Explanation:
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What property is used to distinguish the layers of the atmosphere?
2 answers:
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Hi there,
for this question we have:
Signal 2.0 MHz = Emitted so we can call it
and we need the Reflected =
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz =
so we have:
- =
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz =
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=>
= (lambda)(
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day
for this question we have:
Signal 2.0 MHz = Emitted so we can call it
and we need the Reflected =
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz =
so we have:
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz =
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=>
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day