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:False
Explanation:
Anything with evidence it can test.
The correct answer is: Option (D) length, speed
Explanation:
According to Faraday's Law of Induction:
ξ = Blv
Where,
ξ = Emf Induced
B = Magnetic Induction
l = Length of the conductor
v = Speed of the conductor.
As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed
I think analog but I could be wrong
Answer:
255 Hz
Explanation:
With 5 beats per second with the 250 Hz fork, we know the unknown fork is either 250 - 5 = 245Hz or 250 + 5 = 255 Hz
With 15 beats per second with the 270 Hz fork, we know the unknown fork is either 270 - 15 = 255Hz or 270 + 15 = 285 Hz (most people would have a hard time discerning 15 beats per second... 5 per second is hard enough)
As 255 is the common frequency, it is the one selected.
