A speaker creates uniformly spherical sounds w/ 500 watts of power a)What is the intensity, I, of the sound at a distance of 20 meters from the speaker? What is the sound intensity level, β, of the sound at a distance of 20 meters from the speaker?

b) What is the intensity, I, of the sound at a distance of 10 meters from the speaker? What is the sound intensity level, β, of the sound at a distance of 10 meters from the speaker?

c) How many deciBels do you increase walking from 20 meters away from the speaker to 10 meters away from the speaker?

c) What power output would the speaker have to put out in order to create 100 dB at a distance 20 meters from the speaker?

A) I = 0.09947 W , β = 109 db , B) β = 116 db , β = 116 db , c) Δβ = 7 dB,

D)  P = 50.27 W

Explanation:

A) The intensity of a spherical sound wave is

          I = P / A

where A is the area of the sphere where the sound is distributed

          A = 4π R²

we substitute

          I = P / 4πR²

let's calculate

          I = 500 / (4π 20²)

          I = 0.09947 W

to express this quantity in decibels we use relate

         β = 10 log (I / I₀)

The detectivity threshold is I₀ = 1 10⁻¹² W / m²

         β = 10 lob (0.09947 / 10⁻¹²)

         β = 10 (10.9976)

         β = 109 db

B) intensity at r = 10m

          I = 500 / (4π 10²)

          I = 0.3979 W / m²

          β = 10 log (0.3979 / 10⁻¹²)

          β = 10 (11.5997)

          β = 116 db

C) the change in intensity in decibles is

          Δβ = β₁ - β₂

          Δβ = 116 - 109

          Δβ = 7 dB

D) let's find the intensity for 100 db

          I = I₀ 10 (β / 10)

          I = 10⁻¹² 10 (100/10)

          I = 10⁻² W / m²

Thus

           P = I A

           P = I 4π R²

           P = 10⁻² 4π 20²

           P = 50.27 W

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this

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