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3 years ago

What is the comparison of Wind turbine and other energy sources?

2 answers:

mote1985 [20]3 years ago

7 0

AnWind power or wind energy is the use of wind to provide mechanical power through wind turbines to turn electric generators for electrical power. Wind power is a popular sustainable, renewable source of power that has a much smaller impact on the environment compared to burning fossil fuels. Wind farms consist of many individual wind turbines, which are connected to the electric power transmission network. Onshore wind is an inexpensive source of electric power, competitive with or in many places cheaper than coal or gas plants. Onshore wind farms have a greater visual impact on the landscape than other power stations, as they need to be spread over more land and need to be built away from dense population. Offshore wind is steadier and stronger than on land and offshore farms have less visual impact, but construction and maintenance costs are significantly higher. Small onshore wind farms can feed some energy into the grid or provide power to isolated off-grid locations. The wind is an intermittent energy source, which cannot be dispatched on demand. Locally, it gives variable power, which is consistent from year to year but varies greatly over shorter time scales. Therefore, it must be used together with other power sources to give a reliable supply. Power-management techniques such as having dispatchable power sources (often gas-fired power plant or hydroelectric power), excess capacity, geographically distributed turbines, exporting and importing power to neighboring areas, grid storage, reducing demand when wind production is low, and curtailing occasional excess wind power, are used to overcome these problems. As the proportion of wind power in a region increases the grid may need to be upgraded. Weather forecasting permits the electric-power network to be readied for the predictable variations in production that occur. In 2019, wind supplied 1430 TWh of electricity, which was 5.3% of worldwide electrical generation, with the global installed wind power capacity reaching more than 651 G…swer:

Explanation:

egoroff_w [7]3 years ago

3 0

Wind is a more efficient power source than solar. Compared to solar panels, wind turbines release less CO2 to the atmosphere, consume less energy, and produce more energy overall.

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A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

krek1111 [17]

Answer:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

6 0

3 years ago

Argue as to why drivint on icy or snowy roads can be dangerous

Vika [28.1K]

Icy/Snowy roads have less friction than normal roads. This means that the wheels are less likely to stay positioned because of traction, and you will spin out of control

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3 years ago

Ls -2 a solution of 4x +3= -5?.<br>

torisob [31]

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

4x+3=−5

Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

4x=−8

Step 2: Divide both sides by 4.

4x / 4 = −8 / 4

x=−2

Hope it helps,

Please mark me as the brainliest

Thank you

5 0

3 years ago

One billiard ball is shot east at 2.00 m/s. A second, identical billiard ball is shot west at 1.00 m/s. The balls have a glancin

dimulka [17.4K]

Answer:

Velocity is 1.73 m/s along 54.65° south of east.

Explanation:

Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 2i + m x (-1)i = m i

Final momentum = m x v + m x 1.41 j = mv + 1.41 m j

Comparing

mi = mv + 1.41 m j

v = i - 1.41 j

Magnitude of velocity

$v=\sqrt{1^2+(-1.41)^2}=1.73m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0$

Velocity is 1.73 m/s along 54.65° south of east.

5 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago