A sound absorber attenuates (decreases) the s | vedclass

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Question

$\mathrm{L}_{1}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}\right) ; \quad \mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\

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Vedclass · Answer

$\mathrm{L}_{1}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}ight) ; \quad \mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}ight)$

$	herefore \mathrm{L}_{1}-\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}ight)-10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}ight)$

or $\quad \Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}ight) \quad$ or $\quad 20 \mathrm{dB}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}ight)$

or $10^{2}=\frac{I_{1}}{I_{2}}$ or $I_{2}=\frac{I_{1}}{100}$

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

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