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 decibels of the air conditioner is 50, therefore its intensity of sound, I₁, is given by
$10 log \frac{I_{1}}{I_{0}}=50 \\\\ log \frac{I_{1}}{I_{0}}=5$

The decibels of the washing machine is 65. Its intensity of sound, I₂, is given by
$10log \frac{I_{2}}{I_{0}}=65 \\\\ log \frac{I_{2}}{I_{0}}=6.5$

Therefore
$\frac{I_{1}}{I{0}}=10^{5} \\\\ \frac{I_{2}}{I_{0}}=10^{6.5} \\\\ \frac{I_{2}}{I_{1}}= \frac{10^{6.5}}{10^{5}}=10^{1.5}= 31.6$

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

Answer 2

Answer:

the answer is 31.6 on PLATO