A reduction of 20 dB means reducing the intensity by a factor of `10^2` so we expect the radial distance to be 10 times larger, namely 30 m. A further reduction of 90 dB may correspond to an extra factor of `10^(4.5)` in distance, to about `30xx30000m`, or about 1000 km. We use the difinition of the decibel scale and the inverse suqare law for sound intensity. From the difinition of sound level,
`beta=10log((1)/(10^(-12)(W)/(m^2)))`
We can compute the intensities corresponding to each of the levels mentioned as
`I=[10^((beta)/(10))]10^(-12)(W)/(m^2)`
They are `I_(120)=1(W)/(m^2)`,`I_(100)=10^(-2)(W)/(m^2)` and `I_(10)=10^(-11)(W)/(m^2)`.
a. The power passing through any sphere around the source is
`P=4pir^2l`
If we ignore absoption of sound by the medium,
conservation of energy requires that
`r_(120)^(2)I_(120)=r_(100)^(2)I_(100)=r_(10)^(2)I_(10)`
then, `r_(100)=r_(120)sqrt((I_(120))/(I_(100)))=(3.00m)sqrt((1(W)/(m^2))/(10^(-2)(W)/(m^2)))=30.0m`
b. `r_(10)=r_(120)sqrt((I_(120))/(I_(10)))=(3.00m)sqrt((1(W)/(m^2))/(10^(-11)(W)/(m^2)))=9.49xx10^5m`
At 949 km away the faint 10 dB sound would not be indetifiable among many other soft and loud sounds produced by other sources across the continent. And 949 km of air is such a thick screen that it may do a significant amound of sound absorption.