Chapter 15, Problem 42A

(a)

To determine

To calculate: Speed of sound in air.

Answer to Problem 42A

Speed of sound in air at $30°\text{C}$ = $\underset{\_}{349\text{ m/s}}$

Explanation of Solution

Given:

Temperature = $30°\text{C}$

Calculation:

As per the relation between speed of sound and temperature, speed of sound increases by $0.6\text{ m/s}$ for every $1°\text{C}$ rise in temperature.

Speed of sound in air at $20°\text{C}$ = $343\text{ m/s}$

Therefore, for a $10°\text{C}$ rise in temperature, speed will increase by $0.6\times 10=6$ units.

Hence, speed of sound in air at $30°\text{C}$ = $343+6=349\text{ m/s}$

Conclusion:

Speed of sound in air at $30°\text{C}$ = $\underset{\_}{349\text{ m/s}}$

(b)

To determine

To calculate: The time taken for the baseball fan to hear the crack of the bat.

Answer to Problem 42A

Time taken for the baseball fan to hear the crack of the bat = $\underset{\_}{0.4\text{ s}}$ .

Explanation of Solution

Given:

Distance between the baseball fan and home plate = $152\text{ m}$

Formula used:

Time taken is given by the formula:

  $t=\frac{d}{v}$

Where $d$ is distance and $v$ is velocity.

Calculation:

Time taken is calculated as:

  $\begin{array}{l}t=\frac{d}{v}\\ =\frac{152\text{ m}}{349\text{ m/s}}\\ =0.4\text{ s}\end{array}$

Conclusion:

Time taken for the baseball fan to hear the crack of the bat = $\underset{\_}{0.4\text{ s}}$ .