A905 kg test car travels around a 3.25 km circular track. if the magnitude of the centripetal force is 2140 n, what is the car's speed?

Consider that a centripetal force can be written as:
Fc=mv2r

where r is the radius.

so in your case: circular distance=2πr

3.25×103=2⋅3.14⋅r
remember to change into meters.

r= 3.25 ×1036.28 = 517m
2,140 = 905*v^2*517 
v=√1.1×106905 = 34.9 m/s ≈ 35m/s = 126 km/hr

the amount of current in the coil

a straight line sloped