How far away is the rock face? (Wavelengths)

In summary, the wavelength of the sound wave is 4.71 m and it travels at a speed of 1530.75m/s. The sound is reflected back to the submarine in 8.50s, which means it takes 4.25s to reach the underwater rock face and back. By using the formula v = λ/T, we can calculate that the rock face is approximately 6.51 x 10^3m away from the submarine.
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1. A submarine sonar system sends a burst of sound with a frequency of 325 Hz. The sound waves bounces off an underwater rock face and returns to the submarine in 8.50s. if the wavelength of the sound wave is 4.71 m, how far away is the rock face? The answer is 6.51 x 10^3m.



I keep trying this question different ways but the only thing that I believe I'm doing correct so far is:
v=f[tex]\lambda[/tex]
v=(325Hz)(4.71m)
v=1530.75m/s
I also know that 8.50s must be divided in two, so that it will equal 4.25s.
Than I go to solve how far away is the rock...
v=[tex]\lambda[/tex]/T
[tex]\lambda[/tex]=vT
[tex]\lambda[/tex]= (1530.75m/s)(4.25s)
[tex]\lambda[/tex]= 6505
Never mind i figured out that i made a calculator error. Sorry ):
 
Last edited:
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Glad it worked out. Welcome to Physics Forums!

By the way, if you use [noparse][itex] instead of [tex][/noparse], it won't put the symbol on a new line all by itself. Or, for "λ", you can just copy-and-paste from the symbols list at the bottom of this post.
 

1. How do you measure the distance to a rock face in wavelengths?

To measure the distance to a rock face in wavelengths, you would need to know the wavelength of the light being used and the angle of incidence of the light hitting the rock face. Then, you can use the formula d = λ/2tanθ, where d is the distance to the rock face, λ is the wavelength, and θ is the angle of incidence.

2. What is the relationship between distance and wavelength?

The relationship between distance and wavelength is inversely proportional. This means that as the distance increases, the wavelength decreases. This relationship is described by the formula d = λ/2tanθ, where d is the distance, λ is the wavelength, and θ is the angle of incidence.

3. Why is it important to know the distance to a rock face in wavelengths?

Knowing the distance to a rock face in wavelengths is important because it can help us determine the properties of the rock face, such as its texture and composition. It can also help us calculate the angle of incidence and reflectance, which are important for studying the behavior of light and how it interacts with different surfaces.

4. Can the distance to a rock face be measured in any other units besides wavelengths?

Yes, the distance to a rock face can be measured in other units such as meters, feet, or kilometers. However, using wavelengths as a unit of measurement is beneficial in certain situations, especially when studying the behavior of light, as it allows for more precise calculations and comparisons.

5. Is the distance to a rock face always measured in wavelengths?

No, the distance to a rock face can be measured in various units, depending on the purpose of the measurement. For example, if the distance is being measured for navigation or construction purposes, it would be more practical to use units like meters or feet. However, for scientific purposes, measuring the distance in wavelengths may be more useful and accurate.

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