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Andrei [34K]
2 years ago
14

An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni

c? (assume the velocity of sound is 344 m/s.)
Physics
1 answer:
Marysya12 [62]2 years ago
8 0

When air is blown into the open pipe,

L = \frac{nλ}{2}

where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation

⇒λ=\frac{2L} {n}

Note here that n=1 is for fundamental, n=2 is first harmonic and so on..

⇒ third harmonic will be n=4

Given L=6m, n=4, solving for λ we get:

λ=\frac{(2)*(6)}{4} =3m

Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:

c=f.λ Or f= \frac{c}{λ}

⇒f=\frac{344}{3}

≈115 Hz

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Answer:

3. none of these

Explanation:

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For disk 1:

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For disk 2:

K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2

so the total energy is

K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2

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Answer:

C = 3.77*10⁻¹⁰ F = 377 pF

Q = 1.13*10⁻⁵ C

Explanation:

Given

D = 8.0 cm = 0.08 m

d = 0.95 cm = 0.95*10⁻² m

k = 80.4  (dielectric constant of the milk)

V = 30000 V

C = ?

Q = ?

We can get the capacitance of the system applying the formula

C = k*ε₀*A / d

where

ε₀ = 8.854*10⁻¹² F/m

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then

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Now, we use the following equation in order to obtain the charge on each plate when they are fully charged

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⇒  Q = (3.77*10⁻¹⁰ F)*(30000 V)

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2 years ago
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