Two charges, q1 and q2, are separated by a certain distance r. if the magnitudes of the charges are halved and their separation
1 answer:
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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.
Answer:
a)11.25 J
b)Number of revolution = 1
Explanation:
Given that
Radius ,r= 0.8 m
m= 0.3 kg
Initial speed ,u= 10 m/s
final speed ,v= 5 m/s
a)
Initial energy
KEi= 15 J
Final kinetic energy
KEf=3.75 J
The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J
b)
The minimum value to complete the circular arc
Now by putting the values
V= 2.82 m/s
So kinetic energy KE
KE=1.19 J
ΔKE= KEi - KE
ΔKE= 15- 1.19 J
ΔKE=13.80 J
The minimum energy required to complete 2 revolutions = 2 x 11.25 J
= 22.5 J
Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.
Number of revolution = 1