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exis [7]
3 years ago
15

What best describes the difference between the second sound wave and the first

Physics
1 answer:
Marysya12 [62]3 years ago
3 0

Second sound is a quantum mechanical phenomenon in which heat transfer occurs by wave-like motion, rather than by the more usual mechanism of diffusion. This leads to a very high thermal conductivity. It is known as "second sound" because the wave motion of heat is similar to the propagation of pressure waves in air (sound).


Normal sound waves are fluctuations in the density of molecules in a substance; second sound waves are fluctuations in the density of particle-like thermal excitations (rotons and phonons[1]). Second sound can be observed in any system in which most phonon-phonon collisions conserve momentum. This occurs in superfluids,[2] and also in some dielectric crystals[3][4][5] when Umklapp scattering is small. (Umklapp phonon-phonon scattering exchanges momentum with the crystal lattice, so phonon momentum is not conserved.)

wikipedia sourced to


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A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1
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Explanation:

Given:

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  • no. of turns in the coil, n=520
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(a)

we know, according to Faraday's Law:

emf=n.\frac{d\phi}{dt}

where:

d \phi= change in associated magnetic flux

\phi= B.A

where:

A= area enclosed by the coil

Here

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A=\pi\times 0.0395^2

A=0.0049\ m^2

\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049

So, emf:

emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]

emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]

emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]

emf= 0.0306+3.516\times 10^{-4}\ t^3

(b)

Given:

t_0=5.25\ s

Now, emf at given time:

emf=4.7755\times 10^{-2}\ V

∴Current

I=\frac{emf}{R}

I=\frac{4.7755\times 10^{-2}}{560}

I=8.5\times 10^{-5} A

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