Answer:
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Explanation:
Answer:
the frequency of the second harmonic of the pipe is 425 Hz
Explanation:
Given;
length of the open pipe, L = 0.8 m
velocity of sound, v = 340 m/s
The wavelength of the second harmonic is calculated as follows;
L = A ---> N + N--->N + N--->A
where;
L is the length of the pipe in the second harmonic
A represents antinode of the wave
N represents the node of the wave

The frequency is calculated as follows;

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.
Since the circuit is incomplete or not closed, no current flows in the circuit. as per ohm's law , Voltage is directly proportional to current and is given as
V = Voltage = i R where i = current , R = resistance
as no current flows in the circuit, i = 0
the resistance R can not be zero. hence
V = 0 (R)
V = 0 Volts
so the magnitude of the Voltage is zero Volts