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4 years ago

8

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

ers. Her eardrums have a diameter of 8.4 mm. How much sound energy is transferred to each eardrum in 1.0 second?

1 answer:

Klio2033 [76]4 years ago

8 0

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

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Answer:

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Explanation:

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