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kolezko [41]
3 years ago
6

The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21ms. The total mass of the train is 825 g.

find the average power delivered to the train during its acceleration.
Physics
1 answer:
Rudiy273 years ago
7 0

Explanation:

Average power = change in energy / change in time

P = ΔE / Δt

P = (½ mv²) / t

P = (½ (0.825 kg) (0.620 m/s)²) / (0.021 s)

P = 7.55 Watts

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To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

R_{eq}= \frac{V}{I}

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

P=VI \rightarrow I = \frac{P}{V}

I =0.608 A

Applying Ohm's law

R_{eq} = \frac{120V}{0.608A}

R_{eq} = 197.4\Omega

Therefore the equivalent resistance of the light string is 197.4\Omega

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Answer:

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Explanation:

moments about any convenient point will sum to zero.

I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.

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W[55 - 50) - 0.040(9.8)[ 95 - 55] = 0

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What best describes the relationship between the frequency of an oscillation and the period of the oscillation?
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Explanation:

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Also, f=\dfrac{\omega}{2\pi}

So,

T=\dfrac{2\pi}{\omega}

The time taken to complete one oscillation is called the period of the oscillation and the number of oscillation is called the frequency if an oscillation.

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Answer is b) I think
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3 years ago
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S27253129 ,,, message me please, I can't ask you my homework question in the comments :c
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What’s the problem what do u need help on
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