PLEASE. A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of friction is 0.25. What is the force of friction on the box? What is the acceleration of the box?

Force of Friction = -50 (N)

Acceleration = 2.45 (m/s^2)

Explanation:

Force of Friction = f = N * coefficient of friction

N = m * g

=> Force of Friction = m * g * coefficient of friction

f = 20 * 10 * -0.25 = -50 (N)

(1) F net = F + f

(2) F = m * a

(1) F net = 99 - 50 = 49

(2) a = 49 / 20 = 2.45 (m/s^2)

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