the loudness, l, measured in decibels (db), of a sound intensity, i, measured in watts per square meter, is…

the loudness, l, measured in decibels (db), of a sound intensity, i, measured in watts per square meter, is defined as l = 10log\frac{i}{i_0}, where i_0 = 10^{-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 braylees music than jessicas?\no \frac{1}{3} times louder\no 3 times louder\no 30 times louder\no 90 times louder

the loudness, l, measured in decibels (db), of a sound intensity, i, measured in watts per square meter, is defined as l = 10log\frac{i}{i_0}, where i_0 = 10^{-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 braylees music than jessicas?\no \frac{1}{3} times louder\no 3 times louder\no 30 times louder\no 90 times louder

Answer

Answer:

C. 30 times louder

Explanation:

Step1: Calculate loudness of Jessica's music

Use the formula $L = 10\log\frac{I}{I_0}$. Given $I_{Jessica}=10^{-9}$ and $I_0 = 10^{-12}$, then $L_{Jessica}=10\log\frac{10^{-9}}{10^{-12}}=10\log(10^{-9 + 12})=10\log(10^{3}) = 10\times3=30$ decibels.

Step2: Calculate loudness of Braylee's music

Given $I_{Braylee}=10^{-3}$ and $I_0 = 10^{-12}$, then $L_{Braylee}=10\log\frac{10^{-3}}{10^{-12}}=10\log(10^{-3+12})=10\log(10^{9})=10\times9 = 90$ decibels.

Step3: Find the difference in loudness

To find how many times louder Braylee's music is than Jessica's, calculate $\frac{L_{Braylee}-L_{Jessica}}{10}=\frac{90 - 30}{10}=6$. But we can also calculate the ratio of intensities directly. $\frac{I_{Braylee}}{I_{Jessica}}=\frac{10^{-3}}{10^{-9}}=10^{-3+9}=10^{6}$. In terms of decibel - difference, $L_{Braylee}-L_{Jessica}=90 - 30=60$ decibels. Since each 10 - decibel increase means 10 times louder, $60\div10 = 6$ intervals of 10 decibels, so $10^6\div10^3 = 10^{3}$ times in intensity and in terms of decibel - perception, $\frac{90 - 30}{10}=6$ intervals of 10 decibels which means 30 times louder (because $10^{\frac{90 - 30}{10}}=10^{6}$ in intensity ratio and considering the decibel scale relationship). Another way: The difference in decibels $\Delta L=10\log\frac{I_{Braylee}}{I_0}-10\log\frac{I_{Jessica}}{I_0}=10\left(\log\frac{I_{Braylee}}{I_0}-\log\frac{I_{Jessica}}{I_0}\right)=10\log\frac{I_{Braylee}/I_0}{I_{Jessica}/I_0}=10\log\frac{I_{Braylee}}{I_{Jessica}}$. We know $\Delta L = 90 - 30=60$ decibels. Then $60 = 10\log\frac{I_{Braylee}}{I_{Jessica}}$, so $\log\frac{I_{Braylee}}{I_{Jessica}} = 6$, and $\frac{I_{Braylee}}{I_{Jessica}}=10^{6}$. In terms of the multiple of loudness perception (since every 10 - decibel increase is 10 times louder), $\frac{90 - 30}{10}=6$ intervals of 10 decibels, and the multiple of loudness is 30 times louder.