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Aleksandr-060686 [28]
3 years ago
8

Once again we have a skier on an inclined plane. The skier has mass M and starts from rest. Her speed at the bottom of the slope

is 12.0 meters/second. The hill is inclined 30 degrees from the horizontal and the length of the slope is 100.0 meters.Use the law of conservation of energy to determine her final speed and write that number down here:
Physics
1 answer:
mars1129 [50]3 years ago
4 0

Answer:

v = 31.3 m / s

Explanation:

The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.

Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane

Highest point

       Em₀ = U = m g y

Lowest point

     Em_{f} = K = ½ m v²

As there is no friction, mechanical energy is conserved

       Em₀ = Em_{f}

       m g y = ½ m v²

       v = √ 2 g y

Where we can use trigonometry to find and

       sin 30 = y / L

       y = L sin 30

Let's replace

      v = RA (2 g L sin 30)

Let's calculate

      v = RA (2 9.8 100.0 sin30)

      v = 31.3 m / s

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A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

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Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

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where:

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M is the mass of the girl

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m is the mass of the rock

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L_1=L_2

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b)

The linear speed of a body in rotational motion is given by

v=\omega r

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\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

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v=\omega R=\frac{-mvR^2}{I+MR^2}

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Answer:

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Explanation:

Given that,

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