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A sound differs by 6 dB from a sound of intensity equal to `10(nW)/(cm^2)`. Find the absolute value of intensity of the sound. |
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Answer» Intensity level `=log_(10)((I_2)/(I_1))bels=10log((I_2)/(I_1))dB` here, `I_1=10xx10^-9xx10^4=10^-4(W)/(m^2)` `6=10log((I_2)/(10^-4))` or `I_2=(10^(0.6))/(10^4)=(3.98)/(10^4)=4.0xx10^-4(W)/(m^2)` |
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