The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.
<h3>
Velocity of the wave</h3>
The velocity of the wave is calculated as follows;
v = √T/μ
where;
- T is tension
- μ is mass per unit length = 2 g/m = 0.002 kg/m
v = √(50/0.002)
v = 158.1 m/s
<h3>First harmonic or fundamental frequency of the wave</h3>
f₀ = v/λ
where;
f₀ = v/2L
f₀ = 158.1/(2 x 0.6)
f₀ = 131.8 Hz
<h3>Second harmonic of the wave</h3>
f₁ = 2f₀
f₁ = 2(131.8 Hz)
f₁ = 263.6 Hz
<h3>Third harmonic of the wave</h3>
f₂ = 3f₀
f₂ = 3(131.8 Hz)
f₂ = 395.4 Hz
Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.
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Answer:
D
Explanation:
If it is of very high intensity it will be 85-100%
Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed
Vf^2=Vi^2+2ad
Vf=2ad square rooted
Vf=2(9.81)(44) square root it
Vf=29.3m/s
Answer:
3.536*10^-6 C
Explanation:
The magnitude of the charge is expresses as Q = CV
C is the capacitance of the capacitor
V is the voltage across the capacitor
Get the capacitance
C = ε0A/d
ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m
A is the area = 0.2m²
d is the plate separation = 0.1mm = 0.0001m
Substitute
C = 8.84 x 10-12 * 0.2/0.0001
C = 1.768 x 10-8 F
Get the potential difference V
Using the formula for Electric field intensity
E = V/d
2.0 × 10^6 = V/0.0001
V = 2.0 × 10^6 * 0.0001
V = 2.0 × 10^2V
Get the charge on each plate.
Q = CV
Q = 1.768 x 10-8 * 2.0 × 10^2
Q = 3.536*10^-6 C
Hence the magnitude of the charge on each plate should be 3.536*10^-6 C
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