Aloop-the-loop has a circular arc, with a marble that can run along a track and traverse the entire inside of the loop. when the marble is precisely at the top of the inside loop, it has its minimum speed. ~v what do we know about the direction of the net force on the marble at this point, the top of the loop? a. the net force is instantaneously in the direction of the marble’s velocity b. the net force is instantaneously in the opposite direction to the marble’s velocity ~v. c. the net force is vertically downward. d. the net force is vertically upward

c. The net force is vertically downward.

Explanation:

At the top of  the loop, the only external force that keeps the ball moving around the loop, is the centripetal force.

Now, this centripetal force, is not a " new" force, it's just the vector sum of the two external forces (neglecting friction) , that act simultaneously  on the marble, making it to change its speed, in magnitude and direction: the gravity force (which it is always downward), and the normal force(which is always perpendicular to the contact surface, preventing that the marble comes trough the surface), in this case between the marble and the track, which, at the top of the loop, points down too.

So, the net  force, exactly at the top of the loop, is vertically downward.

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).