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vazorg [7]
4 years ago
5

The light from many stars can be seen from earth. but there is a time delay between the time the light is emitted from the star

and when we see it here on earth. which of the following statements best explains why there is a delay?
a. the stars light is not bright enough to reach earth immediately
b. stars twinkle so they give off light at regular intervals.
c. Earths atmoshpere is to thick for light to reach the surface as soon as it is
emitted
d. stars are light-years away from earth, so light takes time to reach our planet
Physics
1 answer:
mrs_skeptik [129]4 years ago
4 0
Stars are located at a distance which are measured in terms of light years. Light year is an Astronomical unit used to measure distance between distant Celestial bodies. 
1 light year = 9460730472580<span>800 metres 
But no star is located at a distance of 1 light year. Some stars are located at millions of light years and light travels ~ 3 x 10</span>⁸ m/s. Thus light takes time to reach our atmosphere. 
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Explanation:

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g = G\dfrac{M}{R^2}

At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity g_h at this height is

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Simplifying this, we get

g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g

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Let \vec u and \vec v be the vectors, and let x=\|\vec u\|=\|\vec v\| be their common magnitude.

The resultant \vec u + \vec v is \sqrt 2 times larger in magnitude than either vector alone, so \|\vec u+\vec v\| = \sqrt2\,x.

Recall the dot product identity

\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)

where \theta is the angle between the vectors \vec a and \vec b. In the special case of \vec a=\vec b, we get

\vec a \cdot \vec a = \|\vec a\|^2 \cos(0^\circ) \implies \|\vec a\| = \sqrt{\vec a\cdot\vec a}

Now, to get the angle between \vec u and \vec v, we have

\vec u \cdot \vec v = \|\vec u\| \|\vec v\| \cos(\theta) \implies \cos(\theta) = \dfrac{\vec u \cdot \vec v}{x^2}

To compute the dot product, we take the dot product of the resultant with itself.

(\vec u+\vec v) \cdot (\vec u + \vec v) = \|\vec u + \vec v\|^2

Solve for \vec u\cdot\vec v.

(\vec u\cdot\vec u) + 2(\vec u\cdot\vec v) + (\vec v\cdot\vec v) = \|\vec u + \vec v\|^2

\|\vec u\|^2 + 2(\vec u\cdot \vec v) + \|\vec v\|^2 = \|\vec u+\vec v\|^2

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2(\vec u\cdot\vec v) = 0

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