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Vika [28.1K]
2 years ago
7

A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid

e and brings it up to its proper turning rate of one complete rotation every 5.5 s. what is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?
Physics
1 answer:
ioda2 years ago
4 0
<span>Radius = 4.6 m
 Time for one complete rotation t = 5.5 s.
 Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
 Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
 Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
  Force of the cat Fc = 6m, m being the mass.
 Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
 equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116</span>
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3 years ago
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Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

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Learn more about elastic collisions at: brainly.com/question/7694106

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kolbaska11 [484]
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