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raketka [301]
3 years ago
6

Power windmills turn in response to the force of high-speed drag. For a sphere moving through a fluid, the resistive force, FR i

s proportional to r 2 v 2 , where r is the radius of the sphere and v is the fluid speed. The power developed, P = FR v, is proportional to r 2 v 3 . The power developed by a windmill can be expressed as P = d r2 v 3 , where r is the windmill radius, v is the wind speed, and d = 4.3 W s3 /m5 . For comparison, a typical home needs about 3.0 kW of electric power. Note: This representation ignores system efficiency, which is about 25%. For a home windmill with a radius of 1.59 m, calculate the power delivered to the generator if the wind velocity is 12.4 m/s.\
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

The answer is 20727w

Explanation:

The formula is below;

P = d r^2 v^3 *efficiency

In the question, it is stated that the registration ignores efficiency so we are going to ignore efficiency in the equation and use it this way;

P = d r^2 * v^3

d =4.3, r = 1.59, v =n 12.4

Therefore, P = 4.3 X 1.59^2  X 12.4^3 = 20727W

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A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

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However, for a coil the magnetic flux \phi is equal to

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For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

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\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

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putting this value in above equation to get m

this approximately corresponds to m = 0.3 or 30%

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