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3 years ago

Helpppppppppppppppppppppppppppppp

Mathematics

1 answer:

jek_recluse [69]3 years ago

5 0

The first one hopes this helps

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Answer-

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

Solution-

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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...

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