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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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Let F =  required force, N

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3 years ago
N 1800kg car has an<br> of 3.8m/s? What is it<br> on the car?<br> acceleration<br> force acting
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Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

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An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2. The coil rotates in a uniform magnetic f
Reptile [31]

Answer:

331.75 V

Explanation:

Given:

Number of turns of the coil, N = 40 turns

Area, A = 0.06 m²

Magnetic Field, B = 0.4 T

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                           Maximum induce emf, E₀ = NABω

but ω = 2πf

                           Maximum induce emf, E₀ = NAB(2πf₀)

                           Maximum induce emf, E₀ = 2πNABf₀

Where;

N is number of turns of the coil

A is area

B is magnetic field

ω is the angular velocity

f is the frequency

                                     E₀ = 2 × π × 40 × 0.06 × 0.4 × 55

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The maximum induced emf is 331.75 V

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3 years ago
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