Answer to Question #345890 in Physics for mich

Question

. A power station conveyor lifts 500 tons of coal per hour to a height of 80 feet.  What average  

     horsepower is required? 

 

Answer 1

$$P=\frac{W}{t}=\frac{mgh}{t}$$ $$={\rm\frac{500*10^3\: kg*9.8\: N/kg*80*0.305\: m}{3600\: s}}$$ $$=3.32*10^4\:\rm W=44.5\: hp$$