Answer:
μk = (Vf - Vc)/(T×g)
Explanation:
Given
Vi = initial velocity of the car
Vf = final velocity of the car
T = Time of application of brakes
g = acceleration due to gravity (known constant)
Let the mass of the car be Mc
Assuming only kinetic frictional force acts on the car as the driver applies the brakes,
The n from Newtown's second law of motion.
Fk = Mc×a
Fk = μk×Mc×g
a = (Vf - Vc)/T
Equating both preceding equation.
μk×Mc×g = Mc × (Vf - Vc)/T
Mc cancels out.
μk = (Vf - Vc)/(T×g)
I think its a scrambled word. I think its air movement
This problem is to let you practice using Newton's second law of motion:
Force = (mass) x (acceleration)
-- The airplane's mass when it takes off (before it burns any of its load of fuel) is 320,000 kg.
-- The force available is (240,000 N/per engine) x (4 engines) = 960,000 N.
-- Now you know ' F ' and ' mass '. Use Newton's second law of motion to calculate the plane's acceleration.
the main properties of the main wave propertioes include wavelength amplitude, cruest an trough
When acceleration is constant, the average velocity is given by
where 
and 
are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by
where 
are the final/initial displacements, and 
are the final/initial times, respectively.
Take the car's starting position to be at 
. Then
So we have
You also could have first found the acceleration using the equation
then solve for 
via
but that would have involved a bit more work, and it turns out we didn't need to know the precise value of 
anyway.
