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Vanyuwa [196]
3 years ago
13

A rock group is playing in a bar. Sound

Physics
1 answer:
Sloan [31]3 years ago
8 0

Answer:

5292.64 m

Explanation:

dB \rightarrow \textrm{sound level}\\ I \rightarrow \textrm{sound intensity}\\ I_0 \rightarrow \textrm{threshold sound intensity}\\ x \rightarrow \textrm{distance of corresponding to threshold intensity of hearing}

Taking threshold intensity as 1\times 10^{-12} W/m^{2} and since it's a constant then sound intensity for 66.7 dB will be

66.7\;\rm dB = 10\; log_{10}\;\left(\dfrac{I}{I_0} \right)\\ \dfrac{I}{10^{-12}} = 10^{6.67}\\ I = 4.67735\times 10^{-6}\;\rm W/m^2\\ \boxed{I \approx 4.7\times 10^{-6}\;\rm W/m^2}

Also, since sound intensity is inversely proportional to the square of the distance of the source then the distance can be given by

\dfrac{I_1}{I_2} = \dfrac{r^2_2}{r^2_1}\\ \dfrac{4.7\times 10^{-6}\;\rm W/m^2}{10^{-12}\;\rm W/m^2} = \dfrac{x^2\;\rm m}{(5.96\;\rm m)^2}\\ x =\sqrt{28012000}\\ \boxed{x \approx 5292.64;\rm m}

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The wavelength of the radio photon  is 268 m.

Electromagnetic waves travel with the speed of 3×10⁸m/s in vacuum. The speed of a wave <em>c i</em>s related to its frequency<em> f</em> and wavelengthλ as follows:

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Substitute 3×10⁸m/s for <em>c</em> and 1120×10³Hz for <em>f</em>.

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As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? As a rock sinks deepe
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