Question

Find intersity of sound in dB if its intensity in Watt// m^(2) is 10^(-10).

Answer 1

Answer:

Here I = given intensity in watt/m^2

the formula is: intensity in dB = 10*log(I/10^-12)

thus, intensity= 10* log(10^-10/10^-12)     dB

thus, intensity= 10*log(10^2)  dB

intensity = 10*2 dB

intensity = 20dB

Explanation:

Answer 2

Intensity of sound in db = 10 × 2 = 20 db

Given:

Intensity of sound = $10^{-10}$ Watt/ m²

To find:

Intensity of sound in dB.

Formula used:

$\beta=10 \log _{10} \cdot \frac{I}{I_{0}}$

$\beta$ = Intensity of sound in db

$I_0$ = Reference Intensity of sound = $10^{-12}$   Watt/ m²

$I$ =  Intensity of sound.

Explanation:

$I$ =  Intensity of sound = $10^{-10}$ Watt/ m²

$I_0$ = Reference Intensity of sound = $10^{-12}$   Watt/ m²

$\beta=10 \log _{10} \cdot \frac{I}{I_{0}}$

$\beta=10 \log _{10} \cdot \frac{10^{-10}}{10^{-12}}$

Intensity of sound in db = 10 × $10 \log _{10} \cdot 10^2$

Intensity of sound in db = 10 × 2 = 20 db

To learn more...

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2)If intensity of sound is increased by factor of 30 by how many decibels is the sound level increased

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