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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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g A hydraulic press has a safety feature which consists of a hydraulic cylinder with a piston at one end and a safety valve at t
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Answer:

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Explanation:

Area of a circle = \pir^{2}

where r is the radius of the circle.

The cylinder has a radius of 0.02 m, its area is;

A_{1} = \pir^{2}

  = \frac{22}{7} x (0.02)^{2}

  = \frac{22}{7} x 0.0004

  = 1.2571 x 10^{-3}

Area of the cylinder is 0.0013 m^{2}.

The safety valve has a radius of 0.0075 m, its area is;

A_{2} = \pir^{2}

    = \frac{22}{7} x (0.0075)^{2}

    = \frac{22}{7} x 5.625 x 10^{-5}

    = 1.7679 x 10^{-4}

Area of the valve is 0.00018 m^{2}.

From Hooke's law, the force on the safety valve can be determined by;

F = ke

F_{2}  = 950 x 0.0085

  = 8.075 N

Minimum force, F_{1}, required can be determined by;

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

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F_{1} = \frac{0.0013 *8.075}{0.00018}

    = 58.32

The minimum force that must be exerted on the piston is 58.32 N.

8 0
3 years ago
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