A rock group is playing in a bar. Sound emerging from the door spreads uniformly in all directions. The intensity level of the music is 116 dB at a distance of 2.82 m from the door. At what distance is the music just barely audible to a person with a normal threshold of hearing? Disregard absorption. Answer in units of m.

1779299.7m

Explanation:

From formulas in acoustic sound, we know that sound intensity is inversely proportional to the square of the distance away.

Thus;

I2/I1 = r2²/r 1²

So,

∆L = 10 log (I2/I1)

Where ∆L is the intensity of music and r1 and r2 are distances away.

∆L=10log 10(r1²/r2²)

∆L=10log 10(r1/r2)²

∆L= - 20log 10(r1/r2)

r2 = r1•10^(-∆L/20)

From the question,

∆L = 116 Db

r1 = 2.82m

Thus,

r2 = 2.82 x 10^(116/20)

r2 = 2.82 x 630957.34 = 1779299.7m

it breaks and dirupture other materials and other sencews from the bottom

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