Question

A 58-dB sound wave strikes an eardrum whose area is 5.0Ã10â5m^2. The intensity of the reference level required to determine the sound level is 1.0Ã10^â12W/m^2. 1yr=3.156Ã10^7s.

a. How much energy is absorbed by the eardrum per second?
b. At this rate, how long would it take the eardrum to receive a total energy of 1.0 J?

Answer 1

Answer: (a) Energy = 3.15 * 10-11 W

(b) time = 1000 years

Explanation:

ok to begin,

(a )  for the output intensity

58 = 10 * log(I/(1 *10^-12))

I = 6.31 *10^-7 W/m^2

energy absorbed = I * Area

energy absorbed = 6.31 *10^-7 * 5 *10^-5

energy absorbed = 3.15 * 10^-11 W

the energy absorbed is 3.15 * 10-11 W

part B)

let the time taken be given as t

1 = 3.15 * 10^-11 * t

t = 3.17 *10^10 s = 1000 years

the time taken is 3.17 *1010 s or 1000 years

i hope this helps