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shepuryov [24]
3 years ago
5

Lightning produces a maximum air temperature on the order of 104K, whereas a nuclear explosion produces a temperature on the ord

er of 107K. Find the order of magnitude of the wavelength radiated with greatest intensity by each of these sources. Name the part of the EM spectrum where you would expect to radiate most strongly.
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

tex]2.898\times 10^{-7}\ \text{m}[/tex] ultraviolet region

2.898\times 10^{-10}\ \text{m} x-ray region

Explanation:

T = Temperature

b = Constant of proportionality = 2.898\times 10^{-3}\ \text{m K}

\lambda = Wavelength

T=10^4\ \text{K}

From Wein's law we have

\lambda=\dfrac{b}{T}\\\Rightarrow \lambda=\dfrac{2.898\times 10^{-3}}{10^4}\\\Rightarrow \lambda=2.898\times 10^{-7}\ \text{m}

The wavelength of the radiation will be 2.898\times 10^{-7}\ \text{m} and it is in the ultraviolet region.

T=10^7\ \text{K}

\lambda=\dfrac{2.898\times 10^{-3}}{10^7}\\\Rightarrow \lambda=2.898\times 10^{-10}\ \text{m}

The wavelength of the radiation will be 2.898\times 10^{-10}\ \text{m} and it is in the x-ray region.

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The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
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Answer:

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Explanation:

Given;

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length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

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ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

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Diffuse reflection have a great day

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