A machine part consists of a thin, uniform 4.00kg bar that is 1.50m long, hinged perpendicular to a similar vertical bar of mass 3.00kg and length 1.80m . The longer bar has a small but dense 2.00kg ball at one end.

By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90 degrees to make the entire part horizontal? Find the magnitude of horizontal displacement. Find the magnitude of vertical displacement.

100 pounds is 100 pounds no matter the object or item being weighed

35.98  million  mile my friend

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)