Question

A sound is recorded at 19 decibels. What is the intensity of the sound?

Answer 1

Answer:

$10^{-10.1}$

Explanation:

We know Intensity of sound is given by the formula

10㏒$\frac{I}{K}$=D        

(where I is intensity, K is a constant(K=$10^{-12}$), and D is decibels)

by putting giving data in formula we get intensity(I)=$10^{-10.1}$

Answer 2

Answer:

The intensity of the sound will be 7.94 × 10^-11 W/m^2 .

Explanation:

Decibels = dB = 19  

The formula to calculate intensity of sound is:  

dB = 10 log (I/K)    ……… (i)      

Where, K is the lowest threshold intensity of sound that a person can hear at 1000 Hz frequency and its value is 10^-12 W/m^2.

By putting values in equation (i),

19 = 10 log (I/10^-12)    

Anti-log (1.9) = I/10^-12

I = (Anti-log (1.9))( 10^-12)

I = 7.94 × 10^-11 W/m^2