Question

A car horn produces a sound intensity level of 90 db at 10 ft away, find the intensity level at 80ft way

Answer 1

it should be 7200 according to me

Answer 2

Given:

Initial intensity of sound $I=90 db$

Initial distance $d=10ft$

Final distance $d'=80ft$

To find:

Final intensity $I'=?$

Formula:

$\frac{I}{I'}=\frac{d'^{2} }{d^{2} }$

Solution:

Step 1 of 1

Intensity of sound is defined as the ratio of power of sound to the area sound is distributed.

Intensity of sound is related to distance by inverse square law.

It is stated that intensity of sound is inversely proportional to the square of the distance of observer.

$I$∝$d^{2}$

So, if the intensity at 10 ft distance is 90 db, then the intensity at 80 ft away will be

$\frac{I}{I'}=\frac{d'^{2} }{d^{2} }\\\frac{90}{I'}=\frac{80^{2} }{10^{2} }$

Then, the intensity at 80 ft away will be

$I'=\frac{(90)(100)}{6400} \\I'=1.41 db$

So, the intensity at 80 ft away is 1.41 db.

Final Answer:

The intensity at 80 ft away is 1.41 db.