F = ma
F = (6kg) (3.5m/s2)
F = 21 N
Here we will the speed of seagull which is v = 9 m/s
this is the speed of seagull when there is no effect of wind on it
now in part a)
if effect of wind is in opposite direction then it travels 6 km in 20 min
so the average speed is given by the ratio of total distance and total time


now since effect of wind is in opposite direction then we can say



Part b)
now if bird travels in the same direction of wind then we will have


now we can find the time to go back



Part c)
Total time of round trip when wind is present


now when there is no wind total time is given by


So due to wind time will be more
5.8x10^3
3.02x10^8
4.5x10^5
8.6x10^10
True? I think jdjsjexosjkxnd
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.