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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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Answer:

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Explanation:

Given the following data;

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Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

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Please comment below if you have any questions.

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