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Elis [28]
4 years ago
9

In the diagram below, a stationary source located at point S produces sound having a constant frequency of 512 hertz. Observer A

, 50 meters to the left of S, hears a frequency of 512 hertz. Observer B, 100.0 meters to the right of S, hears a frequency lower than 512 hertz.
A) Observer A is stationary, and observer B is moving away from point S.

B) Observer A is stationary, and observer B is moving toward point S.

C) Observer A is moving away from point S, and observer B is stationary.

D) Observer A is moving toward point S, and observer B is stationary.
Physics
1 answer:
Nikolay [14]4 years ago
4 0
This has to do with Doppler effect. If the source or the observer moves, they the observer will hear a different frequency. Since the source is stationary, we only need to know whether the observer is moving.

Since Observer A does not hear a change in frequency, he a must be stationary. So we can ignore C and D

Since Observer B hears a different frequency, Observer B must be moving. Here's the tricky part. We need to find whether B is moving towards or away the source. Since he hears a lower frequency, he is walking away from the source. So the answer is A.

You can conform with the Doppler equation.

Answer: A
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The kinetic energy when it returns to its original height is 100 J

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

u² = v² -2·g·s

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P.E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.

Learn more about kinetic energy at:

brainly.com/question/25959744

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The following reaction is balanced: N2 + 3H2 → 2NH3 True or False?
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What is the velocity of a car that travels from mile marker 32 on I-10 to mile marker 312 on I-10 in a time of 2 hours and 45 mi
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8 0
3 years ago
Suppose the B string on a guitar is 24" long and vibrates with a frequency of about 247 Hz. You place your finger on the 5th fre
pashok25 [27]

To solve this problem it is necessary to apply the related concepts to string vibration. This concept shows the fundamental frequency of a string due to speed and length, that is,

f = \frac{v}{2L}

Where

v = Velocity

L = Length

Directly if the speed is maintained the frequency is inversely proportional to the Length:

f \propto \frac{1}{L}

Therefore the relationship between two frequencies can be described as

\frac{f_2}{f_1}=\frac{l_1}{l_2}

f_2 = \frac{l_1}{l_2}(f_1)

Our values are given as,

l_1 = 24"\\f_1 = 247Hz\\l_2 = 18"

Therefore the second frequency is

f_2 = \frac{l_1}{l_2}(f_1)\\f_2 = \frac{24}{18}(247)\\f_2 = 329.33Hz

The frequency allocation of 329Hz is note E.

8 0
3 years ago
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