Answered

a 1,800 kg car is parked on a road that has an elevation angle of 7?. suppose the coefficient of static friction of the kinds of rubber and asphalt involved is 0.65. which is approximately the force of static friction between the tires and the road?



Answer :

A car weighing 1,800 kg is parked on a road with an elevation angle of 7, and Fn = 17526.4 N

Explain about the force?

The definition of force in physics states that when an object with mass is pushed or pulled, it alters its velocity. A body's resting or moving status can be changed by an external agent known as force. It has both size and motion.

It takes force to push or pull something. Pushing, tugging, picking, hitting, lifting, running, and bending are examples of actions that involve the application of force. The movement or stopping of a body, as well as changes to an object's shape or motion, are a few examples of actions that show the usage of force.

In mechanics, a force is any action that aims to maintain, alter, or deform a body's motion.

   =mgcosФ

   =1,800 (9.81)cos7

   =17526.4 N

To learn more about force refer to:

https://brainly.com/question/12970081

#SPJ4

Other Questions