Question

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as where and is the least intense sound a human ear can hear. Jessica is listening to soft music at a sound intensity level of 10-9 on her computer while she does her homework. Braylee is completing her homework while listening to very loud music at a sound intensity level of 10-3 on her headphones. How many times louder is Braylee's music than Jessica's?

Answer 1

Answer:

Barlyee's music is 3 times louder than Jessica's.

Step-by-step explanation:

We know that, loudness is measured as,

$L = 10 \times log \frac{I}{I_o}$

So let us measure the loudness of music that Jessica is listening to,  

$L = 10 \times log \frac{10^{-9}}{10^{-12}}=10 \times log {10^3}= 10 \times 3 log 10= 30 dB$

So Jessica is listening to music at 30dB.

And the loudness of music that Barylee is listening to,

$L = 10 \times log \frac{10^{-3}}{10^{-12}}= 10 \times log {10^9}= 10 \times 9 log 10= 90 dB$

So Barylee is listening to music at 90dB.

$90=3 \times 30$

Therefore, the music that Barylee is listening to, is 3 times louder than the music Jessica is listening to.

Answer 2

Answer:

Taking the course on edge? Don't sweat it brotha.

Step-by-step explanation:

The answer is B. 3 times louder