To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
Where
Angular acceleration
Angular velocity
t = Time
Our values are
Replacing at the previous equation we have that the angular velocity is
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
Therefore the angular acceleration of a point on the outer edge of the tires is 
The power developed is 500 W ( to the nearest Watt)
Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.
Therefore,
Substitute the given values of force , displacement and time
Thus the Power can be rounded off to the nearest value of 500 W
answer
option d is the correct answer
explanation
as we know frequency is equal to 1 /t
f= 457 Hz
t=1
SO, 1/457
=0.0022sev
Kinetic energy, KE, is modeled by the formula
, where m is the mass in kg and v is the velocity in m/s.
In this scenario, mass and one-half are constant but the velocity changes.
You can see that by squaring twice the velocity, that is equal to four times the original KE. Therefore, the answer is 4k.
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
