two people hear a tornado siren, but one listener is 71.1 times farther away from the source of the sound than the other. what is the intensity level of the sound (in decibels) for the more distant listener relative to that for the nearer listener? relative intensity:

The intensity level of a sound, in decibels, is described as 10 times the logarithm to the base 10 of the ratio of the intensity of this sound to a reference intensity.

By using the inverse square law,

As one listener is 71.1 times farther away from the source of the sound than the other.

Then Increasing the distance by 71.1 times reduces the intensity by

(71.1)^2 = 5055.21

The intensity level of the sound (in decibels) for the more distant listener relative to that for the nearer listener is:

= 10log(ratio of intensities)

= 10log(1/5055.21)

= 10log(0.00019782)

=10 × -3.7028

= -37.03 dB (The negative sign represents the sound is less intense for the farther listener)

So, the intensity level of the sound (in decibels) for the more distant listener relative to that for the nearer listener is -37.03 dB.

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https://brainly.com/question/10287755

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