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KatRina [158]
3 years 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?
Physics
1 answer:
ddd [48]3 years 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

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A car is running at a velocity of 50 miles/hour and the driver accelerates the car by 10miles/hour.How far the car travels from
dezoksy [38]
Initial velocity u = 50 miles/hour
acceleration a = 10 miles/hour
Time t = 2 hours
Distance travelled S = ut + (at^2)/2
Substituting the values in the second equation of motion,
S = 50*2 + (10 * 2 *2)/2
S = 100 + 20
S = 120 miles
Therefore the distance travelled by the car in the next two hours is 120 miles
4 0
3 years ago
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Is a truck braking to avoid an accident moving at constant
frutty [35]

Answer:

its not moving at a constant velocity because it is slowing down

Explanation:

3 0
2 years ago
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Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are
Sergio039 [100]

Answer:

21678.47223\ lbf-in^2

383.1109\ lbf-in^2

Explanation:

d = Diameter of column = 0.5 inch

A_c = Area of concrete = 119.4\ in^2

The strain in the system is conserved

\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s

Now

F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf

F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf

Stress is given by

\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2

The stress in the steel is 21678.47223\ lbf-in^2

\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2

The stress in the steel is 383.1109\ lbf-in^2

4 0
3 years ago
Si la aceleración de gravedad en la superficie del planeta Mercurio es de 3,7 m/s2, entonces ¿cuál sería el peso de una persona
erica [24]

Answer:

W = 222 N.

Explanation:

The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface? "

Mass of the person, m = 60 kg

The acceleration due to gravity on the surface of gravity is 3.7 m/s²

We need to find the weight of a person on the surface of Mercury.

Weight of an object is given by :

W = mg

So,

W = 60 kg × 3.7 m/s²

W = 222 N

Hence, the person will weigh 2222 N on the surface of Moon.

4 0
3 years ago
If someone walkes 1000m <br> in 20min, what is their speed?
slava [35]

Answer:

Distance - 1000m

Time - 20min

Speed - ?

Use the formula of distance ÷ time = speed.

s = d/t

s = 1000m/20min

s = 50 m/min

Hope this helps, thank you !!

6 0
2 years ago
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