studiot

I find this first response to anything I have said to you totally negative and really disappointing.

+1

Such a thing was actually built and did work, but was controversial. Because it did pour huge currents into the Earth, just as Tesla said. https://en.wikipedia.org/wiki/Project_Sanguine

Mathematically a surface has at least two dimensions. Time only has one. Which is part of one important reason you cannot travel in time. Where are you starting your Maths from?

No is the simple answer. A better answer is that the energy of the tap is called activation energy. Activation energy is often required for (energy releasing) events to proceed. https://www.google.co.uk/search?source=hp&ei=Pd-tW_aRC5KZgQbnk4nQDQ&q=activation+energy&oq=activation+energy&gs_l=psy-ab.3..0i131k1l2j0l8.984.5582.0.7010.17.13.0.4.4.0.266.1786.1j9j2.12.0....0...1c.1.64.psy-ab..1.16.1918....0.QGu9yNXcKhA

Interestingly there is a current homework question about a pendulum receiving a small amout of energy. Question 69 here. Perhaps you should try it? You should not consider such phenomena in terms of energy input and output. They are not clearly defined terms in this case. As to resonance, a pendulum is the simplest device to understand the workings of the phenomenon, because it can be made to occur at human timescales. Most resonance phenomena are too fast for humans to follow directly.

Strangely enough Maths hangs together. It is not a collection of separate bags of tricks to get you through an exam. That sort of Maths won't help you at all in Particle Physics. PP invokes some very fundamental maths that you don't meet until you have enough under your belt to cope with it (ie at university) You need a good understanding of probability and statistics (they are different). I don't know what your budget is but here are a couple of inspiring books yopu should be able to get hold of at reasonable cost. Both explore the relationship between (modern) maths and physics in an accessible way but include/introduce some quite advanced maths topics along the way From Calculus to Chaos (An introduction to dynamics) David Acheson Oxford University Press The Mathematical Mechanic (Using physical reasoning to solve mathematical problems) Mark Levi Princeton University Press

Be prepared for a big shock. In the UK the first high school maths exams is taken at 15/16 and nowadays called GCSE I thought I knew all about Maths (and other things) when I took mine. I thought the maths for the second exam A level - pre university evel - was just more difficult versions of the same. What a shock I had. University Maths as a secondary subject is just more difficult A levels, University Maths for Maths is something else entirely. So as Klaynos says, Lap up your lessons and learn them well and one more thing Ask plenty of questions here at SF about subject matters you are not sure of. Go well in your future studies.

Can someonetell me how this came about? How did this (legitimate as far as I know despite the name) member last vist on August 7, yet post twice today? I also noticed that the last member to be banned (yesterday) was lsited online again today. Since this latter type of event has occurred several times lately, are the two matters linked?

OK so a bucket of incompressible liquid such as water has a free surface (with the air). The pressure at that free surface is equal to the air pressure there. Below each point on that free surface the pressure in the column of water has two components, clled the static head and the dynamic head so you need to know which one you are looking for. This free surface takes on the shape of a parabaloid of revolution if the bucket of water is spun. The static head is due to the weight of the water above any point in the fluid and varies with depth. The dynamic head is due to the tangential velocity of the water, which decreases with increasing radius from the centre, but remains constant with depth.

Actually this question is simpler than I first thought, but your diagram is all wrong as I said. However physical reasoning about energy is definitely the best way to solve this. I make the answer to be A, [math]\cos \theta = \frac{1}{6}[/math] No complicated maths is needed.

Is this homework? You need to describe it properly and show your working. If not you still need to describe it properly.

Look carefully AD is the y axis or the line x = 0. AB is the line y = 2 DB is the line y = 1/2X2 I said this was the area of interest, ie the plate.

The clue is in the word 'draw'. I assume you understand what the tangent and slope iare. Sso when you have a measured curve and manually draw a tangent to it at some point the tangent will be a straightr line with the same slope all the way along it as far as you care to extend it. To calculate that slope you can either use a protractor or most people do it by taking two measurements with a ruler - The Y coordinate change and the X coordinate change, and then dividing one by the other. Clearly the larger each of these is the less % error your measurements will make. So the phrase is conveniently large, that you can measure with your ruler. Does this help?

A learning point here. Your analysis route was quite complicated. When they teach series and parallel circuits, using resistors the fact that in a series circuit the same current flows in all series elementrs is highlighted. Capacitors do not have conductive current, but capacitors in series have the same charge. This is often overlooked. But it can be used to develop simple and useful relationships.

For 70 attached is a plan of action. The area of interes forms a curved triangle which I have labelled ABD. The centroid of ABD is labelled T which has an x coordinate [math]{\overline X _T}[/math] The area under the graph of [math]y = \frac{{{x^2}}}{2}[/math] forms another curved troangle I have labelled DBC. This has centroid V with x coordinate Note that these two curved triangles fit together to form a 2 x 2 square (I have labelled this ABCD) This square has centroid S with standard values for its centroidal position. Now the trick is to realise that the moment of this square must equal the sum of the moments of its two componeent curved triangles. This should be a good start for you. As to question 69, Are you sure you have presented the problem correctly? The half metre length of string must go slack at the intial throw and only become tensioned when the mass has fallen back. The questions states that theta is measured to the left, ie when the mass is hanging on 1.5 m of string and swings to the left, clear of the peg. This would make the radius of circular swing 1.5 m.

Ok I have read through your working and here are my thoughts. 1) It is nice and - a pleasure to read, unlike some. 2) You have correctly deduced that for dielectrics havinf the same cross sectional area in series the electric field strengths (or potential gradients) are inversely proportional to their relative permittivities. Well done. 3) You have correctly used this to establish that the air layer will be the first to breakdown and subsequent working is correct to calculate that voltage. However 2) You need to be carful with terminology: permeability refers to magentic effects. The correct electric term is permittivity. 3) Apart from the silly sign in the power of 10 I agree with your arithmetic so far as it goes and you have found the sign error yourself. 4) But the question was What is the maximum voltage that can be impressed and you have not addressed what would happen after breakdown of the air. Would the capacitor fail? In fact I suggest that the air layer would fail (ie become ionised and of low resistance) and the capacitor would continue to function with only the porcelain dielectric acting. Thus you can recalculate the voltage where the porcelain would fail by considering the air a short circuit. Does this give the answer they seek?

I have just put my thought down in this thread both about the orientation and size of posted pictures. Neither too large nor to small and the correct way up. greyscale also saves space. Could some instructions along these line be drafted so that we could point to them?

This is a more suitable size of file 300k instead of 3000k. Greyscale also helps if you don't need the colout. And please put it up the right way. I have just spent the time I could have been reading it doing this for you. I will be back later when I have read it.

You are more polite than I am. +1

Perhaps you should post in the poetry section ? We could then get the resident poetry specialist to dechipher it. Let us start with this one. Looking back I have to say that I made a mistake because you specified a positron and I wrote about a proton. My apologies. So I now see why you keep asking about "in an atom". An isolation box is not an atom. Of course there are no positrons in an atom. An isolation box is just a theoretical way or device for including one positive and one negative charge to say that the net charge is zero. When I asked you to state the Law of conservation of charge I wanted you to understand that the Law does not say simply charge is conserved. That is not good enough. The Law also includes a description of the place where the charge is conserved - the isolation box. Otherwise you could simply say that charge in one place (or within one 'box') is not conserved because the charge moves outside the box or because you include a charge that was somewhere else when the process took place. So when we talk about the conservation charge for an electron - positron annihilation on Earth we start (before the annihilation) with one electron and on positron - net charge zero - in the box (Earth) After the annihilation we must still keep the same box (Earth) We cannot suddenly include a positron on the Sun. Since we now have no charges in our box the net charge is still zero. If we had one electron, one positron and one proton (to start) then the net charge would be +1+1-1 = +1 After the annihilation the net charge would be 0+1 = +1 Again charge would be preserved. Remember also that there is no requirement for the charges listed to be connected in any aprticular atom or even to remain in that atom or molecule if they are. The Law is not about atoms. For instance consider the ionisation of common salt - sodium chloride. Yo start off with one neutral molecule of sodium chloride. The net charge is zero. Following ionisation you have two ions one positive sodium ion and one negative chloride ion. The net charge is still zero. If this is this any better we can move on to 'mass charge' and 'shadow stuff'.

O great God Moderator I pray to you to save us from the evil of this thread and close it.

How does this meet the Uncertainty Principle?

If you are going to do any sort of degree you are going to need to do some degree level thinking. So I suggest you get your thinking cap on and re-read what I said about the common core.

Hi, It's good to think seriously about the future. Here are a few things you need to know. Most colleges/universities that teach engineering degrees have a 'common core' for the first year for mechanical, aeronautical, automotive, production, civil, building services and so on. One bonus of this is that it is easy to switch after the first year if you decide to change the emphasis of your studies, when you specialise. The studies will include the basics of mechanical science, materials science, electrical science, engineering mathematics, engineering computing and so on. The computing element will be about the use of computers. Specialist computing courses will be about the design of the software that engineers (and others) use, and/or the design of the computers themselves. It is not usual to get several first degrees (Batchelors) for several reasons. The usual route is first degree then a higher degree - a Masters (which may be in accounting or business and taken after a few years industrial experience) Remember much of the art of engineering is about best use of money. The other reason is that full/top professional qualifications for most engineering disciplines used to be available at Batchelor degree level, but have now been elevated to require Masters. Does this help? Please ask if you have further questions.