if a copper coin with a volume of exactly 50 cubic cm is dropped into the volume of water 4cubic cm,how much water will be overf
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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
Answer:
force becomes one - ninth
Explanation:
According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.
Here both the charges remains same but the distance is variable.
So, we can say that
.... (1)
Where d be the distance between the tow charges
As the distance between two charges increases by factor of three, let the new force be F'.
.... (2)
Divide equation (2) by equation (1), we get
Thus, the force becomes one - ninth times the initial force.