Interactive solution 9.1 presents a model for solving this problem. the wheel of a car has a radius of 0.380 m. the engine of the car applies a torque of 456 n·m to this wheel, which does not slip against the road surface. since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. moreover, the car has a constant velocity, so this countertorque balances the applied torque. what is the magnitude of the static frictional force?

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x = τ1

= the static frictional force (N)

= τ1 /r

  = 456 N/0.380 m

  = 1200 N

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

found in the nucleus

has mass of one amu