If the moon's acceleration due to gravity caused by its gravitational field is one-sixth that of the earth, what is its acceleration at a point in space 3r (three moon radii) from its center? (use the earth's value of "g" as your significant figure reference; i. e., use 2 sf.) m/s^2

To make it easy, let's call the moon's surface gravity 'Q'.  We know that Q is 1/6 of 'g', but let's just hold onto that for a minute.   Let's first work out what it is at 3 moon radii from the moon's center, and once we have that, relate it back to the Earth.

We know that the strength of gravity is inversely proportional to the square of the distance between the centers of the two objects.  On the moon's surface, you're 1 Moon radius from the center.  At 3 Moon radii from the center, you're 3 times as far from the center, so the gravity out there is (1/3²) = 1/9 of the gravity on the surface.

So at 3 Moon radii from the surface, the Moon's gravity is ( Q/9 ) .  

So far, so good. Now, we know that Q is (1/6) x (Earth 'g').

Moon's gravity at 3 Moon radii = Q/9

Substitute g/6 for Q.

Moon's gravity at 3 Moon radii = (g/6) / 9 .

Moon's gravity at 3 Moon radii = g/54

(9.8 m/s) / 54 = 0.18 m/s²

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.

the ima (higher than 1m) will be greater than the ima with the height of 1m.