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djverab [1.8K]
3 years ago
8

A friend of yours who has not taken an astronomy class looks at your textbook and really likes the picture of the Pleiades, a cl

uster of stars surrounded by a bluish reflection nebula. She wants to know what causes that beautiful blue glow. To explain it to her, you want to compare the process that causes the blue glow to something that is in your friend's everyday experience. Which of the following terrestrial phenomena is the result of the same type of process that makes a reflection nebula in space?
A. the blueish glow of the flame on a gas stove
B. the blue light you see reflected from a blue sweater
C. the blue colour of the Earth's sky
D. the blue-white colour of the hottest stars
E. the blue feeling you get when you don't do well on an astronomy exam
Physics
1 answer:
Andrei [34K]3 years ago
8 0

Answer:

<em>C. the blue colour of the Earth's sky</em>

<em></em>

Explanation:

The Pleiades is a cluster of sister stars that are among the closest star cluster to earth.

The reflection nebula of the Pleiades is due to the scattering of the blue light from the hot blue luminous stars that dominate the star cluster. Th blue light is scattered from dust molecules, thought to be predominantly carbon compound like diamond dusts, and other compounds like iron.

The blue colour of the Earth's sky is the closest terrestrial phenomenon to the reflection nebula. On a clear cloudless day, molecules in the air scatter the blue component of light more than the other component colours of white light, giving the sky its characteristic blue coluor.

The common characteristics of the luminous nebula and the Earth's blue sky is that they both have their light scattered by the presence of small particles.

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Answer:

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Explanation:

We are given that

Initial circumference=C=168 cm

\frac{dC}{dt}=-15cm/s

Magnetic field,B=0.9 T

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r=\frac{C}{2\pi}

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\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

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\frac{168}{2\pi}=C

r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

t=8 s

B=0.9

E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

E=103.1 V

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