studiot

Not to me it doesn't indicate the above conclusion. What it indicates is that c is an asymptotic value to some curve. There are many other such examples in applied maths. All travelling entities follow the same laws of motion; At one end of the velocity curve (the low velocity end) the curve may be approximated by a straight line (as with many curves), which is equivalent to saying that the laws of motion are Newtonian (or Galilaean). At the other (high) end the velocity curve becomes asymptotic to a line defining c and Einstinian (or Lorentz) laws of motion hold. That's all there is to it - It's the same for all cats.

I liked your analogy delta1212 +1

I am glad you are beginning to see the light, but I sorry to tell you that my comments only addressed the beginning of the problem and you have quite a way to go to fully sort it. It would make your life a lot easier for future problems if you tried out the questions I was suggesting, in the order I suggested them. There are three and only three forces acting on the body. All three of these may be replaced by a single resultant. Two of these are a right angles. Your question asks for the coefficient of friction, which is the ratio of two of them. It is, however very, very important to realise that you either have the resultant acting or the system of three forces, but not both. It is common to show the resultant dashed or in another colour or distinguished in some other way because of this. The resultant is the sum of all the forces acting (in this case 3) it replaces them as an alternative. This leads to the theorem that if the resultant is zero then the forces are in equilibrium, or that the acceleration of the body is zero, by Newton’s second Law. You need this fact to complete the problem. Turning to your use of x and y directions. It is conventional to use x for the horizontal direction and y for the vertical. So if an examiner sees them, that is what he will think. The acceleration given is parallel to the plane, not vertical ie not along the x or y axis. Can you tell me what is the acceleration perpendicular (normal) to the plane? You then have two accelerations, one parallel to the plane and one perpendicular to the plane and you can write two Newton’s second law equations to obtain the two forces you require to find the coefficient of friction. How are we doing ?

So Duhem didn't understand Maxwell, which is probably why he ends up as a footnote if mentioned at all. Duhem had many mistaken ideas. http://en.wikipedia.org/wiki/Pierre_Duhem You seem to have a remarkable reluctance to acknowledge my contributions to your thread. Why is this?

Perhaps the word resultant force instead of net force might make it more clear?

Since there is only one temporal dimension, any proposal can only look at it fractally in a Cantor's dust sort of way. https://www.google.co.uk/search?hl=en-GB&source=hp&q=cantor+dust+fractal+dimension&gbv=2&oq=cantors+dust&gs_l=heirloom-hp.1.3.0i10j0i22i30l9.1265.3640.0.6422.12.12.0.0.0.0.156.1408.1j11.12.0....0...1ac.1.34.heirloom-hp..0.12.1408.5-hF1xKtqq4 does this help?

Yes and the body is accelerating So it is not in equilibrium So your free body diagram should not be an equilibrium diagram This is what swansont (and your book) means by 'net force' So can you write Newton's second law for the body in the direction of down the plane? Remember that Newton's second law is a vector equation, so the force and the acceleration have direction.

Consider the two equations x = 3 2x = 6 These equations are really one and the same equation, since we can get the second by multiplying the first through by 2. We say they are dependent. Dependent equations may or may not be solvable. The ones above are since there is one equation and one unknown but x+y=3 2x+2y=6 cannot be solved since there are now two unknowns and only one equation. The second is still double the first and still dependent. Now consider x=3 x2= 1 These two equations are not consistent since either can be solved, but there is no solution that satisfies both.

Hi Nicholas If you do more maths, you will find that it is not always so easy. But yes you have detailed two approaches to solutions. Did you also have any thoughts on the actual question?

Imatfaal you may wish to know that Chikis is also working on this question in this thread, or at least the presentation. (This is not a complaint.) http://www.scienceforums.net/topic/84559-how-do-i-show-the-underline-when-am-solving-equations/

The equations are dependent if one is a multiple of the other That is if there is one number I can multiply one equation through to get the other. What happens if you multiply the second equation by 3? Can you now solve the system?

Where does this force come from. That is : what provides it or why does it exist? And how does this fit in with Newton's Laws? That is : which of Newton's laws says that a body 'has a force because it is moving'? Swansont has already provided one strong hint. IMHO it is more important that you properly understand this point than that you get the correct numerical answers to this problem (which you will do easily once you get the point).

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

Have you tried the underline icon on the Toolbar in the full editor? Go to More Reply Options Click on U

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