Question

Jet engine of an airplane has the sound intensity level 130 dB

at distance of 30 meters. At what distance the sound is decreased
to normal speaking voice level of 60 dB?

Answer 1

The sound will be decreased to a normal speaking voice level of 60 dB at a distance of approximately 0.009 meters from the jet engine.

Step-by-step explanation:

To determine the distance at which the sound is decreased to a normal speaking voice level of 60 dB, we can use the inverse square law for sound intensity. According to the inverse square law, the sound intensity decreases as the inverse square of the distance from the source.

Let's assume that the initial sound intensity level of the jet engine at 30 meters is 130 dB. We can set up the equation:

130 dB - 60 dB = 10 log((30 m)² / (x m)²)

Simplifying the equation, we get:

70 = 10 log(30² / x²)

7 = log(900 / x²)

10^7 = 900 / x²

10^7 * x² = 900

x² = 900 / (10^7)

x = sqrt(900 / (10^7))

Therefore, the sound will be decreased to a normal speaking voice level of 60 dB at a distance of approximately 0.009 meters, or 9 millimeters from the source.