<u>electrial</u><u> </u>
<u> </u><u>In an electrical circuit, electric current flows through the closed path</u>
<u>magnetic</u><u> </u><u>circuit</u><u> </u>
<u>In the magnetic circuit, magnetic flux flows through the closed path</u>
Answer:
The procedure is the plan for how you will conduct your experiment.
Explanation:
Answer:
a) 5.22 m/s
b) 31.4 %
Explanation:
f = rotating speed = 15 rpm = 15/60 =0.25 rps
m = Mass flow rate of air = 42000 kg/s
v = Tip velocity = 250 km/h = 250/3.6 = 69.44 m/s
W = Work output = 180 kW
A = Swept area of wind turbine
r = Radius of wind turbine
η = Efficiency



∴ The average velocity of the air is 5.22 m/s


∴ Conversion efficiency of the turbine is 0.314 or 31.4 %
Let m₁ = 3.0 kg and v₁ = + 8 m/s (so right is positive), and m₂ = 1.0 kg and v₂ = 0. The total momentum of the two balls before and after collision is conserved, so
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' = + 5 m/s and v₂' are the velocities of the two balls after colliding, so
(3.0 kg) (8 m/s) = (3.0 kg) (5 m/s) + (1.0 kg) v₂'
Solve for v₂' :
24 kg•m/s = 15 kg•m/s + (1.0 kg) v₂'
(1.0 kg) v₂' = 9 kg•m/s
v₂' = (9 kg•m/s) / (1.0 kg)
v₂' = + 9 m/s
which is to say, the second ball is given a speed of 9 m/s to the right after colliding with the first ball.
“Wave-B is every bit as invisible as Wave-A is.”