By knowing how intensity relates to distance, we will see that at 4 meters of the engine the intensity is 80.
How sound intensity relates to distance?
The sound spreads in a solid angle like all waves, so the intensity of the sound will decrease with the distance squared, this means that:
[tex]I = \frac{P}{4*pi*r^2}[/tex]
Where P is the sound power, we know that when the student is at 8m from the engine, the intensity measured is 20 (it does not tell the units) so we have:
[tex]20 = \frac{P}{4*3.14*(8m)^2} \\\\P = 20*4*3.14*(8m)^2 = 16,078.8 m^2[/tex]
Now, if you move 4 meters closer to the engine, your new position will be:
r = 4m
Then we have:
[tex]I = \frac{16,078.8 m^2}{4*3.14*(4m)^2} = 80[/tex]
If you want to learn more about sound intensity, you can read:
https://brainly.com/question/26376755
