What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

Explanation:

Acceleration of gravity on the surface of

Neptune is given

g' = G M/R^2,

where M is the planet's mass.

M g' will be the apparent weight registed at the poles.

G is the universal constant of gravity

The "apparent weight" at the equator will be

reduced from M g' by the centripetal force M

R w^2, where w is the rotational angular

velocity in radians/s), which you can deduce

from the 16 hour length of a day.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement