You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin
1 answer:
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Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone,
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
Answer:
The current in the wire is 31.96 A.
Explanation:
The current in the wire can be calculated as follows:
<u>Where</u>:
q: is the electric charge transferred through the surface
t: is the time
The charge, q, is:
<u>Where</u>:
n: is the number of electrons = 7.93x10²⁰
e: is the electron's charge = 1.6x10⁻¹⁹ C
Hence, the current in the wire is:
Therefore, the current in the wire is 31.96 A.
I hope it helps you!