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k0k
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A submarine sonar system sends a burst of sound with a frequency of 325Hz. The sound wave bounces off an underwater rock face and returns to the submarine in 8.50s. If the wavelength of the sound is 4.71m, how far away is the rock face? (Ans: 6.51km )
--
v=fλ
λ=wavelength
f=frequency
v=velocity
--
v=fλ
v=325Hz X 4.71m
=1530.75m/s
v=d/t
d=vt
= 1530.75m/s X 8.50s
=13011.375m
=13011.375m/2
=6505.6875m
=6.51km.
What exactly am I doing wrong here? ..
--
v=fλ
λ=wavelength
f=frequency
v=velocity
--
v=fλ
v=325Hz X 4.71m
=1530.75m/s
v=d/t
d=vt
= 1530.75m/s X 8.50s
=13011.375m
=13011.375m/2
=6505.6875m
=6.51km.
What exactly am I doing wrong here? ..
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