SOLUTION: A 905 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?

Algebra ->  Length-and-distance -> SOLUTION: A 905 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?      Log On


   

Question 1187976: A 905 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?
Answer by Bala290(1) About Me  (Show Source):
You can put this solution on YOUR website!
Cicumference=3.25*1000=3250m
F= 2140N, and M=905km
first, we find radius
circumference=2πr
3250=2πr
r=3250/2π
r=517.25m
Fc=%28mv%5E2%29%2F%28r%29
2140=905v%5E2%2F%28517.25%29
905v%5E2=%28517.25%29%2A%282140%29
905v%5E2=1106915
v%5E2=%281106915%29%2F%28905%29
v%5E2=1223.11
v=sqrt%281223.11%29
v=34.97
v≈35m/s