Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.