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maria [59]
3 years ago
9

PLEASE HURRY IM TIMED!!!

Mathematics
1 answer:
andriy [413]3 years ago
5 0

<u>Answer-</u>

<em>A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.</em>

<u>Solution-</u>

Given that, loudness measured in dB is

L=10\log \frac{I}{I_0}

Where,

I   = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's  = 10⁻⁴

Then,

L_1=10\log \frac{I_1}{I_0}=10\log \frac{10^{-10}}{10^{-12}}=10\log \frac{1}{10^{-2}}=10\log 10^{2}=2\times10\log 10=20

L_2=10\log \frac{I_2}{I_0}=10\log \frac{10^{-4}}{10^{-12}}=10\log \frac{1}{10^{-8}}=10\log 10^{8}=8\times10\log 10=80

\therefore \frac{L_1}{L_2} =\frac{20}{80} =\frac{1}{4}

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