Suggested languages for you:

Americas

Europe

Q17PE

Expert-verified
Found in: Page 629

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Show that an intensity of $${10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}$$ is the same as $${10^{ - 16}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}$$ ?

The intensities are equal.

See the step by step solution

## Step1: Given Data

The intensity of the first sound is $${10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}$$

The another intensity is $${10^{ - 16}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}$$.

## Step 2:  Concept of sound intensity

The sound intensity is the power passing perpendicular to the given per second. The intensity has the unit Watt per meter square.

## Step 3: Calculation of the intensities

The intensity of the first sound is,

\begin{align}{10^{ - 12}}\;W/{m^2}\\ &= \frac{{{{10}^{ - 12}}\;{\rm{W}}}}{{100 \times 100\;{\rm{c}}{{\rm{m}}^{\rm{2}}}}}\\ &= {10^{ - 12}} \times {10^{ - 4}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}\\ &= {10^{ - 16}}\;{\rm{W/c}}{{\rm{m}}^{\rm{2}}}\end{align}

Hence, both the sounds have the same intensity.