A box with its contents has a total mass of 40 kg. It is dropped from a very high building. After reaching terminal speed, what is the magnitude of the air resistance force acting upward on the falling box

The magnitude of air = 392N

Explanation:

We use Newton's 2nd law. The sum of the vertical forces must be equal to zero because at terminal speed , the acceleration is zero. Solving for the air resistance force,F(air ) gives:

EFvertical = mg - F(air)= ma

F(air) = mg = 40 × 9.8 = 392N

392N

Explanation:

Newton's second law of motion states that "The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object."

the sum of vertical forces has to be equal to zero because by the time the terminal speed has been attained, the acceleration is zero. Now, we solve for air resistance force.

summation of F(vertical) = mg - F(air) = ma

a = 0 m/s²

thus, F(air) = mg

F(air) = 40kg*9.8m/s²

F(air) = 392N

the inner core.

good luck: )

klicker