Answer:
Star A is brighter than Star B by a factor of 2754.22
Explanation:
Lets assume,
the magnitude of star A = m₁ = 1
the magnitude of star B = m₂ = 9.6
the apparent brightness of star A and star B are b₁ and b₂ respectively
Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: 
The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.
We need to find the factor by which star A is brighter than star B. Using the equation given above,



Thus,

It means star A is 2754.22 time brighter than Star B.
Answer:
the second one!
Explanation:
the question is well, the question, a hypothesis is an educated guess on what you think will be the outcome
I think it was either Robert Hook or Anton Van Leevwen Hoek
Answer:
The change in kinetic energy (KE) of the car is more in the second case.
Explanation:
Let the mass of the car = m
initial velocity of the first case, u = 22 km/h = 6.11 m/s
final velocity of the first case, v = 32 km/h = 8.89 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(8.89² - 6.11²)
= 20.85m J
initial velocity of the second case, u = 32 km/h = 8.89 m/s
final velocity of the second case, v = 42 km/h = 11.67 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(11.67² - 8.89²)
= 28.58m J
The change in kinetic energy (KE) of the car is more in the second case.