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larisa [96]
3 years ago
10

A guitar string with a linear mass density of 2.0 g/m is stretched between two vertical rods that are 65 cm apart. The string ho

lds a standing wave, fixed at both ends, with three antinodes when it oscillates at a frequency of 430 HzWhat is the frequency of the fifth harmonic of this string?
Physics
1 answer:
Rudiy273 years ago
7 0

Answer:

717 Hz

Explanation:

<u>solution:</u>

The wave with three antinodes has m = 3. Thus, f_3 = 3f_1 and so the fundamental frequency of the string is  

f_1 =f_3/3

    =430 Hz/3

    =143 Hz

Thus, the frequency of the fifth harmonic is

f_5 = 5*f_1

      = 5*143 Hz

     = 717 Hz

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Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate
vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

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Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

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Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>

5 0
3 years ago
Explain where most of the mass of an atom is located. Also, explain why some particles that make up the atom do not contribute m
Naddika [18.5K]
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Answer:

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