Answer:
As b ∝ (L/r²) and
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
hence,
the Sun appear brighter in the sky
Explanation:
The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).
thus, mathematically,
b ∝ (L/r²)
now,
given
L for sirius is 23 times more than the sun i.e 23L
now,
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
thus,
using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.
hence, the sun appears brighter
Explanation :
The forces acting on hot- air balloon are:
Weight, (W)
Force due to air resistance, (F)
Upthrust force, (U)
Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.
In this case, the hot-air balloon descends vertically at constant speed.
so, 
and 
so,
....................(1)
when it is ascending let the weight that it is releasing is R, so
..........(2)
solving equation (1) and (2)

2F is the weight of material that must be released from the balloon so that it ascends vertically at the same constant speed.
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)