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tatuchka [14]
2 years ago
15

What is the name of the perceived change in a sound wave’s frequency due to motion between the observer and the sound source?

Physics
2 answers:
andrew11 [14]2 years ago
6 0

Answer: The answer is Doppler effect.

Explanation:

When there is a relative motion between the source and the observer then there will be change in the frequency of the wave. This phenomenon is called Doppler effect.

For example,when the source and the observer are moving each other then the frequency increases in this case. When the source and the observer are moving away from other then the frequency decreases in this case.

Therefore, Doppler effect is the name of the perceived change in a sound wave's frequency due to motion between the observer and the sound source.

Shkiper50 [21]2 years ago
4 0

Answer: Hello mate!

the name is "doppler effect"

As you may know, the frequency (and hence the wavelength) of the sound determines the pitch.

If you move towards the source of the sound, the wavelenght seems shorter, which means that the frequency is higher, and also the pitch.

In the other case, if you are moving away from the sound, the wavelength seems larger and the frequency smaller, so the pitch is lower.

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A 100kg crate slides along a floor with a starting velocity of 21ms If the force due to friction is 8N how long will it take for
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A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.

We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

<h3>F = ma</h3>

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Divide both side by 1000

a = \frac{-8}{100}

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Therefore, the deceleration of the box is –0.08 ms¯²

Finally, we shall determine the time taken for the box to come to rest. This can be obtained as follow:

Deceleration (a) = –0.08 ms¯²

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<h3>Time (t) =.? </h3>

<h3>v = u + at</h3>

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0 = 21 – 0.08t

Collect like terms

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Divide both side by –0.08

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Learn more: brainly.com/question/14446351

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Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
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Answer:

a) x = 0.200 m

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q_1 = 21.0\mu C

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by relation for electric field we have following relation

E = \frac{kq}{x}^2

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FROM FIGURE

x is the distance from left point charge where electric field is zero

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