Answer:
fb = 240.35 Hz
Explanation:
In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.
Open tube:
(1)
vs: speed of sound = 343m/s
L: length of the open tube = 0.47328m
You replace in the equation (1):
Closed tube:
L': length of the closed tube = 0.702821m
Next, you use the following formula for the beat frequency:
The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz
Current at all points of a series circuit must be the same, because there's no place in the circuit where electrons are being manufactured, and no place where they're leaking out and falling on the floor. The nimber of electrons that leaves the loop is the same number that entered it.
I'm not sure what is nmeant by "p.d. remains different" .
Answer:
|Δf| = 37.3 kHz
Explanation:
given,
peak velocity = 4 m/s
speed of the sound = 1500 m/s
frequency = 7 MHz
= 37.3 kHz
|Δf| = 37.3 kHz
hence, frequency shift between the opening and closing valve is 37.3 kHz
Answer:
v = 3.27 m/s
Explanation:
KE = 1/2 mv^2
695 J = 1/2 (130kg)(v^2)
695 J / (1/2 x 130kg) = v^2
v^2 = square root of 10.69
v = 3.27 m/s
