InterviewSolution
Saved Bookmarks
| 1. |
Calculate the power of an engine required to lift `10^5`kg of coal per hour from a mine 360 m deep. |
|
Answer» Correct Answer - A Power of the engine required, `P = (W)/(t) = (mgh)/(t) = ((10^5kg)(10m//s^2)(360m))/(60xx60s) = 10^5W =100kW` |
|