A1230kg car pushes a 2160kg truck that has a dead battery. when the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4590n . take the friction between the truck's tires and the ground to be 470n .

what is the magnitude f_cont of the force that the car exerts on the truck? use the coordinate system shown in part b, where upward is the positive y direction and to the right, the direction f_cont points, is the positive x direction.

The magnitude of the force that the car exerts on the truck ( F_C on T ) is 3095 N

Acceleration is rate of change of velocity.

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

This problem is about Newton's Law of Motion.

Given:

mass of car = = 1230 kg

mass of truck = = 2160 kg

force of car = = 4590 N

friction of truck = = 470

Unknown:

Solution:

→ Equation 1

In accordance with Newton's 3rd law, then

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Mug , Friction , Coefficient , Static

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answer: (431,2)

step-by-step explanation: