Answer: I don't know spanish
Explanation:
What What What What...... error
MARK ME BRAINLIEST PLEASE!!!!!
The total energy TE = mgh + 1/2 mU^2; where h = 20 m, g = 9.81 m/sec^2, and U = 10 mps. When the ball reaches max height H, all that TE will be potential energy PE = mgH = TE.
So there you are. TE = mgh + 1/2 mU^2 = mgH = TE from the conservation of energy. Solve for H.
1) H = (gh + 1/2 U^2)/g = h + U^2/2g = ? meters where everything on the RHS is given. You can do the math.
2) As the ball drops from H to h, it picks up KE as the potential energy mgH is converted when the potential energy is diminished to mgh, where h < H. So PE - pe = ke = mg(H - h) = 1/2 mv^2 so solve for v = sqrt(2g(H - h)) and, again, everything is given. You can do the math.
3) Same deal as 2) except now its V = sqrt(2gH) because all the PE = mgH = 1/2 mV^2 = KE when it is about to hit the ground. You can do the math.
A, D is the correct answers
Answer:
4.1%
Explanation:
Given that water specific heat c = 4186 J/kgC. If water is flowing with mass rate of 200kg/s and temperature is dropping from 200C to 25C. Then the thermal energy rate from the water should be
or 146510kW
Since the turbine is only able to extract 6000kW of power, then the thermal efficiency is:
or 4.1%
