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?

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  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.

13875138answer:

step-by-step explanation:

use a calculator!

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation: