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Sagot :
The distance from observer A of intensity of sound 59 db is 28.64 m and the distance from observer B of intensity of sound 83 db is 11.36m
Explanation:
Let's solve this problem in parts
let's start by finding the intensity of the sound in each observer
observer A β = 59 db
β = [tex]10 log\frac{I_a}{I_o}[/tex]
where I₀ = [tex]1 \times 10^{-12}[/tex] W / m²
[tex]I_a = I_b \times 10 ^{ \frac{\beta}{10} }[/tex]
[tex]I_a = 1\times 10 ^{-12} \times 10 ^{ \frac{59}{10} }[/tex]
[tex]I_a[/tex] = [tex]2.51 \times 10 ^{-6}[/tex] W / m²
Similarly for Observer b [tex]I_b = 3.16 \times 10 ^ {-4}[/tex]
now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed
P = I A
therefore for the two observers
the area of a sphere is
[tex]A= 4\pi \times r^2[/tex]
we substitute the above formula, we get
Let us call the distance from the observer A be to stereo speaker = x, so the distance from the observer B to the stereo speaker = 40- x; we substitute
[tex]I_a (35 -x) ^2 = I_b x^2[/tex]
after solving the above equation we get x = 28.64 m
This is the distance of observer A
similarly The distance from observer B is 35 - x
= 40 - 28.64
= 11.36m
To know more about intensity of sound with the given link
#SPJ4
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