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3 years ago

How does Brandon's sound intensity level compare to ahmad's?

Mathematics

1 answer:

sergejj [24]3 years ago

7 0

we know that

$L=10log(\frac{I}{I0} )$

Find the value of L for Brandon

$L=10log(\frac{10^{-10}}{10^{-12}} )$

$L=10*2=20$

<u>Find the value of L for Ahmad</u>

$L=10log(\frac{10^{-4}}{10^{-12}} )$

$L=10*8=80$

therefore

<u>the answer is</u>

Brandon's sound intensity level is (1/4) the level of Ahmad

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