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 
Being made mostly of gas is NOT a
characteristic of an inner planet. The correct answer between all the choices
given is the last choice or letter D. I am hoping that this answer has
satisfied your query and it will be able to help you in your endeavor, and if
you would like, feel free to ask another question.
Answer:
Explanation:
The force of friction equals the sine component of the force due to gravity
Have them give up there phone for 15 minutes
talk to them when there in a good mood
watch a movie then have a disscution about it
The formula of the kinetic energy is:

where m is a mass of the object, v is speed of the object at the moment of time. So we have:

The answer is
2000 Joules.