Question #2e09c

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Question #2e09c

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Answer 1 · 2017-04-18T11:21:55

$3.98 \times 10^{- 4} W m^{- 2}$ $3.98xx10^-4Wm^-2$

Explanation:

Threshold of Hearing (TOH) is $0 d B$ $0dB$ and taken as $= 1 \times 10^{- 12} W m^{- 2}$ $=1xx10^-12 Wm^-2$

Let $86.0 d B = x W m^{- 2}$ $86.0dB=xWm^-2$

Implies that
$10 log (\frac{x}{1 \times 10^{- 12}}) = 86.0$ $10 log(x/(1xx10^-12))=86.0$
$\Rightarrow log (\frac{x}{1 \times 10^{- 12}}) = 8.60$ $=> log(x/(1xx10^-12))=8.60$
Taking $anti log$ $"anti "log$
$\Rightarrow \frac{x}{1 \times 10^{- 12}} = 3.98 \times 10^{8}$ $=> x/(1xx10^-12)=3.98xx10^8$
$\Rightarrow x = 3.98 \times 10^{8} \times 1 \times 10^{- 12}$ $=> x=3.98xx10^8xx1xx10^-12$
$\Rightarrow x = 3.98 \times 10^{- 4} W m^{- 2}$ $=> x=3.98xx10^-4Wm^-2$

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Question #2e09c

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