Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.50. The top block has a mass of 2.1 kg, and the bottom block's mass is 4.7 kg. If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the top block begins to slip

33.32N

Explanation:

we would be using newton's second law of motion to solve this question.

The force on the block before it slides down is 2.1 *g * 0.5 =

2.1 * 9.8 * 0.5 = 10.29N

Maximum acceleration of both blocks would then be f/m = 10.29/2.1 = 4.9m/s²

Maximum force applied to the bottom box f, = ma

F = (2.1 + 4.7) * 4.9

F = 6.8 * 4.9

F = 33.32N

Therefore, the maximum force F can have before the top block begins to slip is 33.32N

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.