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Aliun [14]
2 years ago
12

A physical therapy exercise has a person shaking a 5.00 kg weight up and down rapidly. if the barbell is moving at 4.50 m/s, wha

t is the magnitude of the force required to stop it in 0.333 seconds?
Physics
1 answer:
Ivanshal [37]2 years ago
7 0

The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

<h3>Magnitude of required force to stop the weight</h3>

The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;

F = ma

F = m(v/t)

F = (mv)/t

F = (5 x 4.5)/0.333

F = 67.6 N

Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

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r(t) models the water flow rate, so the total amount of water that has flowed out of the tank can be calculated by integrating r(t) with respect to time t on the interval t = [0, 35]min

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