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KatRina [158]
2 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]2 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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An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must
Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

0.2\times 1=0.2\ W

Total power required is 1.6 W

Number of Light Emitting Diodes would be

n=\dfrac{1.6}{0.2}\\\Rightarrow n=8

The number of Light Emitting Diodes is 8.

Power would be

P=8\times 1=8\ W

The power that is required to run the Light Emitting Diodes is 8 W

7 0
2 years ago
Come si compongono due forze che agiscono in diversi punti di un corpo rigido? Oof
bagirrra123 [75]

Answer:

Explanation:

I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.

7 0
2 years ago
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob
givi [52]

Answer:

Charge, Q=1.56\times 10^{-8}\ C

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

E=\dfrac{kQ}{r^2}

Q is the charge on an object

Q=\dfrac{Er^2}{k}

Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}

Q=1.56\times 10^{-8}\ C

So, the charge on the object is 1.56\times 10^{-8}\ C. Hence, this is the required solution.

7 0
2 years ago
Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment
riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a  look at the free bodily diagrammatic representation.

The first diagram will aid us in answering  question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ  = \frac{360+240}{450}

tan θ  = \frac{600}{450}

tan θ  = 1.333

θ  = tan⁻¹ (1.333)

θ  = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

M__B}=(Fcos \theta *240)-(Fsin \theta *450)

where Force(F) = Force in the cord AC = 1350 N and θ  = 53.13° ; we have:

M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)

M__B}= 194400.463-485999.348

M__B}=-291598.885 N-mm\\

M__B}=-291.598 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B (M__B}),

we have:

M__B}=-Fcos \theta *360 - Fsin \theta * 0

M__B}=-Fcos \theta *360 - 0

M__B}=-Fcos \theta *360

where Force(F) =  1350 N and θ  = 53.13° ; we have:

M__B}= -1350*cos53.13^0*360

M__B}= -291600 N-mm

M__B}= -291.6 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

6 0
2 years ago
Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed
makkiz [27]
Using the given formula with v0=56 ft/s and h=40 ft 
h = -16t2 + v0t  
40 = -16t2 + 56t 
16t2 - 56t + 40 = 0  
Solving the quadratic equation:  
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
 We have two possible solutions  
t1 = (56+24)/32 = 2.5 
t2 = (56-24)/32 = 1  
So initially the ball reach a height of 40 ft in 1 second.
3 0
2 years ago
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