A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 240 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?

The value of F= - 830 N

Since the force is negative, it implies direction of the force applied was due south.

Explanation:

Given data:

Mass = 1000-kg

Distance, d = 240 m

Initial velocity, v1 = 20.0 m/s

Final velocity, v2 = 0 (since the car came to rest after brake was applied)

v2²= v1² + 2ad (using one of the equation of motion)

0=  20² + (2 x a x  240)

0= 400 + 480 a

a = - 400/480

a = - 0.83 m/s²

Then, imputing the value of a into

F = ma

F = 1000 kg x ( - 0.83 m/s²)

F= - 830 N

The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

i think it's 35 meters