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3 years ago

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A 20-kg child starts at the center of a playground merry-go-round that has a radius of 1.5 mm and rotational inertia of 500kg⋅m2

and walks out to the edge. The merry-go-round has a rotational speed of 0.20 s−1 when she is at the center.
What is its rotational speed when she gets to the edge?

1 answer:

Lemur [1.5K]3 years ago

6 0

Answer:

w = 0.189 rad/ s

Explanation:

This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants

Initial

L₀ = I₀ w₀

Final

$L_{f}$ = I w

L₀ = L_{f}

I₀ w₀ = I_{f} w

.w = I₀/I_{f} w₀

The initial moment of inertia is

I₀ = 500 kg. m2

The final moment of inertia

$I_{f}$ = 500 + m r²

I_{f} = 500 + 20 1.5

I_{f} = 530 kg m²

Initial angular velocity

w₀ = 0.20 rad / s

Let's calculate

w = 500/530 0.20

w = 0.189 rad / s

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Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

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Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

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Answer:

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Where K = Coloumbs constant =

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The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

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F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

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F = +0.060N

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Read more on Brainly.com - brainly.com/question/14592748#readmore

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Note that I'm reading the given details as

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