A sound absorber attenuates (decreases) the s | vedclass

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(A) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the ref

Vedclass · Accepted Answer

$100$

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(A) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.Given the change in sound level $\Delta L = L_1 - L_2 = 20 \ dB$.Using the formula $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$,$20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$,$2 = \log_{10} \left( \frac{I_1}{I_2} \right)$,$\frac{I_1}{I_2} = 10^2 = 100$.Thus,the intensity decreases by a factor of $100$.

$A$ sound absorber attenuates (decreases) the sound level by $20 \ dB$. The intensity decreases by a factor of

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