The intensity level of the sound produced by loud rock music is 120 dB.The power incident on an eardrum of area 0.4 cm² from this sound level is A)4 * 10^-6 W. B)1 * 10^-5 W. C)4 * 10^-5 W. D)8 mW. E)2 mW.

Question

Quizplus · Accepted Answer

We can use the formula for sound intensity:
I = P/A

Where I is the sound intensity, P is the power, and A is the area.

We are given the sound intensity (I) as 120 dB. We can convert this to watts per square meter (W/m^2) using the formula:

I = 10log(I/I0)

Where I0 is the threshold of hearing, which is 1e-12 W/m^2.

I = 10log(120/1e-12) = 10^6 W/m^2

Now we can find the power (P) incident on the eardrum using the formula:

P = I*A

We are given the area (A) as 0.4 cm^2, which we need to convert to square meters:

A = 0.4 cm^2 * (1 m/100 cm)^2 = 4e-6 m^2

P = (10^6 W/m^2)(4e-6 m^2) = 4e-5 W = 4*10^-5 W = 4*10^-8 kW

Therefore, the answer is C, 4*10^-5 W (or 4*10^-8 kW).

The Intensity Level of the Sound Produced by Loud Rock