studiot

Like this? 2c-10= a+5 c + 5 = a-5 subtract c - 15 = 10

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This is correct. Also it says that Ade has more than Chidi that is a is greater than c So how do you make a = 25 and c = 35? Actually I think you have dropped another sign somewhere since the numbers are correct but reversed.

Let's consider the case, where if Chidi gives out N5 to Ade, Ade will now have two times what Chidi has. That gives us the equation: c-5 = 2(a+5) I believe the above equation is correct. Yes that is your second equation. See edit below You have two unknowns, a and c and two independent equations so you can now solve them. Edit, sorry no the 2 is in the wrong place ie on the wrong side of the equation. Ade has twice what Chide has (after the transfer of the 5N)

For any rotodynamic machinery there is an engineering trade off between speed, efficiency, power output and torque. Considerations of this lead to the the large radius of the paddle wheel, generating a large moment. Having said that suppose the vanes (this is the correct fluid dynamics word for the paddles) have wetted area a in the sream flow of velocity V and move with tangential velocity v in the direction of the stream. The energy for this velocity v is extracted from the kinetic energy of the stream by momentum exchange. The force on the plate equals the rate of change of momentum of the stream. The mass of fluid striking a stationary plate per second is m = paV where p is the density of the water. Thus the force = change of momentum per second = mV = (paV)V = paV2 Since our wheel is moving we must use the relative velocity between the vane and the stream ie (V-v) So the force on the vane is F = maV(V-v) So the work done per second in moving the vane is maV(V-v)v = (V-v)v per kg of fluid. The energy in the stream = 0.5V2 per kg of fluid. So the efficiency, e, is the work done on the vane per kg divided by the energy of the fluid per kg [math]e = \frac{{\left( {V - v} \right)v}}{{0.5{V^2}}} = \frac{{2\left( {V - v} \right)v}}{{{V^2}}}[/math] Differentiate this with respect to vanve velocity, v and set to zero to find a max [math]\frac{{de}}{{dv}} = V - 2v = 0[/math] Leading to [math]v = \frac{V}{2}[/math] Back substituting [math]\max efficiency = \frac{{2\left( {V - \frac{V}{2}} \right)\frac{V}{2}}}{{{V^2}}} = \frac{1}{2}[/math] That is an undershot paddle wheel can never be more than 50% efficient. This can be improved by shaped vanes as in the Poncelot Wheel https://www.google.co.uk/search?hl=en-GB&source=hp&q=poncelot+wheel&gbv=2&oq=poncelot+wheel&gs_l=heirloom-hp.3..0i10.953.3969.0.4437.14.14.0.0.0.0.125.1486.5j9.14.0....0...1ac.1.34.heirloom-hp..0.14.1486.MUPDmWEQ2nA Best efficiencies are achieved by running the wheel fully submerged, either in a casing as with the Pelton Wheel https://www.google.co.uk/search?q=pelton+wheel&hl=en-GB&gbv=2&oq=&gs_l= or by using a vertical axis wheel such as a Kaplan turbine. https://www.google.co.uk/search?q=kaplan+turbine&hl=en-GB&gbv=2&oq=&gs_l=

There is considerable recorded knowledge and wisdom about the use of river flow to provide energy. This came from the times when watermills were used to provide mechanical energy. Externet is correct, gearing is essential to provide correct rotational speeds. Further considerations are that most river levels are not constant, but some by several meteres, some by a few tens of metres. This makes 'dipping a paddle in' more difficult. Conventional technology also has that overshot wheels are significantly better than the undershot (dipped in) variety. Like most technology there is the capital investment to consider and the available return. This is what led to watermill technology dying out originally, it was much less cost effective than later (steam) technology. If this is a (school) project, re-evaluating the economics in today's world may be a good subject.

Yes, I'm trying to make the example as simple as possible. With binary variables and a single term there are two (or three possible sequences, depending whether you include the null sequence (with no terms) or not) that meet this. Both of these sequences are defined as random, as noted. The either can be arrived at in either a deterministic way or a way that depends upon chance. There is little more to be said about the deterministic route since it determines the outcome. But there is a twist to the chance route worth discussing if you are interested.

