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Kazeer [188]
3 years ago
14

A baseball player at practice is pushing a tackling dummy across the field. He initially has to push the 130 kg dummy with 600 N

ewton’s of force to get it to start moving and maintain a force of 475 to keep it moving at a steady state. What are the coefficient of friction static and kinetic with the dummy on the ground
.
Physics
1 answer:
Leona [35]3 years ago
5 0

Answer:

Static= 600/(130*9.8)=0.47

Kinetic=475/(130*9.8)=0.37

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To measure the amount of mass in an object, which piece of equipment would be used?
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What happens to the sum of the ball's kinetic energy and potential energy as the ball rolls from point A to point E? Assume ther
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Answer:

C. The sum remains the same.

Explanation:

The sum of the kinetic and potential energy remains the same as the all rolls from point A to E.

We know this based on the law of conservation of energy that is in play within the system.

The law of conservation of energy states that "energy is neither created nor destroyed within a system but transformed from one form to another".

  • At the top of the potential energy is maximum
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The light rays in the illustration below do not properly focus at the focal point. this problem occurs with?
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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]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

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

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

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

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

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

v=\omega r

where

\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

Therefore, her linear speed is:

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

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3 years ago
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