- #1

WhyHello

- 1

- 0

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.

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 ):

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: