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

1 the 4 is sig

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.