Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
In factor it is desired to lift 2000 kg of metal through a distance of 12 m in 1m .find the minimum horsepower of the engine to be used.
velocity with which it would lift => 0.5 * 2000 * v^2and v = 0 + 10*1 = 10 m/s.Therfore net work done would be = > 0.5 * 2000 * 10^2 + 2000* 10 * 12 => 340,000 JouleHorse Power = > 340,000/746 = > 455.764 HP.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -