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A thunderclap sends a sound wave through the air and the ocean below. The thunderclap sound wave has a constant frequency of [tex]$20 \, \text{Hz}$[/tex]. What is the wavelength of the sound wave in water? (The equation for the speed of a wave is [tex]v = f \times \lambda[/tex].)

[tex]\[
\begin{tabular}{|l|c|c|c|c|}
\hline
& Water & Diamond & Glass & Air \\
\hline
\begin{tabular}{c}
Speed of \\
sound \\
[tex]$( \text{m/s} )$[/tex]
\end{tabular}
& 1,493 & 12,000 & 5,640 & 346 \\
\hline
\end{tabular}
\][/tex]

A. [tex]$54.3 \, \text{m}$[/tex]
B. [tex]$74.7 \, \text{m}$[/tex]
C. [tex]$64.5 \, \text{m}$[/tex]
D. [tex]$17.3 \, \text{m}$[/tex]



Answer :

GinnyAnswer
To solve the problem of finding the wavelength of a sound wave in water given certain parameters, we can use the wave speed equation. The equation is:

[tex]\[ v = f \times \lambda \][/tex]

where:
- \( v \) is the speed of the sound wave in the medium (in this case, water),
- \( f \) is the frequency of the sound wave,
- \( \lambda \) is the wavelength of the sound wave.

Given data for water:
- The speed of sound in water \( v = 1493 \, \text{m/s} \)
- The frequency of the sound wave \( f = 20 \, \text{Hz} \)

We need to find the wavelength \( \lambda \). Rearrange the equation to solve for \( \lambda \):

[tex]\[ \lambda = \frac{v}{f} \][/tex]

Now, substituting the given values into the equation:

[tex]\[ \lambda = \frac{1493 \, \text{m/s}}{20 \, \text{Hz}} \][/tex]

After performing the division:

[tex]\[ \lambda = \frac{1493}{20} \][/tex]
[tex]\[ \lambda = 74.65 \, \text{m} \][/tex]

Thus, the wavelength of the sound wave in water is \( \lambda = 74.65 \, \text{m} \).

From the given options:
A. \( 54.3 \, \text{m} \)
B. \( 74.7 \, \text{m} \)
C. \( 64.5 \, \text{m} \)
D. \( 17.3 \, \text{m} \)

The closest matching option is \( 74.7 \, \text{m} \).

Therefore, the correct answer is:
[tex]\[ \boxed{74.7 \, \text{m}} \][/tex]

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