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Sedbober [7]
3 years ago
9

What is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2?

Mathematics
1 answer:
horsena [70]3 years ago
8 0

Given: Sound Intensity ( I ) = 9.9 \times  10^{-5}\frac{watts}{m^2}

We know,  the approximate threshold of human hearing is at 1kHz. In watts/m^2 it's value is =\frac{9.9 \times  10^{-5}}{10^{-12}}⁻¹² W/m².

So, we can say,  Reference sound intensity(I_0) = 10⁻¹² W/m².

We have formula for "sound intensity level LI in dB" when entering sound intensity.

LI = 10×log (I / Io)  in dB

Plugging values of I and Io in formula.

LI = 10 × log (\frac{9.9 \times  10^{-5}}{10^{-12}})

LI = 10 × log (99000000)

= 10 × 7.99564

LI = 79.96 dB.

Therefore,  79.96 dB is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2.

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