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4 years ago

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During experiments on laser etching of microchips for IBM, a nuclear engineer places a discshaped sample 5 mm in diameter in a v

acuum chamber with surface temperature of 300 K. A 500 mW laser is then used to irradiate the sample uniformly. The sample surface can be considered ‘gray’ with an emissivity of 0.9 and can be considered ‘thin’ so that the disk temperature is uniform. In addition, a non-participating gas at a temperature of 27⁰C is used to cool the sample with a convection coefficient of 50 W/(m2K). Find the steady-state temperature of the disk

1 answer:

Andru [333]4 years ago

7 0

Answer:

Steady state temperature is approximately 801K

Explanation:

Detailed explanation and calculation is shown in the images below.

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Which of the following results from the fact that skip signals refracted from the ionosphere are elliptically polarized?

zhenek [66]

Answer:

B

Explanation:

The fact that skip signals refracted from the ionosphere are elliptically polarized is a result of either vertically or horizontally polarized antennas may be used for transmission or reception.

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3 years ago

Select the correct answer. Ruby wants to create a cube puzzle game. For that she has to create a cube. Which drafting tool do yo

Alexxx [7]

Answer:

D. a triangle and a T-Square

Explanation:

A T-Square is the best drawing tool to create squares. You would need a squares to create cubes.

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3 years ago

Can be used to eliminate rubbing friction of wheel touching frame. 1.Traction 2.Thrust washer

Vilka [71]

Answer:

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3 years ago

The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published

Iteru [2.4K]

Answer:

a

The rate of radiation of the energy is $E_r = 1.523747635*10^9 W/m^2$

b

The irradiation is $G =46.177\ kW/m^2$

c

The amount of energy absorbed is $E_B = 461.772 KJ$

d

The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

From the question we are told that

The temperature is $T = 7200K$

The diameter of the ball is $d = 1.5 km = 1.5 *1000 = 1500m$

Hence the radius = $= \frac{1500}{2} = 750m$

The total energy radiated can be mathematically represented as

$E = \sigma A T^4$

Where $\sigma$ is the Stefan-Boltzmann constant $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}$

A is the area of a sphere = $\pi d^2 = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2$

Substituting values we have

$E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4$

$=1.077*10^{15} W$

Now the state of the energy is mathematically represented as

$Rate \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4$

$= 5.67*10^{-8} * 7200^2$

$= 1.523747635*10^9 W/m^2$

A sketch illustrating the b part of the question is shown on the first uploaded image

looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

from the diagram x can be calculated as follows

$x = \sqrt{40^2 + 16^2}$

$= 43.0813 Km$

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

Now the irradiation G is mathematically represented as

$G = \frac{E}{4 \pi r^2}$

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

$G= \frac{1.077*10^15}{4 \pi (431081.3^2)}$

$G =46.177\ kW/m^2$

Generally the amount of energy absorbed can be mathematically represented as

$Amount \ of \ energy \ absorbed \ (E_B ) = G * t$

Where t is the time taken

Therefore $E_B = 46.177 *10 = 461.77 KJ$

6 0

3 years ago

On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch

rewona [7]

Answer:

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

Explanation:

Distance of earth from sun = $R_{2} = 147.4 \times 106 Km$

Spacecraft perihelion = $R_{2} = 120\times106Km$

gravitational parameters are now given as

$\mu_{sun} = 132.7\times 10^{9}$

$\mu_{earth} = 398600$

radius of earth = 6378 Km

Heliocentric spacecraft velocity at earth sphere of influence =

$V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }$

$V_{D}^{v} =28.43\frac{km}{s}$

Heliocentric velocity of earth = $v_{earth} = 30.06\frac{km}{sec}$

$V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}$

assume

$r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km$

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

$v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}$

geometric space craft velocity in its circular parking orbit

$v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } } = 7.784 \frac{km}{s}$

ΔV = $=v_{p} -v_{c} = 3.337 \frac{km}{s}$

7 0

4 years ago