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2 years ago

8

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe

rature 1000K and emissivity 0.6:

1 answer:

stiks02 [169]2 years ago

3 0

Answer:

6.0 × $10^{11}$ W/ $m^{2}$

Explanation:

From Wien's displacement formula;

Q = e A $T^{4}$

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = $\frac{Q}{A}$ = e $T^{4}$

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × $(1000)^{4}$

= 0.6 × 1.0 × $10^{12}$

= 6.0 × $10^{11}$ $\frac{W}{m^{2} }$

Therefore, the emissive intensity coming out of the surface is 6.0 × $10^{11}$ W/ $m^{2}$ .

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A sound wave traveling at 340 m/s is generated by a 480 Hz tuning fork.

Shtirlitz [24]

Answer:

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Explanation:

Given the following data;

Speed of wave = 340 m/s

Frequency = 480 Hz

To find how long is the sound wave, we would determine its wavelength;

Mathematically, the wavelength of a waveform is given by the formula;

Wavelength = velocity/frequency

Wavelength = 340/480

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2 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

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Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

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$\tau=NIAB\ \sin\theta$

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$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

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$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

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2 years ago

what is the name of the tool that allows you to copy two or more shapes into a single part? when working with 3D

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Explanation:

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2 years ago

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2 years ago

Read 2 more answers

A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can

GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

$v$ = speed at which jumper jumps at all time

initial distance jumped is given as

$R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}$

final distance jumped is given as

$R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}$

Dividing final distance by initial distance

$\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}$

$\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}$

$R_{2} =7.38$

distance lost is given as

d = $R_{1} - R_{2}$

d = 7.4 - 7.38

d = 0.02 m

8 0

2 years ago