Chapter 15, Problem 45A

a.

To determine

To calculate: The depth of water.

Answer to Problem 45A

The depth of water = $\underset{\_}{\text{1333}\text{.71 m}}$ .

Explanation of Solution

Given:

Temperature of seawater = $25°\text{C}$

Time taken for the reflection off the mud to reach the ship = $1.74\text{ s}$ .

Formula used:

The distance for an echo is given by,

$d=\frac{1}{2}vt$

Where $v$ is velocity and $t$ is time taken.

Calculation:

Speed of sound in seawater( $25°\text{C}$ ) = $1533\text{ m/s}$

Distance is calculated as:

$\begin{array}{l}d=\frac{1}{2}vt\\ \text{ =}\frac{1533\text{ m/s}\times 1.74\text{ s}}{2}\\ \text{ = }1333.71\text{ m}\end{array}$

Conclusion:

The depth of water = $\underset{\_}{\text{1333}\text{.71 m}}$ .

b.

To determine

To calculate: Thickness of the mud.

Answer to Problem 45A

Thickness of the mud = $\underset{\_}{581.25\text{ m}}$ .

Explanation of Solution

Given:

Speed of sound in mud = $1875\text{ m/s}$

Time taken for the reflection off the bedrock to reach the ship = $2.36\text{ s}$ .

Formula used:

The distance for an echo is given by,

$d=\frac{1}{2}vt$

Where $v$ is velocity and $t$ is time taken.

Calculation:

Difference between the time for the reflection to reach the ship off the bedrock and to reach the ship off the mud = $2.36-1.74=0.62\text{ s}$

Therefore, sound takes $0.62\text{ s}$ to travel through the mud twice.

Distance is calculated as:

$\begin{array}{l}d=\frac{1}{2}vt\\ \text{ =}\frac{1875\text{ m/s}\times 0.62\text{ s}}{2}\\ \text{ = 581}\text{.25 m}\end{array}$

Conclusion:

Thickness of the mud = $\underset{\_}{581.25\text{ m}}$ .