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tekilochka
17 days ago
10

Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A p

erson stands 4.1 m away from one speaker and 4.8 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 343 m/s.
Physics
1 answer:
Keith_Richards [3.2K]17 days ago
7 0
The frequency is calculated to be 735 Hz. Given: Person's distances to speakers are r₁ = 4.1 m and r₂ = 4.8 m. The path difference calculates to d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m. In terms of destructive interference, we derive: λ = v/f, where v is the sound speed of 343 m/s. Using n = 1 gives f = 245 Hz, and for n = 3, f = 735 Hz. Thus, the second lowest frequency for destructive interference is 735 Hz.
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\frac{1}{2}v^2 = s\Delta T

the resulting temperature change is expressed as

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\Delta T = \frac{25^2}{2\times 387}

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3 0
1 month ago
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Sav [3153]

Answer:

        h = 12.8 cm

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The initial parameters are as follows:

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         mg = ky... equation 1

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                kyh =  0.5 x k x h^{2}

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Yuliya22 [3333]

Answer:

Explanation:

a )

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Total moment of inertia for 3 blades is:

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ml²

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Details provided include:

m = 5500 kg

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Substituting these values produces:

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f = 11 rpm (revolutions per minute)

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