A 1500 kg car has an applied forward force of 5000 N and experiences an air resistance of 1250 N. What is the car's acceleration?

Answers

Answer 1

Answer:

[tex]2.33\ m/s^2[/tex]

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn = m.a

Where a is the acceleration of the object.  The net force is the sum of the individual vector forces applied to the object.

The m=1500 Kg car has two horizontal forces applied: the forward force of 5000 N that causes the movement and the air resistance force of 1250 N that opposes motion.

The net force is Fn = 5000 N - 1500 N = 3500 N

To find the acceleration, we solve the equation for a:

[tex]\displaystyle a=\frac{Fn}{m}[/tex]

[tex]\displaystyle a=\frac{3500}{1500}[/tex]

[tex]\boxed{a = 2.33\ m/s^2}[/tex]

The car's acceleration is [tex]a = 2.33\ m/s^2[/tex]


Related Questions

Shows a car travelling around a bend in the road. The car is travelling at a constant speed. There is a resultant force acting on the car. This resultant force is called the centripetal force. (i) In which direction, A, B, C or D, does the centripetal force act on the car? Tick ( ) one box. A B C D (1) (ii) State the name of the force that provides the centripetal force.

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete but the missing figure is in the attachment below.

When an object is travelling around a circular path, there is a force that tends to draw that object towards the center of the circular path and keep the object moving in the curved path, that force is called the centripetal force. From this description, it can be deduced that the direction of the centripetal force that acts on the car (in the attachment below) is D.

The name of the force that provides this centripetal force is frictional force. This is the force that prevents the car from slipping off the road; keeping it moving in the curved path.

- During a certain period, the angular position of a rotating object is given by: = − + , where  is in radian and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the rotating object at = Sec.

Answers

The question is not complete. The complete question is :

During a certain period of time, the angular position of a rotating object is given by [tex]$\theta =2t^2 +10t+5$[/tex], where θ is in radians and t is in seconds.  Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.

Solution :

Given :

Displacement or angular position of the object, [tex]$\theta =2t^2 +10t+5$[/tex]

∴ Angular speed is   [tex]$\omega = \frac{d \theta}{dt}$[/tex]

                                 ω = 10 + 4t

And angular acceleration is [tex]$\alpha = \frac{d \omega}{dt}$[/tex]

                                              α = 4

a). At time, t = 0.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(0)^2 +10(0)+5$[/tex]

                                               = 5 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(0)

                                     = 10 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

b). At time, t = 3.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(3)^2 +10(3)+5$[/tex]

                                               = 53 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(3)

                                     = 22 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

                                   

                                           

Which of the following is NOT one of the essential components of an exercise program?

Answers

Answer:

i dont know dude but ask someone who is in your grade lol

Explanation:

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