A small-sized 155-kg mass is located 1.50 m from a small-sized 275-kg mass, with both masses fixed in place. Where should you place a third small-sized mass so that the net gravitational force on it due to the original two masses is zero

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

Answerthe distance between the two ends of the galaxy is 150,000 light years;