Mathematics: The sound level measured in a room by a person watching a movie on a home theater system varies from 60 dB during a quiet part to 100 dB during a loud part. Approximately how many times louder is the latter sound

The answer is Bandwagon.Explanation:I just took the quiz on APEX :)...

The sound level measured in a room by a person watching

nana55959

29.01.2022

The sound in a loud part of the room is 10000 times louder than sound in a quiet part of the same place.

Step-by-step explanation:

The acoustic intensity sound is a logarithmic function whose form is:

$L = 10\cdot \log_{10}\left(\frac{I}{I_{o}} \right)$

Where:

- Acoustic intensity sound, measured in decibels.

$I_{o}$ - Reference sound intensity, measured in watts per square meter.

- Real sound intensity, measured in watts per square meter.

Sound intensity is now cleared:

$10^{\frac{L}{10} } = \frac{I}{I_{o}}$

The ratio of the sound intensity in a loud part to the sound intensity in a quiet part is:

$\frac{I_{100}}{I_{60}} = \frac{10^{\frac{100\,dB}{10} }}{10^{\frac{60\,dB}{10}}}$

$\frac{I_{100}}{I_{60}} = \left(10^{100\,dB-60\,dB}\right)^{\frac{1}{10} }$

$\frac{I_{100}}{I_{60}} = (10^{40\,dB})^{\frac{1}{10} }$

$\frac{I_{100}}{I_{60}} =10^{4}$

The sound in a loud part of the room is 10000 times louder than sound in a quiet part of the same place.

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