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kondaur [170]
2 years ago
4

Solving Expressions Analytically Consider the following equation, which describes the speed of sound a in an ideal gas: a=kRT−−−

−√. The Mach number M describes the ratio of a velocity v to the acoustic velocity or speed of sound a: M≡va. The Mach number can also be written in terms of the stagnation temperature T0, M=2k−1(T0T−1)−−−−−−−−−−−−−√. Combine these equations and solve symbolically for the temperature T in terms of all other quantities except M. Your answer should be stored in a variable T_result, which will be a list containing one or more sympy expressions. Prefer T0 instead of T_0 in this case.
Physics
1 answer:
lawyer [7]2 years ago
8 0

Answer:

T = (2k + 1)/(v^2kR + T0)

Explanation:

There are 3 equations

a = √kRT ------ eqn (i)

M = va -------- eqn (ii)

M = √2k - 1(T0T - 1) ---- eqn (III)

Substitute the expression of a in eqn (i) in eqn (ii)

eqn (ii) becomes M = v√kRT.

Equate this equation with eqn (iii) because M = M

v√kRT = √2k - 1(T0T - 1)

square both sides to eliminate the square root

v^2(kRT) = 2k - 1(T0T - 1)

v^2kRT = 2k - T0T + 1

v^2kRT + T0T = 2k + 1

T(v^2kR + T0) = 2k + 1

T = (2k + 1)/(v^2kR + T0)

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Explanation:

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All the above phenomena are due the electromagnetic wave spectrum. Electromagnetic waves travel at a constant speed of 3 x 10^8 m/s in a vacuum. Within the spectrum, the different types of electromagnetic waves exists in different band range of frequencies and wavelengths unique to each of the waves, and the energy they carry. When these waves enter a non-vacuum medium, their speed change, depending on the nature of the material of the medium, and the frequency or the wavelength of the incoming wave.

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2 years ago
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Explanation:

Below is an attachment containing the solution

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A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

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Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

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E = m_{p}c^{2}

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This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

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\Delta t_{o} = \Delta t

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3 0
2 years ago
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<span>m = mass (kg) </span>
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<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
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IgorC [24]

Answer:

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