A sound absorber attenuates the sound level b | 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.Let $L_1$ and $L_2$ be the initial and final sound levels,and $I_1$ and $I_2$ be the corresponding intensities.The change in sound level is given by $\Delta L = L_1 - L_2 = 20\, dB$.Using the formula: $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_0} \right) - 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Substituting the given value: $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Dividing by $10$: $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.Taking the antilog: $\frac{I_1}{I_2} = 10^2 = 100$.Therefore,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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