acceleration of the ball as the change in velocity over a unit of time.

a = \frac{\Delta v}{\Delta t} = \frac{-8 - 10}{0.012} = \frac{-18}{0.012} = -1500 m/s^2

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Answer:

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

Explanation:

As we know that there is no external torque on the system of two disc

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So we will have

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now we have

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also we have

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now from above equation we have

(\frac{1}{2}MR^2)\omega_o  = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now we have

\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

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3 years ago
A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p
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Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

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