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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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3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

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E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

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Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

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we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

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3 years ago
How much work is being done if a force of 75 Newtons is used to push a box a distance of 100 meters?
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A 0.250 kg fan cart accelerates at 24 cm/ s^2 for 4.5 seconds. what is the net force (fan thrust minus drag and friction) on the
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Using the equation

v = v₀ + a t

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v = 1.08 m/s

6 0
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