studiot

What makes you think there is a single most fundamental branch of mathematics? Why should this be so, after all the Hindus thought that the world was carried on four elephants and even neolithic man placed his lintels on two supports. There are many different ways you could divide maths into branches. One such is maths is the study of number and shape. This was certainly true early maths, although the study of Geometry was indeed codified first, this could not have happened without some numebr theory. Here is an interesting viewpoint from one of the world's most famous mystical poets. Coleridege was also responsible for this poem about the first proposition on Euclid's first book of fomal geometry. http://blogs.ams.org/mathgradblog/2013/06/05/euclid-coleridge-poem-2/ More recently we have categorised the subject into two branches Analysis and Synthesis. Synthesis is interesting because it leads to the idea of cmputability, a modern notion. In modern times a good source for your research would be the Cambridge University text "Computability and Logic" by Boolos and Jeffrey go well

Just out of interest what do you mean by this? To balance the forces (is force balance on the GCSE? syllabus) in an infinite ocean it is a relatively simple calculation as CharonY suggests, although you must know a good deal of underlying physics to do this. But to measure the viscosity using stokes law requires some quite sophisticated corrections for real world apparatus.

I am having real trouble reconciling GCSE with first year university, which is where a student would normally meet the Stokes falling ball viscometer. Google provides many hits on stokes visometer here is a university lab from one. http://www.engr.uky.edu/~egr101/ml/ML3.pdf This gives the fully corrected formula. A simpler one is viscoscity, f =[math]\frac{{2{r^2}g}}{{9\nu }}({\rho _s} - {\rho _L})[/math] ps and pL are the sphere and liquid densities respectively r is the sphere radius v is the terminal velocity. I can't see that you could be expected to produce a derivation of even this simplified equation for GCSE

Actually that was the question. The kilograms were an example, as shown by the opening words "for example". The infinitesimal in the example was dm, which dimensionally correctly refers to mass. Since you have revised your question (1) from how many numbers make 1 kg? to How many infinitesimals (I will take that as dm) make 1 kg? and I said I would answer a properly posed question here is my answer. (1) An uncountable number. Perhaps I should point out here that is because the domain is R3. For the series I posted on the other hand the answer is a countable number. I already did answer your second question, quite specifically. You still have not answered any of mine.

Actually it is on topic since it addresses the OP. You have already been told by a moderator that your approach is off topic. Although you have not answered my question, I will answer yours (1) The question is flawed since the units are different on either side of the equation. Couch your question correctly and I will try again. (2) For the purpose of many physical models (already described) yes that is true. It is not a belief system, but a matter of definition.

Are you saying that the following are not true and should not therefore be taught to final high school / tech college / first year undergrads as they have been for more than the last hundred years? [math]\bar x = \frac{{\iiint {\rho xdxdydz}}}{{\iiint {\rho dxdydz}}}[/math] [math]{I_{xx}} = \int {\rho ({y^2} + {z^2})dv} [/math]

No, physics was not implied, maths was explicity called for since the OP not only posted in the maths section as opposed to the physics one, but further chose an area of pure maths over applied maths. In any case your argument that an infinite number of physical points cannot make up 10 kg is suspect. A vast area of physics, including most of classical physics is predicated upon the premise that you can indeed either infinitely divide a finite piece of matter or alternatively assemble a finite piece from an infinite number of parts. This underlies classical statics and dynamics, continuum mechanics, and even the angular momentum of quantum particles is derived from continuum mathematical analysis. Can you tell me any reason why, if each of the numbers in my series posted above was a mass coefficient, I could not assemble 1.6 kg from the first series and 1.2 kg from the second and any other value by suitable scaling?

I don't have a full solution, but try the substitution [math]n = \sqrt 2 p[/math] Which leads to the condition [math]p = \sqrt {\frac{{1 + {m^2}}}{4}} [/math] Which should be easier to handle.

This site may help with your general study. Look around beyond the particular page linked. http://sciencepark.etacude.com/chemistry/law.php Yes you are on the right track with your thoughts, that atom arrangements are changed during a chemical reaction, forming new compounds (molecules) from the old ones. Nothing is gained or lost.

Electrolysis? Split Elements? Perhaps you are talking about bond energies of compounds? Would you like to rephrase your question?

Here is areference that may assist http://en.wikipedia.org/wiki/Exergonic_reaction

Remember that the terms exothermic and endothermic refer to heat energy only, and does not include entropy effects directly. There are more general energy balances, Chemists tend prefer the Gibbs Free Energy, which does include entropy in the TdS term.

Your question brings up an interesting bit of history because Archimedes was the first to successfully study this question and he wrote down his mathematical treatment in the middle of the third century BC. Unfortunately this document (called The Methodf) was lost in antiquity and only rediscovered in 1910, when and old parchment was cleaned. Anyway suppose we consider this thought experiment: Take a ruler and pencil and draw a thin line 25mm long. Draw another line right along side the first line so that you cannot see any gaps between them. Continue drawing lines for about three hours. You will then have a rectangular area on your paper. No imagine sharpening you pencil and repeating the experiment. It will now take you many more lines to draw the same rectangle, say six hours work. Sharpen again and repeat. Perhaps you can see where this is going. The thinner the line the more you need to create the area until. Until the line is so thin the number is ver very large indeed. This is what is meant by tending to infinity. This was exactly the process by which Archimedes derived his famous mensuration formulae, and the process by which we add up a very large number of very small contributions to create a whole. It is well known that we can add up an infinite number points to obtain a finite total. Mathematically that is what taking limits is about. However you ask how can an infinite number of points add up to different values? Well the simplest way to see this is to look at and compare a couple of infinite series. [math]\sum {_1^\infty \frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{5^2}}}.............} [/math][math] = 1.645[/math] [math]\sum {_1^\infty \frac{1}{{{1^3}}} + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + \frac{1}{{{4^3}}} + \frac{1}{{{5^3}}}.............} [/math][math] = 1.202[/math] You can see by direct term by term comparison that these two series have the same (infinite) number of terms, but their sums to infinity are different.

I sugest you reconsider your answer to part (i), which I have underlined. Setting aside the inefficiency of light conversion (typically less than 10%), and given the configuration mentioned what % of the light from the source reaches the reflector? This should then help with ideas for part (ii)

When certain individuals are firing on the range the safest place to stand is by the target!

I agree, Doe anti mean in some way inverted? A mirror image perhaps? Take a simple plane mirror. It appears to 'invert', left to right but not top to bottom Of opposite 'sign'? mass has no sign Some version of chirality perhaps? Just as we cannot assign an absolute coordinate system to space we cannot then assign a chiral structure either.

The horizontal distance between A and the impact point is x, not 50+x. It is 50+x from the launch point to the impact point. I get 50+x = 98.29825t horizontal flight y = xtan42 = 0.9x Triangle y = 68.829t - 4.9t2 vertical flight Three equations, three unknowns. Easy solution over to you

You are correct that the origin of magnetism in material matter arises from the motion of its electrons. This is because electrons carry an electric charge. Any moving electric charge generates a magnetic field, and the electrons are always in motion. The electric current in a copper wire is due to electrons in motion and this also generates a magnetic field due to this, however not all material objects carry electric current and yet respond to magnetic fields and some actually produce their own field. Except for ferromagnetic materials (Iron, nickel and a few other substances) no substance exhibits magnetic effects unless it is situated in an externally generated field. Ferromagnetic materials can form what are known as permanent magnets. That is they remain 'magnetised' when there is no external magnetic field. So they can generate a magnetic field in their own right. Copper is not a ferromagnetic material and cannot do this. The magnetic effect in copper wires is due to electrons that are 'free'. That is they are not part of any particular atom and move along the wire, whilst (obviously) the copper atoms stay in place. Electrons belonging to particular atoms are very important since not only do they determine the chemical properties of that atom, they also determine its magnetic properties. The electrons in an atom are circling round the nucleus of the atom in quite specific manner, not "loosely round an object". Because they are travelling in orbit around the nucleus they generate a (small) magnetic field. In the atoms of many substances there are as many electrons going one way as there are going the other so the effects cancel out and there is no net magnetic field. However in iron atoms there are four electrons more going one way than there are going the other around the iron nucleus so these four electrons generate quite a powerful net magnetic field. Finally some atoms have one or two unbalanced electrons orbiting, but the atoms are arranged randomly so that the fields of one atom cancel the fileds of nearby atoms with a net zero result. How are we doing so far?

