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3 years ago

Explain the different types of energy in a working wind turbine

Physics

1 answer:

tatiyna3 years ago

8 0

Answer:

A wind turbine converts the kinetic energy of the wind into mechanical power.

The wind moves the turbine and the turbine produce energy.

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Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is uniform and given by E = (4i+3j)N/C. What

kkurt [141]

Answer:

- 144 V

Explanation:

A = (2, 3) m = 2 i + 3 j

B = (5, 7) m = 5 i + 7 j

Displacement vector = AB = B - A = (3 i + 4 j) m

Electric field, E = (4 i + 3 j ) N/C

V = - E . AB

V = - (4 i + 3 j) . (3 i + 4 j)

V = - (12 + 12) = - 144 V

5 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3° angle. What i

Oksi-84 [34.3K]

Answer:

v_{1fy} = - 0.4549 m / s

Explanation:

6 0

3 years ago

WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of

nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

$A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km$