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3 years ago

A wind turbine has a total input power of 2 500 kW.

Physics

1 answer:

Sonbull [250]3 years ago

6 0

Answer:

Output power = 500 KW

Explanation:

Given the following data;

Efficiency = 20%

Input power = 2500 KW

To find the output power;

$Efficiency = \frac {Out-put \; power}{In-put \; power} * 100$

Substituting into the equation, we have;

LET Output power = OP

$20 = \frac {OP}{2500} * 100$

Cross-multiplying, we have;

$20 * 2500 = OP * 100$

$50000 = OP * 100$

$OP = \frac {50000}{100}$

Output power = 500 KW

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Four power transistors, each dissipating 12 W, are mounted on a thin vertical aluminum plate 22 cm 3 22 cm in size. The heat gen

Sveta_85 [38]

Answer:

The temperature of the Aluminium plate 44.84⁰C

Explanation:

Number of transistors = 4

Since the heat dissipated by each transistor is 12W

Total heat dissipated, Q = 4 * 12 = 48 W

Q = 48 W

Cross sectional Area of the Aluminium plate, A = 2(l * b)

l = Length of the aluminium plate = 22 cm = 0.22 m

b = width of the aluminium plate = 22 cm = 0.22 m

A =2( 0.22 * 0.22 )

A = 0.0968 m²

From the heat balance equation, Q = hAΔT

h = 25 W/m²·K

A = 0.0968 m²

ΔT = T - T(air)

T(air) = 25°C

ΔT = T - 25°C

Q = 25 * 0.0968 * ( T - 25)

Q = 2.42 (T - 25)

Substitute Q = 48 into the equation above

48 = 2.42 (T - 25)

T - 25 = 19.84

T = 25 + 19.84

T = 44.84 ⁰C

6 0

3 years ago

Truck drivers approaching a steep hill that they must climb often increase their speed. What good does this do, if any?

Inga [223]

Um, this doesn't make any sense. By climbing a hill, you are decreasing your momentum and kinetic energy, so it slows you down. The only positive, is after you have climbed the hill, you have more potential energy, and it will be released once you go down the hill, but you will not be as fast as if you ignored the hill.

6 0

3 years ago

Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang

Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

m*g*s*sin(θ) = 0.25*m*R^2*w^2

2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0

3 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

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3 years ago

PLEASE HELP ASAP! THANKS!<br><br> Why is it a rare occasion to see a solar eclipse from your home?

Y_Kistochka [10]

Answer:

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5 0

3 years ago