Question

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as L = 10 log StartFraction I Over I 0 EndFraction, where I 0 = 10 Superscript negative 12 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?
One-third times louder

3 times louder

30 times louder

90 times louder

Answer 1

Answer:

$db_1 = 10 log_{10} (\frac{10^{-9}}{10^{-12}})= 30 db$

And using the same formula we can find the number of decible for the Braylee's music:

$db_2 = 10 log_{10} (\frac{10^{-3}}{10^{-12}})= 90 db$

And since we want to find how many times louder is Braylee’s music than Jessica’s we can find the following relationship:

$\frac{db_2}{db_1} = \frac{90 dB}{30 dB}= 3$

So then we can conclude that the Braylee's music is 3 times louder then the Jessica's sound. And the best option would be:

3 times louder

Step-by-step explanation:

For this case we know that the loudness measured in decibels is given by this formula:

$dB = 10 log_{10} (\frac{I}{I_o})$

Where $I_o = 10^{-12} W/m^2$. Using this formula we can find the decibels for the sound for Jessica:

$db_1 = 10 log_{10} (\frac{10^{-9}}{10^{-12}})= 30 db$

And using the same formula we can find the number of decible for the Braylee's music:

$db_2 = 10 log_{10} (\frac{10^{-3}}{10^{-12}})= 90 db$

And since we want to find how many times louder is Braylee’s music than Jessica’s we can find the following relationship:

$\frac{db_2}{db_1} = \frac{90 dB}{30 dB}= 3$

So then we can conclude that the Braylee's music is 3 times louder then the Jessica's sound. And the best option would be:

3 times louder