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miss Akunina [59]
3 years ago
11

A sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00-m rope to a

post set in the ice. once given a push, the sled revolves uniformly in a circle around the post. If the sled makes five complete revolutions every minute, find the F exerted on it by the rope.​
Physics
1 answer:
Tpy6a [65]3 years ago
7 0

Answer:

The value is  F  =  34.3 \  N

Explanation:

From the question we are told that

    The mass of the shed is  m  =  25 \  kg

    The length of the rope is  l = 5.0 \  m

    The angular speed of the shed is w = 5 rev/minute = \frac{2* \pi * 5}{60 }= 0.524 \  rad/sec

Generally the force exerted on the shed is mathematically represented as

     F  =  m  *  w^2 *  l

=>  F  =  25  *  0.524^2 *  5

=>  F  =  34.3 \  N    

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A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
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Answer:

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Explanation:

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M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

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d = 30 m

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