parker120601
contestada

The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?

Respuesta :

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N   average force applied

onyebuchinnaji

The magnitude of the average total force acting on the object during this time interval is 20 N.

The given parameters:

  • Mass of the object, m = 2.0 kg
  • Initial velocity, u = 30 m/s
  • Final velocity, v = 40 m/s
  • Time of motion, t = 5.0 s

The magnitude of the average total force acting on the object during this time interval is calculated as follows;

[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]

Learn more about resultant force here:  https://brainly.com/question/25239010

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