The word is insolation and Wikipedia suggests the annual max average insolation at the equator is 0.25kW/m2, making a nonsense of your claim ofnearly 30 times that. http://en.wikipedia.org/wiki/Insolation

This was posted here a year ago, resulting in some interesting comments. http://physicsbuzz.physicscentral.com/2013/06/reinventing-wheel.html The physicist in the link was totally skeptical. However I have an open mind to further persuasion. I note, for instance that any side thrust puts great bending stress on an axle and greatly increases wear. The wider the wheel the lower this effect and the skateboard wheels are certainly wide. For bicycles to achive the cubical idea a bike wheel would have to be two foot to two foot six wide! Now this latter has been done with the popular 'ballbarrow', whose main claim is the ease of turning corners with a loaded wheelbarrow.

+1

Which other statement, please be specific, then I can try to answer your question and explain further. How can this be? The process of calculation is longer than the list.

Force is mass times acceleration. Weight is the force a given mass applies due to the acceleration due to gravity. So you are asking if a body can exert a greater force than its own weight. (Since pressure is force divided by area there is no theoretical limit to the possible pressure that can be exerted.) Since gravity is not the only acceleration a body may be subject to, given a greater acceleration than gravity, it can apply a greater force than its own weight by virtue of this acceleration. If the extra acceleration is combined with gravity the total is called apparent weight. There was a recent question here as to why the apparent weight increased in a lift, accelerating upwards. http://www.scienceforums.net/topic/84324-can-someone-help-me-understand-the-apparent-weight-of-a-body-on-a-moving-lift/ Note also that a body may use the principle of the lever to multiply the applied force and that impulse (impact) forces are generally greater than steady ones.

I don't know what make of torch yours was but I used to use Clulites for years on site. Mine were the big bulky sort that definitely were designed to run upright, but Cluson offer a range of rechargeables some of which look as though they have batteries on their sides. Some lead acid batteries are completely sealed and the electrolyte is contained in gel form so they should be able to run in any orientation. https://www.google.co.uk/search?hl=en-GB&source=hp&q=clulite&gbv=2&oq=clulite&gs_l=heirloom-hp.1.0.0l10.1921.3812.0.8000.7.5.0.2.2.0.187.609.2j3.5.0....0...1ac.1.34.heirloom-hp..0.7.671.fmfIJouWjSA

Mike, This is a good site to plot changes of ancient boundaries etc. You can check on each of the major geological ages. http://www.scotese.com/earth.htm

More to the point whose money are they spending since states in this region are not self supporting.

Thank you Tom, old maps are one of my hobbies. Do you have any more information about it? Edit I have found more here. Interesting history that I know nothing of. http://en.wikipedia.org/wiki/Sykes%E2%80%93Picot_Agreement

You made a lot of statements, without support in post#13. Was this in response to my question? Here are your linked lines to which I was referring

I see no point proceeding into the document until I have understood the statement(s) made in the opening sections. You have access to the text, please post the first twenty lines of section 2 with explanation as to how it conforms to conventional definitions.

So it is. Thanks for correcting my oversight. +1

Whilst I am able to open you link and read the content, I am unable to copy and paste any text here for comment. I am, however, concerned about your use of the term dipole moment in the opening of section 2. Dipole moment is defined as the product of separated charge (virtual or otherwise) and the distance between the centres of charge. The strength of any external field does not enter into this, as you seem to state. Perhaps you could explain this part further?

A much simpler geometric (non calculus) method is to subtract the area of the minor segment of the circle from the area of the right triangle as shown in my diagram.

Formally to solve it algebraically we would assign a boolean variable to our expression and work on that. So I would start Let E = "the expression" Then AE =A(.......).. finally A(.......) = A(1) hence E = 1 But the trick of multiplying through by A is not obvious.

Yes OK so here is a boolean algebra solution I am ging to use ' instead of the overbar for ease of typing. expression = AB + A' + B' = A(expression) =AAB + AA' + AB' =A2B + 0 + AB' =AB + AB' =A(B + B') =A(1) = A (expression) Hence expression = 1

try not A and B or not(notA or notB) You need to decide whether you are going for a mathematical solution using boolean algebra or an electronic solution using and and or gates. Do not mix them. I did ask this before?

I'm sorry but I find both the pictorial questions you have been asked downright silly and I would not be happy with any course that taught in this fashion. I am guessing that since D is time in all but option a, which seriously mislables other items, that E is the same client as A, but after a time D. Does that make sense?