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ladessa [460]
3 years ago
7

A certain loudspeaker system emits sound isotropically (uniformly in all directions) with a frequency of 1,054 Hz and an intensi

ty of 6.429 mW/m2 at a distance of 7.945 m. Assume that there are no reflections. What is the intensity in 10-3 W/m2 at 39.3 m? (you don't need to write the power/prefix and unit
Physics
1 answer:
Dmitrij [34]3 years ago
5 0

Answer:

The intensity in 10-3 W/m2 at 39.3 m is 0.26275*10^{-3}\ W/m^2

Explanation:

Given Data:

frequency=1,054 Hz

intensity=6.429 mW/m^2

distance=7.945 m.

Required:

intensity in 10^-3 W/m^2 at 39.3 m=?

Solution:

Intensity at distance r from point source which is emitting sound waves of power P is given by the following formula:

Intensity=I=\frac{P}{4\pi r^2}

Where:

P is the power

r is the distance

\frac{I_1}{I_2}=\frac{\frac{P}{4\pi r_1^2}}{\frac{P}{4\pi r_2^2}}  \\\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2} \\I_2=I_1*\frac{r_1^2}{r_2^2}

r_1= 7.945 m\\r_2= 39.3 m\\I_2=6.429*10^{-3} *\frac{7.945^2}{39.3^2}

I_2=0.00026275\ W/m^2\\I_2=0.26275*10^{-3}\  W/m^2

The intensity in 10^-3 W/m2 at 39.3 m is 0.26275*10^{-3}\ W/m^2

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