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3 years ago

How does the size of a wind turbine affect its energy output?

Physics

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

Larger tubines generate more electricity.

Explanation:

Larger blades allow the turbine to capture more of the kinetic energy of the wind by moving more air through the rotors. However, larger blades require more space and higher wind speeds to operate. This distance is necessary to avoid interference between turbines, which decreases the power output.

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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.

Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

The given parameters;

speed of electron, v = 2.2 x 10⁶ m/s
radius of the circle, r = 4.63 x 10⁻¹¹ m

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

$F = \frac{M_e v^2}{r} \\\\$

where;

$M_e$ is mass of electron = 9.11 x 10⁻³¹ kg

$F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N$

Thus, the central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

Learn more about centripetal force here:brainly.com/question/20905151

8 0

3 years ago

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rollin

kaheart [24]

Answer:

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.

Explanation:

spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.

Let any rolling ball of mass (m ) is traveling with velocity v ,

common effect on ball (N) = mg

because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.

hence,

fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.

7 0

3 years ago

Block 1 (mass 2.00 kg) is moving rightward at 10.0 m/s and block 2 (mass 5.00 kg) is moving rightward at 3.00 m/s. The surface i

DaniilM [7]

Answer:

a) 0.25m

b) 5 m/s

Explanation:

When the spring is compressed both boxes are moving with the same velocity, so applying the principle of linear momentum conservation:

$m1*v_{o1}+m2*v_{o2}=(m1+m2)*v\\v=5m/s$

Now applying the principle of energy conservation:

$K1+K2+U_{g1}-U_e=Kf+U_{g2}\\K1+0-U_e=K2+0\\U_e=K1+K2-kf\\\frac{1}{2}*k*x^2+=\frac{1}{2}*m1*v1^2+\frac{1}{2}*m1*v1^2-\frac{1}{2}*(m1+m2)*v^2\\\\x=\sqrt{\frac{2.00kg*(10m/s)^2+5.00kg*(3.00m/s)^2-7.00kg*(5m/s)^2}{1120N/m}}\\x=0.25m$

We got that the maximum compression is 0.25m.

5 0

3 years ago

I need help with the question 8B

ratelena [41]

The answer of this is C

3 0

3 years ago

A calorimeter contained 350.0 g of water [cp=4.18 J/(g °C)] at 24.0 °C. An electric current was passed through a heater placed i

Shtirlitz [24]

Cp shows the amount of energy needed to raise temperature by one degree for one gram of water.

Formula for calculating cp is:
$cp= \frac{energy}{(mass)*( temperature_{change} ))} \\ temperature_{change}= \frac{energy}{(mass)*( cp))} \\ \\ temperature_{change}= \frac{16700}{(350)*( 4.18))} \\ \\ temperature_{change}=2.73 \\ \\ temperature_{final} =temperature_{initial}+temperature_{change} \\ temperature_{final}=24 + 2.73 \\ temperature_{final}=26.73$

Final temperature is 26.73°C.

4 0

3 years ago