A 9 kg log is being dragged along flat ground by a 120 N force at an angle of 45°. If the log accelerates at a rate of 0.78 m/s2, what is the normal force acting on the log?

9514 1404 393

  3.347 N

Step-by-step explanation:

The vertical component of the dragging force is (120 N)/√2, so the net downward force is the difference between that due to gravity and the upward component of the dragging force.

  (9.8 m/s²)(9 kg) -(120 N)/√2 = 3.347 N

The normal force acting on the log is 3.347 N.

inequality:

620-37.4p ≥ 450

the minimum amount of packages needed to be removed from the elevator is 5.

step-by-step explanation: