An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
1 answer:
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.
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Answer:
E = 2k
Explanation:
Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.
We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.
Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀
tells us that the linear charge density is
λ = q_ {int} /l
q_ {int} = l λ
we substitute
E A = l λ /ε₀
is area of cylinder is
A = 2π r l
we substitute
E =
E =
the amount
k = 1 / 4πε₀
E = 2k