GingerAle

Hi Ron,

If you want to use spaces to align ASCII text in HTML documents, the first requirement is to be smarter than your computer.  LOL

Seriously, using spaces to align ASCII text in HTML documents is often an exercise in futility.

There is no solution for this question as stated. 

The sum of two included angles (BMC + angle B) of a triangle has to be less than 180° The question states the sum of 201° for these angles.

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you referred to "v" as velocity, shouldn't it be speed?

(Because K.E. is scalar not vector, and scalar quantities must be of either: Vector x Vector or scalar x scalar).

That’s an interesting question.

Most equations use velocity to define kinetic energy. Though I have seen speed used. Kinetic energy is produced when mass is in motion without regard to its direction; however, in most closed systems there is at least an implied direction.  

In the case of the wind turbine, the implied direction is at zero (0) degrees relative to the face of blades. If the mass of the wind was vectored at 90 degrees to the face of the blade then none of the kinetic energy would translate to the mass of the blades. 

GA

There is no need to delete your answer. I usually leave my mistakes, unless they are a major distraction to the continuity of the post. They are uinique monuments to exceedingly rare events.

Your solution is well thought out and explained; you just have mistakes in the application. The more complex the subject, the easier it is to make mistakes. Physics becomes very complex, very fast.  We learn from our mistakes; others learn from them too. It’s part of educating ourselves.

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After rereading the question, I realize I made a mistake too, so I’m adding my name to the list of dumb-dumbs.

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The initial energy of (0.5MW) is what passes through the wind turbine system—the input power, not the output power.  Doubling the wind speed increases the input power to (4MW). The input power is then multiplied by the coefficient of efficiency (4.0MW * 0.47407) giving the output power of 1.9MW.  (This is consistent with the OP’s answer, which I dismissed because there wasn’t any associated work product.)  

GA

Neither of you two dumb-dumbs have the correct answer.

The change in power is proportional to the ratio of the cubed velocities.

$\Delta P = \dfrac {V_2^3}{V_1^3}$

$\Delta P = \dfrac {10^3}{5^3} = 8$

Then

$0.5MW *8 = 4 MW$

This is solvable using kinetic energy equations. But you have two errors in your second equation.

Your second equation has an erroneous value for the mass. Also, you multiplied by the coefficient of efficiency instead of dividing. In this case, you should have done neither.

Dividing gives the actual kinetic energy of the air that’s required to produce this power output.  However, if you do this on the second equation you have to do it on the first equation. The results will be the same if the proportions are the same. Despite its mention in the question, the coefficient is not needed to solve this question. In fact, most of the BS written in the context isn’t needed for the question.      

$E_k=\dfrac{1}{2}(40000)(10^2) = 2.0MJ$ – Your result without multiplying by the coefficient of efficiency.

Consider carefully: if you have twice the velocity there will also be twice the mass in the same unit of time.

So...

$E_k=\dfrac{1}{2}(80000)(10^2) = 4.0MJ$

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Defining how kenetic energy relates to power:

$KE = \dfrac{1}{2} M * v^2$ the velocity may be expressed as $\left(\dfrac{\dfrac{m}{s}}{s}\right)$

And power as:

$P = \dfrac{KE}{\Delta t} | \;\; t \;\; \text { is in seconds.}$

Then

$P = \dfrac{1}{2}*\left(\dfrac{{\dfrac{\dfrac{m}{s}}{s}}}{s}\right)$

Then

$P = 0.5 M * v^3$

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A point about your hint:

Hint: (Although the text is talking about power and energy, they are the same here due to "1 second" (Since P=E/t -->P=E if t=1))

Power and energy are related and one can be derived from the other, but they are never the same. Just as kilometers per hour is never the same as kilometers. It does not matter that the time unit is one (1); they have different units: for power it’s watts and Joules per second. For energy it’s Joules and watt-seconds.

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The general equation for power produced by a wind turbine is

$P = 1/2 \pi * r^2 * v^3 * \rho * \eta  \;\;| \;\; \tiny \text {where r is the radius of the turbine blade, v is velocity of the wind, $\rho$ is the air density, and $\eta$ is the efficiency factor.}$ 

With this equation you can reconstruct the basic parameters of the wind turbine. 

This equation was not included, but a lot of BS filler was.

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Both the question and the guest answer look like they could have been written with a quill...

That is, written in a narrative style consistent with the lecture circuit of the nineteenth century.    

The great masters of physics and science, such as Michael Faraday and Joseph Henry gave public lectures on various subjects. But for every true master there were dozens of morons, idiots and charlatans, who either did know the subject or couldn’t teach it with any kind of coherent logic.   

Here’s an example in written form.

This question has illusion of complexity because it’s intuitive for many mathematicians to find the values of the two (2) numbers (a) and (b), but the question only asks for their sum. 

Solution:

\(\small \text{ }\\ \begin{align*} \small x^2 + y^2 &= 327\\ \hspace{1em} \small x + y &= 101\\ \end{align*} \;\\\;\\ \text {Note that }\\ \left(x+y\right)^2 \;\; = \quad x^2 + y^2 +2xy \hspace{3ex}| \scriptsize \text { perfect square expansion formula }\\ \text{then}\\ \left(x + y\right)^2 \;\; = \quad 327 +2(101)\\ \left(x + y\right)^2 \;\; = \quad 529\\ \text{then }\\ \sqrt{\left(x + y\right)^2} \;\; = \quad \sqrt{529}\\ \hspace{1.5em}{\left(x + y\right)} \;\; = \quad23 \; \hspace{2em} \leftarrow \small \text{sum of (x) and (y)}\)

GA

Hi Amy,

The question only asks for the arc (an angular measure) of LPM, in degrees. It gives the arc of LM as 72 degrees. 

All circles have 360 degrees, and all angular measurements of a circle are constant for any radius.

Of course, the linear measure of the arc length is proportional to the radius. 

--. .-

GA

Solution:

If the arc of LM is 72° on the larger circle, then the arc LPM on the larger circle is

360 - 72 = 288°

Am I missing something?   

--. .-

Hi Catmg,

Your text should read: “If they have an odd number of divisors, they're perfect squares.

Also, the question asks:

How many positive integers less than 2008 have an even number of divisors?

So ...

\($2007-\lfloor\sqrt{2007}\rfloor = 1963 \)

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Catmg, you have wonderful math skills. You present excellent solutions to a fairly broad range of math questions. 

GA