The best possible Answer is c
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
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.
Explanation:
We have,
The initial position of an object is zero.
The starting velocity is 3 m/s and the final velocity was 10 m/s.
The object moves with constant acceleration..
The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".
Answer:
angular acceleration is -0.2063 rad/s²
Explanation:
Given data
mass m = 95.2 kg
radius r = 0.399 m
turning ω = 93 rpm
radial force N = 19.6 N
kinetic coefficient of friction μ = 0.2
to find out
angular acceleration
solution
we know frictional force that is = radial force × kinetic coefficient of friction
frictional force = 19.6 × 0.2
frictional force = 3.92 N
and
we know moment of inertia that is
γ = I ×α = frictional force × r
so
γ = 1/2 mr²α
α = -2f /mr
α = -2(3.92) /95.2 (0.399)
α = - 7.84 / 37.9848 = -0.2063
so angular acceleration is -0.2063 rad/s²