Answer:
Because the light only spears to part of the water so it would appear less deep
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 %
Answer:
4.6s
Explanation:
v=u+at
0=22.5+(-9.8)t
-22.5=-9.8t
t=-22.5/-9.8
t=2.295 s
The total time will double
2.295×2=4.59s
=4.6s
Answer:
Explanation:
Given an RL circuit
A voltage source of.
V = 108V
A resistor of resistance
R = 1.1-kΩ = 1100 Ω
And inductor of inductance
L = 34 H
After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor
A. Time the inductor current will reduce to 12% of it's initial current
Let the initial charge current be Io
Then, final current is
I = 12% of Io
I = 0.12Io
I / Io = 0.12
The current in an inductor RL circuit is given as
I = Io ( 1—exp(-t/τ)
Where τ is time constant and it is given as
τ = L/R = 34/1100 = 0.03091A
So,
I = Io ( 1—exp(-t/τ))
I / Io = ( 1—exp(-t/τ))
Where I/Io = 0.12
0.12 = 1—exp(-t/τ)
0.12 — 1 = —exp(-t/τ)
-0.88 = -exp(-t/0.03091)
0.88 = exp(-t/0.03091)
Take In of both sides
In(0.88) = In(exp(-t/0.03091)
-0.12783 = -t/0.030901
t = -0.12783 × 0.030901
t = 3.95 × 10^-3 seconds
t = 3.95 ms
B. Energy stored in inductor is given as
U = ½Li²
So, the current at this time t = 3.95ms
I = Io ( 1—exp(-t/τ))
Where Io = V/R
Io = 108/1100 = 0.0982 A
Now,
I = Io ( 1—exp(-t/τ))
I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))
I = 0.0982(1—exp(-0.12783)
I = 0.0982 × 0.12
I = 0.01178
I = 11.78mA
Therefore,
U = ½Li²
U = ½ × 34 × 0.01178²
U = 2.36 × 10^-3 J
U = 2.36 mJ