KE = 1/2 * m * v^2
m = 44
v = 10
KE = 1/2 * 44 * (10)^2
KE = 22 * 100 = 2200 J or B
Tangential speed = radius of track times angular velocity.v=r omega.(180,000/60x60)=r omega.2pi r = 2,400 metres -> r = 2,400/(2pi)(180,000/60x60)=2,400/(2pi) omega(180,000x2pi/60x60x2,400)= omega****************************************
(A) 
The energy stored by the system is given by

where
P is the power provided
t is the time elapsed
In this case, we have
P = 60 kW = 60,000 W is the power
t = 7 is the time
Therefore, the energy stored by the system is

(B) 4830 rad/s
The rotational energy of the wheel is given by
(1)
where
is the moment of inertia
is the angular velocity
The moment of inertia of the wheel is

where M is the mass and R the radius of the wheel.
We also know that the energy provided is

So we can rearrange eq.(1) to find the angular velocity:

(C) 
The centripetal acceleration of a point on the edge is given by

where
is the angular velocity
R = 0.12 m is the radius of the wheel
Substituting, we find

To come up with an idea of what you want,and in case you want to change it during the process.