Answer:
d
Explanation:
Ya gon find the Kenitic Energy first
K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004
and now the replacement:
0.004=½×0.4V²====> v²=0.02===> V=0.14m/s

<h2><em>calculate</em></h2>
<em>
</em>
<h2><em>reduce </em><em>the </em><em>numbers</em></h2>
<em>
</em>
<h2><em>multiply</em></h2>
<em>
</em>
<h2><em>there </em><em>for </em><em>we </em><em>have </em><em>a </em><em>solution</em><em> to</em><em> the</em><em> </em><em>equation</em></h2>
<em>hope </em><em>it</em><em> helps</em>
<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on</em><em> learning</em>
<em>mark </em><em>me</em><em> as</em><em> brainlist</em><em> plss</em>
Answer:
b) Betelgeuse would be
times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is
the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
(1)
where
is the brightness,
is the star luminosity and
, the distance from the star to the point where the brightness is calculated (measured). Thus:
b)
and
where
is the Sun luminosity (
) but we don't need to know this value for solving the problem.
is light years.
Finding the ratio between the two brightness we get:

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is
. Then

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.
The formula we can use in this case is:
v = v0 + a t
where v is final velocity, v0 is initial velocity, a is
acceleration and t is time
So finding for v0:
v0 = v – a t
v0 = 43.7 – (2.5) 2.7
v0 = 36.95 m/s