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lidiya [134]
2 years ago
5

The label has been scratched off a tuning fork and you need to know its frequency. From its size, you suspect that it is somewhe

re around 250 Hz. You find a 250-Hz tuning fork and a 270-Hz tuning fork. When you strike the 250-Hz fork and the fork of unknown frequency, you hear a modulation (fluctuation of the amplitude) at a frequency of 5 Hz. When you strike the unknown with the 270-Hz fork, you hear a modulation at a frequency of 15 Hz. What is the unknown frequency
Physics
1 answer:
bonufazy [111]2 years ago
3 0

Answer:

255 Hz

Explanation:

With 5 beats per second with the 250 Hz fork, we know the unknown fork is either 250 - 5 = 245Hz or 250 + 5 = 255 Hz

With 15 beats per second with the 270 Hz fork, we know the unknown fork is either 270 - 15 = 255Hz or 270 + 15 = 285 Hz (most people would have a hard time discerning 15 beats per second... 5 per second is hard enough)

As 255 is the common frequency, it is the one selected.

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An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
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<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

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3 years ago
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Marta_Voda [28]

Answer:

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Nope.
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E = hf

Hope this helps!
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rjkz [21]

Answer:

it would have potential energy

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