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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2