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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 burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.
V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

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Co-efficient of kinetic friction, \mu_k = 0.3

Now,

The force against the kinetic friction is given as:

f = \mu_k N = u_k Mg

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

f = 0.3\times108\times9.8

or

f = 317.52N

Now, the net force on to the metal safe is

F_{Net}= F-f

Substituting the values in the equation we get

 F_{Net}= 534N-317.52N

or

F_{Net}= 216.48

also,

 

F_{Net}= M\timesacceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe=\frac{F_{Net}}{M}

or

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or

 

acceleration of the safe=2.00 m/s^2

Hence, the acceleration of the metal safe will be  2.00 m/s²

3 0
3 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

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  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
3 years ago
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Ivahew [28]

Answer:

nope don't think so

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the heat causes the molecules to move faster therefore expanding in watever it the air is in

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