Ask question

Ask question

All categories

1 year ago

13

A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e

nergy. A large wind turbine has 45-m-radius blades. In typical conditions, 92,000 kg of air moves past the blades every second. If the air is moving at 12 m/s before it passes the blades and the wind turbine extracts 40% of this kinetic energy, how much energy is extracted every second?

1 answer:

ddd [48]1 year ago

8 0

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J$

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

$E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J$

The energy extracted by the turbine every second is 2649600 Joules

You might be interested in

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

1 year ago

At what distance above earth would a satellite have a period of 125 min?

Nezavi [6.7K]

Rw^2 = GmM/r^2
<span> Leads to
</span><span> w^2 r^3 = GM
</span><span> (2pi /T) ^2 r^3 = GM
</span><span> 4pi^2 r^3 = GM T^2
</span><span> r^3 = GM T^2 / 4pi^2
</span><span> Work out r^3 then r.
</span> T = 125 min = 125(60) = 7500 s
<span> R = 6.38E6 m
</span><span> m = 5.97E24 kg
</span><span> G = 6.673E-11
</span> r=<span> 8279791.78</span><span> m
Since r = radius R of Earth + height above urface,h
</span><span> h = r - R = </span><span> 8279791.78 - </span>6.38E6 = <span> <span>1899791.78 m
h=</span></span><span> <span>1899.79178 Km</span></span>

5 0

1 year ago

Read 2 more answers

If electromagnetic radiation a has a lower frequency than electromagnetic radiation b the wavelength of a is

Jobisdone [24]

Inversely proportional to its frequency. If electromagnetic radiation A has a lower frequency than electromagnetic B, then compared to B, the wavelength of A is...? - equal - shorter - longer - exactly half the length of

5 0

1 year ago

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

MaRussiya [10]

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

<u>The relation between the change in pressure with height is given as:</u>

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$ = standard value of gravity

<em>Now putting the respective values:</em>

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0

1 year ago

A car drives toward the right over the top of a hill, as shown below. An illustration of car at the top of a hill pointing right

Sav [38]

Answer: X

Explanation:

This situation can be illustrated as a car in circular motion (image attached).

In circular motion the acceleration vector $\vec{a}$ is always directed toward the center of the circumference (that's why it's called centripetal acceleration).

So, in this case the arrow labeled X is the only that points toward the center, hence it represents the car's centripetal acceleration

6 0

1 year ago

Read 2 more answers