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aalyn [17]
3 years ago
8

A man, holding a weight in each hand, stands at the center of a horizontal frictionless rotating turntable. The effect of the we

ights is to double the rotational inertia of the system. As he is rotating, the man opens his hands and drops the two weights. They fall outside the turntable. Then:
Physics
1 answer:
Alex_Xolod [135]3 years ago
7 0

Answer:

 w = 2w₀     the angular velocity of man doubles

Explanation:

In this exercise, releasing the weights reduces the moment of inertia

       I= I₀ / 2

Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved

         L₀ = L

       I₀ w₀ = I w

       I₀ w₀ = I₀ / 2 w

       w = 2w₀

therefore the angular velocity of man doubles

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The answer to your question is: F  = 0.4375 N. The force will be 16 times lower than with the first conditions.

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Data

F = 7 N

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F = \frac{Km1m2}{r2}

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3 0
3 years ago
A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th
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Answer:

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Explanation:

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The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

F.d=\dfrac{1}{2}m(v^2-u^2)

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F = -22713.92 N

F=2.27\times 10^4\ N

So, the magnitude of the force that stops the bullet is 2.27\times 10^4\ N

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