The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.
- The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.
Reasons:
The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

Where;
a₀ = The current size of the Universe
a = The size of the early Universe = 5% of a
Therefore;


0.05 + 0.05·z = 1

- The redshift is of the observed light is, z = <u>19</u>
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Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
I'm going to do the math on the top and then on the bottom and divide at the end.
and now when I divide I will express my answer to the correct number of sig dig's:
6.45 × 10¹⁶ N
Answer: v = 2[m/s]Explanation:This avarage velocity can be found with the ... B. 2 meters/ second. C. 3 meters/second. D. 4 meters/second. 1.
C because it’s not a or B so 50/50 c or d and d is def not the answer so c