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Lisa [10]
3 years ago
10

A 53.5-g Super Ball traveling at 27.5 m/s bounces off a brick wall and rebounds at 21.5 m/s. A high-speed camera records this ev

ent. If the ball is in contact with the wall for 4.05 ms, what is the magnitude of the average acceleration of the ball during this time interval
Physics
1 answer:
Tema [17]3 years ago
4 0

Answer:

<h3>12,098.77m/s²</h3>

Explanation:

Acceleration is the rate of change of velocity of a body. It is expressed as;

acceleration = change in velocity/time

acceleration a = v-u/t

Given

v = 27.5m/s

u = -21.5m/s (rebounds in the opposite direction)

time t = 4.05ms = 0.00405secs

Required

acceleration

Substitute the given values into the formula

a = 27.5-(-21.5)/0.00405

a = 27.5+21.5/0.00405

a = 49/0.00405

a = 12,098.77m/s²

Hence  the magnitude of the average acceleration of the ball during this time interval is 12,098.77m/s²

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<h3>What is the average fixed cost?</h3>

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