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 
Answer:11686.5 joules
Explanation:
elastic constant(k)=53N/m
extension(e)=21m
Elastic potential energy=(k x e^2)/2
Elastic potential energy=(53 x (21)^2)/2
Elastic potential energy=(53x21x21)/2
Elastic potential energy=23373/2
Elastic potential energy=11686.5
Elastic potential energy is 11686.5 joules
Answer:0 J
Explanation:
Given
For first step
change in internal Energy of the system is 
Work done on the system 
For second step
change in internal Energy of the system is 
Work done on the system 
Work done on the system is considered as Positive and vice-versa.
and from first law of thermodynamics

for first step


overall heat added
For overall Process Heat added is 0 J