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:
r = 0.02 m
Explanation:
from the question we have :
speed = 1 rps = 1x 60 = 60 rpm
coefficient of friction (μ) = 0.1
acceleration due to gravity (g) = 9.8 m/s^{2}
maximum distance without falling off (r) = ?
to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force
mv^2 / r = m x g x μ
v^2 / r = g x μ .......equation 1
where
velocity (v) = angular speed (rads/seconds) x radius
angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm
angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds
now
velocity = 6.28 x r = 6.28 r
now substituting the value of velocity into equation 1
v^2 / r = g x μ
(6.28r)^2 / r = 9.8 x 0.1
39.5 x r = 0.98
r = 0.02 m
Answer:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.
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