1.

If the power of a motor is 200 KW. at what speed can it raise a load of 40,000 N?​

Answer» TE required speed is 5 m/sExplanation:Power P = 200 kW = 2 x 10⁵ WForce = 40,000 N = 4 x 10⁴ NSince load is CONSTANTPOWER = FORCE x velocityP = F x v∴ speed v = P/F = 2 x 10⁵/4 x 10⁴ = 20 x 10⁴/ 4 x 10⁴ = 5 m/s


Discussion

No Comment Found

Related InterviewSolutions