A sound absorber attenuates the sound level b | vedclass

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(A) The change in sound level $\Delta L$ in decibels $(dB)$ is given by the formula: $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Given th

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$100$

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(A) The change in sound level $\Delta L$ in decibels $(dB)$ is given by the formula: $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Given that the sound level is attenuated by $20\, dB$,we have $\Delta L = 20\, dB$.Substituting the values into the formula: $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Dividing by $10$,we get: $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.Converting the logarithmic form to exponential form: $10^2 = \frac{I_1}{I_2}$.Therefore,$\frac{I_1}{I_2} = 100$,which means the intensity decreases by a factor of $100$.

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

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