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

B: 3times louder is correct シ

Explanation:

good luck

Answer 2

`db_1 = 10 log_(10) ((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) ((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:

`(db_2)/(db_1) = (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

Explanation:

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

`dB = 10 log_(10) ((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) ((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) ((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:

`(db_2)/(db_1) = (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