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FromTheMoon [43]

3 years ago

6

A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a

surface area of 7.70 m2. The total power it absorbs from the star is 3800 W.
Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K)

1 answer:

Fudgin [204]3 years ago

8 0

Answer:

Temperature will be 305 K

Explanation:

We have given The asteroid has a surface area $A=7.70m^2$

Power absorbed P = 3800 watt

Boltzmann constant $\sigma =5.67\times 10^{-8}Wm/K^4$

According to Boltzmann rule power radiated is given by

$P=\sigma AT^4$

$3800=5.67\times 10^{-8}\times 7.70\times T^4$

$T^4=87.0381\times 10^8$

$T=305K$

So temperature will be 305 K

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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

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$F=18.58\times 10^{-11}\ N$

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coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

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$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

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$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

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3 years ago

CaHeK987 [17]

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3 years ago

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Olegator [25]

Answer:

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8 0

2 years ago