An air-filled pipe is found to have successive harmonics at 945 Hz , 1215 Hz , and 1485 Hz . It is unknown whether harmonics bel
ow 945 Hz and above 1485 Hz exist in the pipe. What is the length of the pipe
1 answer:
Answer:
L = 0.635m
Explanation:
This problem involves the concept of stationary waves in pipes. For pipes closed at one end,
The frequency f = nv/4L for n = 1,3,5....n
For pipes open at both ends
f = nv/2L for n = 1,2,3,4...n
Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.
The film solution can be found in the attachment below.
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Answer:
a. q2 = 16.4μC, positive charge
b. F = 0.900N
c. downward
Explanation:
a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:
(1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r: distance between the charges = 0.300m
q1: charge 1 = -0.550 μC = 0.550*10^-6C
q2: charge 2 = ?
Fe: electric force = 0.900N
The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.
You solve the equation (1) for the second charge ans replace the values of the other parameters:
The values of the second charge is 1.64 μC
b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.
The force exerted on the first charge is 0.900N
c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.
Answer:
Explanation:
First let's find the electric potential using y = 22.5:
Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y: