The sound intensity from a jack hammer breaking concrete is 2.00 W/m2 at a distance of 2.00 m from the point of impact. This is sufficiently loud to cause permanent hearing damage if the operator doesn't wear ear protection. What is the sound intensity for a person watching from "20".0 m away?

Answer :

Answer:

Explanation:

We know that intensity of sound is inversely proportional to square of distance of point from the source .

For two points at distance of R₁ and R₂ , Intensity of sound I₁ and I₂ will vary as follows

I₁ / I₂ = R₂² / R₁²

I₁ = 2 W / m² , R₁ = 2 m , R₂ = 20m , I₂ = ?

Putting the values in the expression above

2 / I₂ = 20² / 2²

I₂ = (2 x 4) / 400

= .02 W / m²

lublana

Answer:

[tex]0.02W/m^2[/tex]

Explanation:

We are given that

Sound intensity,[tex]I_0=2W/m^2[/tex]

Distance,r=2 m

We have to find the intensity of sound for a person watching from 20 m away.

r'=20 m

[tex]\frac{I}{I_0}=\frac{r^2}{r'^2}[/tex]

[tex]\frac{I}{I_0}=\frac{2^2}{(20)^2}[/tex]

[tex]I'=I_0(\frac{4}{400})[/tex]

[tex]I=2\times \frac{100}[/tex]

[tex]I=0.02 W/m^2[/tex]

Hence, the sound intensity for a person watching from 20 m away=[tex]0.02W/m^2[/tex]

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