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serg [7]
3 years ago
5

A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pi

tch. The first-chair player's note is exactly 440 Hz, and 2.70 beats per second are heard. Part A What are the two possible frequencies of the other player's note? Express your answers in hertz separated by a comma.
Physics
1 answer:
Marysya12 [62]3 years ago
7 0

Answer:

The two possible frequencies of the other player's note are  437.3 Hz and 442.7 Hz

Explanation:

Given Data

f₁=440 Hz

fₐ=2.7 beats per seconds

The beat frequency is equal to the difference in the frequencies of the two original waves:

So

f =|(f₁±fₐ)|

First to Solve for (-) Sign we get

f=|(440-2.7)|

f=437.3 Hz

Now for Solve for (+) Sign we get

f=|(440+2.7)|

f=442.7 Hz

The two possible frequencies of the other player's note are  437.3 Hz and 442.7 Hz

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Answer:

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(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

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Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

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=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

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ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

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3 years ago
A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
gavmur [86]
<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

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The increase in length can be found by the relation

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here L is the increased length and L₀ is the original length

α  is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

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