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Licemer1 [7]
4 years ago
9

A 1056-hertz tuning fork is struck at the same time as a note on the piano and you hear 2 beats/second. You tighten the piano st

ring very slightly and then hear 3 beats/second. What is the frequency of the piano string?
Physics
1 answer:
elena-14-01-66 [18.8K]4 years ago
6 0

Answer:

The frequency of the piano string is <em>1059 Hz</em>.

Explanation:

The frequency beat (fb), 2 beats/second, is the absolute difference between the frequency of the tuning fork (1056 Hz) and the frequency of the piano string.

As the piano string gets tightened, the frequency beat becomes 3 beats/second.

Therefore,

fb = fb = fpiano - ftuning fork\\ 3 Hz=fpiano-1056Hz\\ fpiano=1056Hz+3Hz\\ fpiano=1059Hz

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IRISSAK [1]

Answer:

\lambda=25.6nm

Explanation:

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\frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})

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7 0
3 years ago
Supply the missing force necessary to achieve equilibrium. Show your work.
Mumz [18]

<u>Analysing the Question:</u>

We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it

In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N

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Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces

So,  (106 + 106 + 142 + x) = 496

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8 0
3 years ago
A playground merry-go-round of radius r = 2.00 m has a moment of inertia = 250 ⋅ , and is rotating at 10.0 rev/min about a frict
bagirrra123 [75]

Answer:

330kgm^2

Explanation:

We are given that

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Hence, the new moment of inertia of the merry -go-round=330kgm^2

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Answer:

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