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kozerog [31]
3 years ago
11

When astronomers look at distant galaxies, what sort of motion do they see?

Physics
1 answer:
arlik [135]3 years ago
8 0
Hello! You can call me Emac or Eric.

I understand your problem, that question is pretty hard. But I found some information that I think you should read. This can get your problem done quickly.

Please hit that thank you button if that helped, I don’t want thank you’s I just want to know that this helped.

Please reply if this doesn’t help, I will try my best to gather more information or a answer.

Here is some good information that could help you out a lot!


Let’s begin by exploring some techniques astronomers use to study how galaxies are born and change over cosmic time. Suppose you wanted to understand how adult humans got to be the way they are. If you were very dedicated and patient, you could actually observe a sample of babies from birth, following them through childhood, adolescence, and into adulthood, and making basic measurements such as their heights, weights, and the proportional sizes of different parts of their bodies to understand how they change over time.

Unfortunately, we have no such possibility for understanding how galaxies grow and change over time: in a human lifetime—or even over the entire history of human civilization—individual galaxies change hardly at all. We need other tools than just patiently observing single galaxies in order to study and understand those long, slow changes.

We do, however, have one remarkable asset in studying galactic evolution. As we have seen, the universe itself is a kind of time machine that permits us to observe remote galaxies as they were long ago. For the closest galaxies, like the Andromeda galaxy, the time the light takes to reach us is on the order of a few hundred thousand to a few million years. Typically not much changes over times that short—individual stars in the galaxy may be born or die, but the overall structure and appearance of the galaxy will remain the same. But we have observed galaxies so far away that we are seeing them as they were when the light left them more than 10 billion years ago.


That is some information, I do have more if you need some! Thanks!

Have a great rest of your day/night! :)


Emacathy,
Brainly Team.


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You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position
Alecsey [184]

Answer:

positive, positive

You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is positive and the velocity of the rock is positive

Explanation:

Given that the ground is defined as the origin.

The position of the rock is positive since the rock is thrown upward, the position also increases with time until it reaches the maximum height. Also, since the rock is thrown upward with the ground as the origin, the velocity of the rock is positive but the velocity reduces with time (change in height per unit time as the rock moves up is positive)

3 0
4 years ago
A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius
chubhunter [2.5K]

Answer:

a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2

b) See the picture

Explanation:

We can use Gauss's law to find the electric field in all the regions:

EA = qen/e0 where qen is the enclosed charge

Remember that the electric field everywhere outside a sphere is:

E(r) = q/(4*pi*eo*r^2) = Kq/r^2

a)

  1. For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0                       EA = 0/e0 = 0                                                                                                    E = 0
  2. For R < r < 2R: Here the enclosed charge is equal Q                                      E =  Q/(4*pi*eo*r^2) = KQ/r^2      
  3. For r > 2R: Here the enclosed charge is equal 2Q                                              E =  Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2

b)  At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance

7 0
3 years ago
Why does sodium form an ion +1 charge?
Rashid [163]
First of all we know about the ionic bond that the attraction b/w two atoms the atom which lose electrons is become cation and have positive sing which one get electrons it become ionic. so in such sitivation the sodium lose one electons "that's why sodium has 1+ charge in Nacl chemacal bond."
           hope it will help you and please get my answer in brainlist
                        "i should be thankfull of you"
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loris [4]
Just one the earth is the only one like it it's completely unique
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The moment of inertia of a lever is 50 kg m^2 with a mass of 2 kg. What is the<br> radius?
astra-53 [7]

Answer:

huh

Explanation:

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