Question

PLEASE HURRY IM TIMED!!!

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as L=10log l/l0, wherel0 =10^-12 and is the least intense sound a human ear can hear. Brandon is trying to take a nap, and he can barely hear his neighbor mowing the lawn. The sound intensity level that Brandon can hear is 10-10. Ahmad, Brandon’s neighbor that lives across the street, is mowing the lawn, and the sound intensity level of the mower is 10-4. How does Brandon’s sound intensity level compare to Ahmad’s?

Answer 1

a

Explanation:

Answer 2

A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.

Solution-

Given that, loudness measured in dB is

`L=10\log (I)/(I_0)`

Where,

I = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's = 10⁻⁴

Then,

`L_1=10\log (I_1)/(I_0)=10\log (10^(-10))/(10^(-12))=10\log (1)/(10^(-2))=10\log 10^(2)=2*10\log 10=20`

`L_2=10\log (I_2)/(I_0)=10\log (10^(-4))/(10^(-12))=10\log (1)/(10^(-8))=10\log 10^(8)=8*10\log 10=80`

`\therefore (L_1)/(L_2) =(20)/(80) =(1)/(4)`