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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].)

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

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



Answer :

GinnyAnswer
To determine the wavelength of the sound wave in water, we will use the given wave speed equation:
[tex]\[ v = f \times \lambda \][/tex]

Here:
- [tex]\(v\)[/tex] is the speed of sound in water.
- [tex]\(f\)[/tex] is the frequency of the sound wave.
- [tex]\(\lambda\)[/tex] is the wavelength of the sound wave.

From the problem statement, we know:
- The frequency [tex]\(f\)[/tex] of the sound wave is [tex]\(20 \, \text{Hz}\)[/tex].
- The speed of sound [tex]\(v\)[/tex] in water is [tex]\(1,493 \, \text{m/s}\)[/tex].

We need to solve for the wavelength [tex]\(\lambda\)[/tex]. Rearranging the equation [tex]\( v = f \times \lambda \)[/tex] to solve for [tex]\(\lambda\)[/tex], we get:
[tex]\[ \lambda = \frac{v}{f} \][/tex]

Substituting the known values:
[tex]\[ \lambda = \frac{1493 \, \text{m/s}}{20 \, \text{Hz}} \][/tex]

Calculating the value:
[tex]\[ \lambda = 74.65 \, \text{m} \][/tex]

So, the wavelength of the sound wave in water is approximately [tex]\(74.65 \, \text{m}\)[/tex].

Among the given options, the closest value to [tex]\(74.65 \, \text{m}\)[/tex] is:
[tex]\[ \boxed{74.7 \, \text{m}} \][/tex]

Thus, the correct answer is option:
A. [tex]\(74.7 \, \text{m}\)[/tex]

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