Answer:
option D
Explanation:
given,
length of the pipe, L = 0.96 m
Speed of sound,v = 345 m/s
Resonating frequency when both the end is open

n is the Harmonic number
2nd overtone = 3rd harmonic
so, here n = 3
now,

f = 540 Hz
The common resonant frequency of the string and the pipe is closest to 540 Hz.
the correct answer is option D
Answer:
(a) 11.8692 ohm
(b) 12.447 A
(c) 17.6 A
Explanation:
a) inductive reactance Z = L Ω
= L x 2π x F
= 45.0 x 10⁻³ x 2(3.14) x 42
= 11.8692 ohm
b) rms current
= 100 / 8.034
= 12.447 A
c) maximum current in the circuit
= I eff x rac2
= 12.447 x 1.414
= 17.6 A
Well you need to have lots of heat
<span>The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the "Radio waves"
So, option B is your answer
Hope this helps!
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