Hello, you have posted this in homework help, but you don't seem to have a homework question. Are you really asking for a better explanation of magnetism or do you have an associated question. Please clarify.

So the projectile travels a horizontal distance = (50 + x) where x is to be determined ( and a vertical distance of y, also to be determined). Have a look back at your first projectile question and draw a sketch similar to my one there and then look at the equations I started with. Can you find a pair of simultaneous equations from this that you can solve for x and y?

There is another approach, that leads to the same answer, but you said you were considering energy. Since the 200N force is constant it imparts constant acceleration to the box. So if you draw a velocity-time graph; The box starts at time t0 = 0 at zero velocity v0=0 and is then accelerated by the constant force with acceleration a1 for a distance of 10 metres until it enters the friction zone at a velocity of v1. The box is then subject to a constant acceleration a2 (which works out negative) due to the difference between the 200N force and the opposing frictional force for a further distance of 10 metres until it reaches v2. You have enough information to solve the system this way. Have you covered this method in your studies?

part c? Well if this was an exam I wouldn't get may marks since I didn't read the question properly. Red face. You have correctly identified that velocity is a vector and may be resolved into horizontal and vertical components. Further this is also true of acceleration and distance. There is only one acceleration, that due to gravity which is vertical and downwards. There is no horizontal acceleration (ie it is zero horizontally) So we used the formula s = ut + 1/2 f t2 to solve parts a and b. Note I have used s for general distance, and f for acceleration. This allow the use of a, b, c for angles instead of Greek letters and also allows us to put in the distance in the appropriate direction. To solve part c we introduce a second formula v = u + ft where v is the velocity at time t and u is the initial velocity. Horizontally Vx = 15cos(50) + 0*t = 9.6418 m/s Vertically Vy = 15sin(50) + (-9.81)*(2.153) = -9.630 m/s ie downwards. Total velocity V = sqr(Vx2 + Vy2) = 13.627 m/s You got all this but I see you were unsure about the angle denoting the direction of travel. This angle, call it b, is in the fourth quadrant since the travel is down and to the right. We obtain a fourth quadrant angle by calculating a first quadrant angle © from arctan(Vy/Vx), ignoring any signs and subtracting from 360. b = 360 - c This is consistent with the velocity diagram I have shown superimposed on top of the trajectory diagram below. Have I completed the job this time?

I'm glad there's someone else on the planet still prepared to read books.

Talking of books you might like the following (Cambridge 2008) On Space and Time Edited by Professor Majid Essays by Connes, Heller, Majid, Penrose, Polkinghorne, Taylor on the title subject.

Gosh, Gwiyomi, you sure must like doing arithmetic. I make t = 2.153 seconds and x = 20.76 metres So pretty close to your results. For your information about projectile problems, The path of the projectile is fixed by the launch speed and angle alone. If you launch up and to the right as positive (usual x and y axes) then the position at any time t is given by a pair of equations x = Vcos(a)t y = -1/2gt2 + Vsin(a)t where a is the launch angle, and V is the launch speed. That is the x and y coordinates at any time in the flight are given by those two equations. It is only a couple of simple lines to derive these so you should not try to remember them, just how to get them. If you enter your value of y = 2 (and V and a) into the second equation it yields a quadratic in t. The two solutions correspond to going up and coming down. Knowing t you can substitute into the first equation to find the x distance at this time.