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lilavasa [31]
3 years ago
14

In a skating stunt known as "crack-the-whip," a number of skaters hold hands and form a straight line. They try to skate so that

the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of 80.0 kg and is 6.10 m from the pivot. He is skating at a speed of 6.80 m/s. Determine the magnitude of the centripetal force that acts on him.
Physics
1 answer:
4vir4ik [10]3 years ago
7 0

Answer:

The magnitude of the centripetal force that acts on him

Explanation:

Given that,

Mass = 80.0 kg

Distance = 6.10 m

Speed = 6.80 m/s

We need to calculate the magnitude of the centripetal force that acts on him

Using formula of the centripetal force

F_{c}=\dfrac{mv^2}{r}

Where, F = force

m = mass

v = speed

r = distance

Put the value into the formula

F_{c}=\dfrac{80.0\times(6.80)^2}{6.10}

F_{c}=606.4\ N

Hence, The magnitude of the centripetal force that acts on him

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3 years ago
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a = 9.94 m/s²

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\rho(r) = ar^2 - br^3

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a = 9.94 m/s²

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3 years ago
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