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ankoles [38]
3 years ago
7

A wind turbine takes in energy from wind with the goal of converting it into electrical energy. Much of the wind energy is also

converted into the kinetic energy of the spinning turbine and heat lost to the air around it. An illustration of a windmill with a wide arrow away from it labeled kinetic energy of wind 3000 J and it separates into 3 different arrows, one labeled electric energy 750 J, the second labeled heat 750 J and the third which is the widest labeled kinetic energy of turbine blank J. If the goal of the turbine is to put out electric energy, what is the efficiency of the turbine? 25% 50% 75% 100%.
Physics
1 answer:
kondor19780726 [428]3 years ago
5 0

The efficiency of the turbine is 50%.

The given parameters:

  • <em>Kinetic energy of the wind, E = 3000 J</em>
  • <em>Output electrical energy, E(out) = 750 J</em>
  • <em>Energy lost to heat, E(lost) = 750 J</em>
  • <em>Kinetic energy of the turbine, E(in) = ?</em>

The kinetic energy of the turbine which is the input energy is calculated as follows;

E(in) = 3000 - (750 + 750)

E(in) = 1500 J

The efficiency of the turbine is calculated as follows;

E_f_f = \frac{E_{(out)}}{E_{(in)}} \times 100\%\\\\E_f_f = \frac{750}{1500} \times 100\%\\\\E_f_f = 50  \%

Thus, the efficiency of the turbine is 50%.

Learn more about efficiency of turbine here: brainly.com/question/2009210

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The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0
4 years ago
A 0.50 mm-wide slit is illuminated by light of wavelength 500 nm.What is the width of the central maximum on a screen 2.0m behin
Novay_Z [31]

Answer:

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

y=\frac{n\lambda D}{d}

where

\lambda is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

w=2\frac{\lambda D}{d}

where we have

\lambda=500 nm = 5\cdot 10^{-7}m is the wavelength

D = 2.0 m is the distance of the screen

d=0.50 mm=5\cdot 10^{-4}m is the width of the slit

Substituting, we find

w=2\frac{(5\cdot 10^{-7} m)(2.0 m)}{5\cdot 10^{-4} m}=0.004 m

5 0
3 years ago
g A hydraulic press has a safety feature which consists of a hydraulic cylinder with a piston at one end and a safety valve at t
nlexa [21]

Answer:

58.32 N

Explanation:

Area of a circle = \pir^{2}

where r is the radius of the circle.

The cylinder has a radius of 0.02 m, its area is;

A_{1} = \pir^{2}

  = \frac{22}{7} x (0.02)^{2}

  = \frac{22}{7} x 0.0004

  = 1.2571 x 10^{-3}

Area of the cylinder is 0.0013 m^{2}.

The safety valve has a radius of 0.0075 m, its area is;

A_{2} = \pir^{2}

    = \frac{22}{7} x (0.0075)^{2}

    = \frac{22}{7} x 5.625 x 10^{-5}

    = 1.7679 x 10^{-4}

Area of the valve is 0.00018 m^{2}.

From Hooke's law, the force on the safety valve can be determined by;

F = ke

F_{2}  = 950 x 0.0085

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Minimum force, F_{1}, required can be determined by;

\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

\frac{F_{1} }{0.0013} = \frac{8.075}{0.00018}

F_{1} = \frac{0.0013 *8.075}{0.00018}

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The minimum force that must be exerted on the piston is 58.32 N.

8 0
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Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 2.00 cm from the axis to equal
MAXImum [283]

Answer:

Angular velocity of an ultra centrifuge is 17146.42 rad/s.

Explanation:

Given that,

Acceleration of an ultra centrifuge, a=6\times 10^5\ g=6\times 10^5\times 9.8=5.88\times 10^6\ rad/s2

Distance from axis, r = 2 cm = 0.02 m

We need to find the angular speed. We know that the relation between angular speed and angular acceleration is given by :

a=r\omega^2\\\\\omega=\sqrt{\dfrac{a}{r}} \\\\\omega=\sqrt{\dfrac{5.88\times 10^6}{0.02}} \\\\\omega=17146.42\ rad/s

So, the angular velocity of an ultra centrifuge is 17146.42 rad/s.

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