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.
Answer:
End product of photosynthesis.
Explanation:
Photosynthesis is a process that plants undergo in the manufacture of their food. This is done in the presence of sunlight which is trapped by their coloring pigment called chlorophyll and reactants such as Carbon dioxide and Water.
6CO2+6H2O= C6H12O6+ 6O2
The oxygen gas which is a waste product of photosynthesis is released into the atmosphere and used by animals in respiration.
It is an example of balanced force.
hope this helps. good luck
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = ![\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%261%5C%5C1%260%265%5Cend%7Barray%7D%5Cright%5D)
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where 
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that 
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
