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Eduardwww [97]
3 years ago
14

For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the

very first frequency of audible sound for these speakers to produce destructive interference at the listener?The speed of sound is 344 m/sec. (1 point)
A. 166 Hz
B. 331 Hz
C. 662 H2
D. 1323 H
Physics
1 answer:
galina1969 [7]3 years ago
7 0

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x =  52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

d = 2x

d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

f = 1 / T

f = vs / x = 344 m/s / 0.52m

f ≅ 662 Hz

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The work is -67.76 J

Explanation:

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In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.

So, the loss of kinetic energy is \frac{1}{2} *m*(vf^{2} -vi^{2} )

You know:

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<u><em>The work is -67.76 J</em></u>

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