Question

The noise rating of an air conditioner is 50 decibels, and the noise rating of a washing machine is 65 decibels. Determine how many time greater the washing machine's intensity of sound is.

Use the formula , β = 10log I/I 0 where β is the sound level in decibels, I is the intensity of sound you are measuring, and Io is the smallest sound intensity that can be heard by the human ear (roughly equal to 1 x 10^-12 watts/m2).

The washing machine's intensity of sound is (blank) times greater than that of the air conditioner.

Answer 1

the answer is 31.6 on PLATO

Answer 2

The decibels of the air conditioner is 50, therefore its intensity of sound, I₁, is given by

`10 log (I_(1))/(I_(0))=50 \\\\ log (I_(1))/(I_(0))=5`

The decibels of the washing machine is 65. Its intensity of sound, I₂, is given by

`10log (I_(2))/(I_(0))=65 \\\\ log (I_(2))/(I_(0))=6.5`

Therefore

`\frac{I_(1)}{I{0}}=10^(5) \\\\ (I_(2))/(I_(0))=10^(6.5) \\\\ (I_(2))/(I_(1))= (10^(6.5))/(10^(5))=10^(1.5)= 31.6`

Answer:
The washing machine's sound intensity is 31.6 times greater than that of the air conditioner.