An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the
1 answer:
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.
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Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ =
So,
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
- Distance=2.5m
- Time=9.1s
Average Velocity=Total Distance/Total Time