A) 0.189 N
The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

where
G is the gravitational constant
8.7×10^13 kg is the mass of the asteroid
m = 130 kg is the mass of the man
R = 2.0 km = 2000 m is the radius of the asteroid
Substituting into the equation, we find

B) 2.41 m/s
In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

where
v is the speed of the astronaut
Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

Answer:
D. 2^(3/2)
Explanation:
Given that
T² = A³
Let the mean distance between the sun and planet Y be x
Therefore,
T(Y)² = x³
T(Y) = x^(3/2)
Let the mean distance between the sun and planet X be x/2
Therefore,
T(Y)² = (x/2)³
T(Y) = (x/2)^(3/2)
The factor of increase from planet X to planet Y is:
T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)
T(Y) / T(X) = (2)^(3/2)
If you want to change the thermos into an open energy system, you have to remove the lid. Once the lid is removed, the energy is no longer contained inside the thermos bottle. From the bottle, the energy dissipates to the environment.
#39. rain, snow, sleet and hail.
I think the answer would be letter a