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Ber [7]
3 years ago
6

The Andromeda Galaxy is faintly visible to the naked eye in the constellation Andromeda. Suppose instead it were located in the

same direction in space as the center of the Milky Way Galaxy (but still at its current distance). How would it appear to the eye in that case?
a) We could not see it at all.
b) It would look about the same, but would be in the constellation Sagittarius instead of Andromeda.
c) It would be much brighter, because it would be illuminated by the many stars in the center of our galaxy.
d) It would look about the same, but it would be harder to pick out because its cloud-like appearance would make it blend in with the cloud-like appearance of the Milky Way in our sky.
Physics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

a) We could not see it at all.

Explanation:

The most distant object that can be seen is the andromeda galaxy, which we may have a slight view of. The andromeda galaxy is a large galaxy that along with the previous two is also part of the local group. Spiral-type galaxy that is approximately 250,000 light years in diameter (more than twice the diameter of the Milky Way!) And is about 2.9 million light years away from our galaxy. Because of its distance, we have difficulty visualizing this galaxy, we would have this difficulty even if the andromeda galaxy was in the center of the Milky Way, but maintaining its current distance. That is, even if the andromeda galaxy were located in the same direction in space as the center of the Milky Way (but still at its current distance), we could not see it at all.

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A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic
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Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light c= 2.99793\times10^{8}\ m/s

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}

Put the value into the formula

\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}

\Delta l=1\sqrt{1-(0.96)^2}

\Delta l=0.28\ m

We need to calculate the time the stick take to move

Using formula of time

t=\dfrac{\Delta l}{v}

Put the value into the formula

t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}

t=9.72\times10^{-10}\ sec

t=0.97\ ns

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

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