SOLUTION: the electric power generated by a windmill varies jointly with the square of the diameter of the area swept out by the blades and the cube of the wind velocity. If a windmill with

Question 557553: the electric power generated by a windmill varies jointly with the square of the diameter of the area swept out by the blades and the cube of the wind velocity. If a windmill with 6-foot-diameter blades and a 20-mile-per-hour wind generates 10,823 watts, how much power would be generated if the blades were 10 feet in diameter and the wind speed were 12 miles per hour?

i thought that the solution was P(power) = d^2 (diameter)+ v^3 (wind speed), when i substitute in the values given, 10823= (6*2)^2 + 20^3, i came up with 10823=8144, so not the right equation. can anyone offer any help? Thank you!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Hint: If the "power generated by a windmill varies jointly with the square of the diameter of the area swept out by the blades and the cube of the wind velocity", then the equation is (note: everything is multiplied together)
RELATED QUESTIONS
***I NEED THE STEPS TO SOLVE THIS PROBLEM SO THAT I CAN BETTER UNDERSTAND IT...PLEASE... (answered by stanbon)

The power produces by a windmill varies directly with the square of its diameter and the... (answered by Alan3354)

The power produced by a windmill varies directly with the square of its diameter and the... (answered by ikleyn)

Under certain conditions, the power P, in watts per hour, generated by a windmill with... (answered by ankor@dixie-net.com)

Under certain conditions, the power P, in watts per hour, generated by a windmill with... (answered by Alan3354)

Under certain conditions, the power P, in watts per hour, generated by a windmill with... (answered by stanbon)

Translate each statement into a mathematical statement. Use k as the constant of... (answered by Cromlix)

If 1,790 Mega Watts of power, out of 3,000 Mega Watts generated by a power utility... (answered by tommyt3rd)

If 1,790 Mega Watts of power, out of the 3,000 Mega Watts generated by a power utility,... (answered by rfer)