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 
0kg
If the gravitational pull is zero and I multiply by mass I get a zero
Answer:
a-1 Graph is attached. The relation is linear.
a-2 The corresponding height for 68 kPa Pressure is 7.54 m
a-3 The corresponding weight for 68 kPa Pressure is 1394726kg
b The original height of the column is 5.98 m
Explanation:
Part a
a-1
The graph is attached with the solution. The relation is linear as indicated by the line.
a-2
By the equation

Here
- P is the pressure which is given as 68 kPa.
- ρ is the density of the oil whose SG is 0.92. It is calculated as

- g is the gravitational constant whose value is 9.8 m/s^2
- h is the height which is to be calculated

So the height of column is 7.54m
a-3
By the relation of volume and density

Here
- ρ is the density of the oil which is 920 kg/m^3
- V is the volume of cylinder with diameter 16m calculated as follows

Mass is given as

So the mass of oil leading to 68kPa is 1394726kg
Part b
Pressure variation is given as

Now corrected pressure is as

Finding the value of height for this corrected pressure as

The original height of column is 5.98m
In scientific notation", that number would be written as
6.81 x 10⁻⁴ .
You give the ball to the other team