It is defined that a string that is in its two ends, its fundamental frequency and its subsequent frequencies are the product of the whole depending on the number of the harmonic. In other words, harmonics (higher) define the subsequent frequencies under the functions 2f, 3f, 4f, 5f, etc.
Therefore we have that the higher harmonics would be:
1 x 80 Hz = 80 Hz (1st harmonic and Fundalmental Frequency)
2 x 80Hz = 160Hz (2nd harmonic)
3 x 80Hz = 240Hz (3rd harmonic)
4 x 80Hz = 320Hz (4th harmonic)
5 x 80Hz = 400Hz (5th harmonic)
Hence, the frequencies 160Hz and 240Hz are the two higher harmonics of string with a fundamental frequency of 80Hz.
Answer:
a) V = k 2π σ (√(b² + x²) - √ (a² + x²))
,
b) E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))
Explanation:
a) The expression for the electric potential is
V = k ∫ dq / r
For this case, consider the disk formed by a series of concentric rings of radius r and width dr, the distance of each ring to point P
R = √(x² + r²)
The charge on a ring is
σ = dq / dA
The area of a ring is
A = π r
dA = 2π r dr
So the charge is
dq = σ 2π r dr
We substitute
V = k σ 2pi ∫ r dr / √(r² + x²)
We integrate
V = k 2π σ √(r² + x²)
We evaluate from the lower limit r = a to the upper limit r = b
V = k 2π σ (√(b² + x²) - √ (a² + x²))
b) the electric field and the potential are related
E = - dV / dx
E = - k 2π σ (1/2 2x /√(b² + x²) - ½ 2x /√(a² + x²))
E = - k 2π σ x (1 /√(b² + x²) - 1 /√(a² + x²))
Answer:
1/60 mps
Explanation:
We would first have to divide 60 by 60 because there is 60mins per hour to get 1mpm. After that we would have to divide 1 by 60 because there are 60 secs in a min. So our final answer after doing 1/60 would be a fraction.
Answer:
d) What is the force if we doubled both the masses AND we doubled the distance
