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:
Acceleration of that planet is 30
.
Given:
initial speed of hammer = 0 
time = 1 s
distance = 15 m
To find:
Acceleration due to gravity = ?
Formula used:
Distance covered by hammer is given by,
s = ut + 
s = distance
u = initial speed of hammer
t = time taken by hammer to reach ground
a = acceleration
Solution:
Distance covered by hammer is given by,
s = ut + 
s = distance
u = initial speed of hammer
t = time taken by hammer to reach ground
a = acceleration
u = 0
t = 1 s
s = 15 m
a = g
Thus substituting these value in above equation.
15 = 0 + 
g = 15 × 2
g = 30 
Thus, acceleration of that planet is 30
.
Th statement represented above : In a solution, the solvent is present in the larger amount is definitely FALSE. I choose this answer as a correct one as it also could be a solutions of alcohol ot for example <span>walter which means it's not so solid and liquid respectively. Hope it's clear! Regards!</span>
Answer:
It is equal to the overall momentum before collision, so far no external object is involved.
Explanation:
Momentum is always conserved during collision as a rule. This is equal to the product of the mass and velocity. Thank you.
Answer:
19 N
Explanation:
From the question given above, the following data were obtained:
Pressure (P) = 1.9 kPa
Length (L) = 10 cm
Force (F) =?
Next, we shall convert 1.9 KPa to N/m². This can be obtained as follow:
1 KPa = 1000 N/m²
Therefore,
1.9 KPa = 1.9 KPa × 1000 N/m² / 1 KPa
1.9 KPa = 1900 N/m²
Thus, 1.9 KPa is equivalent to 1900 N/m².
Next, we shall convert 10 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
10 cm = 10 cm × 1 m / 100 cm
10 cm = 0.1 m
Thus, 10 cm is equivalent to 0.1 m
Next, we shall determine the area of the square. This can be obtained as follow:
Length (L) = 0.1 m
Area of square (A) =?
A = L²
A = 0.1²
A = 0.01 m²
Thus, the area of the square is 0.01 m².
Finally, we shall determine the force that must be exerted on the sensor in order for it to turn red. This can be obtained as follow:
Pressure (P) = 1900 N/m²
Area (A) = 0.01 m²
Force (F) =?
P = F/A
1900 = F / 0.01
Cross multiply
F = 1900 × 0.01
F = 19 N
Therefore, a force of 19 N must be exerted on the sensor in order for it to turn red.