Via the half-life equation:

Where the time elapse is 11,460 year and the half-life is 5,730 years.

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.
Answer:
b. a massive collapsed star
Explanation:
A black hole in the universe is nothing but a massive collapsed star. When the size of the star crosses a particular limit it cannot holds its mass and it collapses under it own self. This is called supernova. A black hole is actually a region in space where gravity is so strong that even light cannot escape through it. Gravity so strong because the matter has been pressed into a tiny space. hence option b is correct
Assume that the small-massed particle is
and the heavier mass particle is
.
Now, by momentum conservation and energy conservation:


Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

So now, we see that
and
. So therefore, the smaller mass recoils out.
Hope this helps you!
Bye!
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²