studiot

Hello, I don't know if your concern is safety That is can an ordinary antenna, designed for radio waves or even microwaves emit harmful radiation? Or are you hoping to generate such radiation using an antenna? Gamma radiation is the name given to natural radiation that is the result of nuclear processes. X radiation is radiation of the same frequency / wavelength that is artificially generated, but not using an antenna. Old fashioned vacuum tubes, particularly cathode ray tubes, produce (unwanted) X rays as part of the process within the tube. Out of interest here is a spectrum chart showing the relationship between various emissions, after the RSGB Radio Data Reference book.

Strange has already given you a pretty good answer to your question, but since is seems to question something I said, I would like to add the following. I don't recall the thread you refer to, perhaps you could find the link (look in your content in the drop down box under your name), but I suspect that you must have misunderstood the underlined part. Charge is not stored in a field. Energy due to the interaction between a charge and another charge or charges is stored in the electric field that exists between those charges. If that field passes through dielectrics ( and remember that empty space is a dielectric) then the charge densities and directions may be modified. Charge is a physical entity that you can add to other physical entities (the plates in this case). Both physical entities are complete in themselves and have their own existance.

+1, hoola, your answers are improving. In answer to your question: The charge resides on the plates. There should be none in the dielectric, any there would constitute leakage. But charge is not energy and the energy of separation of this charge (if any) is held in the electric field in the dielectric. This appears as a voltage, since voltage is energy per unit charge. Does this help

Good pics SJ, they show my points well. +1

No gas meter reads kilowatt-hours. Gas meters measure the flow of the gas. Usually they measure the total volume of gas that has passed through the meter since it was set in place. As has already been observed this is normally in cubic feet, I do not know of any that are metric, though I suppose the Europeans must have them. Measurements are taken by noting the difference over a specific period. The last digit or dial is only read when testing apparatus or seeking leaks. Once you have the volume of gas that has passes by subtracting an earlier reading from a later one you convert this to a nominal maximum energy by multiplying this volume by the stated calorific value of the gas supplied. All companies by UK law must state this value in their bills. The calorific value varies slightly from day to day. The gas company normally perform this calculation for you and also convert to the metric unit of kilowatt-hours. You should also know that this is the theoretical maximum energy available in the gas. No apparatus is 100% efficient so you will never obtain this figure.

Aw you've made it too easy, I was under the assumption you did not have the circle centres available. You said this was the first of a series. Please make the specifications more precise next time.

I just wonder about the link between the modern early curriculum, particularly in Physics in the UK, which seems to emphasis showy ethereal stuff over (boring) solid foundations. If you compare the numbers who rush to the cosmology / boundaries of science threads, but cannot explain how a bicycle works you might see some connection.

Two find his 'gain' you must consider the difference between two possible courses of action. It is not a question of applying a formula such as gain = sales - costs. So read back over the thread so far and decide what his two possible courses of action are . Obviously he can only carry one course out.

Gosh bicycles in particular seem to cause much confusion, so here is a more formal explanation for those not satisfied by John Cuthber's one. The confusion arises primarily because of the unnecessary complication of the rotational dynamics of the wheels. Consider the bicycle as a free body and apply Newton's First Law. It does not matter whether wheels spin, arms waggle, springs compress or the Lord Mayor blows his trumpet whilst sitting on the bicycle. All is irrelevant to the basic mechanics. The free body has two points of contact with the outside universe. It is only possible to apply external forces there. In the horizontal direction the only forces the external world applies are the frictional forces at each wheel. For a fee body in constant motion (cycling along at constant speed) they must be equal and opposite by Newton's First Law. In maths we can say If the bicycle is travelling from left to right, so the front wheel is at the right hand side and The friction forces are Ff for the front and Fr for the rear If |Fr| > |Ff| the bicycle is accelerating If|Fr| =|Ff| The bicycle is travelling at constant speed If |Fr| < |Ff| The bicycle is decelerating or (braking or freewheeling) Note I have used the moduli of the frictional forces since one must be negative. Can you work out which one?

The point of metals is that they all react with atmospheric oxygen so a chemically pure surface is oxidised within less than a second to few minutes. Some oxides are more stable, persistent and protective than others but all act as a sort of lubricant in thin films. I am in Dundee this week so access to data is limited, however I will se what I can dig up.

Imatfaal, if you are doing any tests you will need to create chemically pure surfaces for your samples. Table 1 is just what is says pure aluminium on pure aluminium etc not aluminium with an oxide layed on aluminium with an oxide layer. For that you need table 2 and 3 and 4 etc. I can only refer you back to Mark's, or any other standard Mechanical Data source. Kay and Layby have some as do Machinery's.

Nice demo, thanks ajb +1

You have entitled this thread relative velocity. Acceleration is also a vector so you could consider relative acceleration, and again answer my question, What causes the (relative) acceleration?

Perhaps a another way to phrase Janus' hint is What causes the ball to accelerate?

Well if I just wanted the roots in a hurry, I would probably go for a numeric method. I haven't fully worked out an analytical solution but some thoughts are You want to get rid of the (x+4) so look for a trig substitution [latex](x+4) = \frac{2}{\pi}(\frac{\pi}{2}x + 2\pi)[/latex] substitute [latex]y = \frac{\pi}{2}x[/latex] [latex]=\frac{2}{\pi}(y + 2\pi)[/latex] and note that [latex]sin\beta(y + 2\pi) = sin(\beta y)[/latex]

Maurice, you really do need to help us here to find the correct level to answer you. I asked about calculus because you said the stretched length is (L+l), which is not calculus. Also you need to answer elfmotat's question Do you mean a transverse wave or do you mean a longitudinal wave. For a wave proceeding down a spring I would guess you mean a longitudinal wave, which is a pressure wave. For such a wave you should be converting the tension to a stress (=pressure) by dividing by the cross sectional area and. Also you should be considering displacement not length for the spring.

Unfortunately this forum won't allow copy and paste of quote with a bog standard windows 7 system, fully up to date by \Microsoft standards. So I can't quote from your post#12 However you may rest assured that the you could not bring two magnets close together without inputting energy (in the form of work) to the system. So you cannot choose to discount or ignore this energy in your accounting. It really is that simple.

You need to tell us if you understand calculus, and the calculus of the wave equation. Or did your professor give you an equation using sine or cos? Can you obtain an equation using a restoring force? you need this for your question if you want to use Hooke's law

Well yes and no. Yes the energy came from your hands. And yes you input a small amount of energy turning the spinners. But no that was not all the energy you input. You have two magnets and it takes energy to bring them from a large separation to the close proximity you show on their pivots. Calculate this and you will find it is the bulk of the energy input, so there is no violation of conservation. Thank you for the video.

Rindler has written several books on relativity/cosmology. Here is the latest http://ukcatalogue.oup.com/product/9780198508359.do

Well let's see 1g of Fe = 1/55.85 moles 20g of CuSO4 = 20/159.62 moles So which one will be completely used up by the displacement reaction and which one will have some reactant left over? Do you know what reaction takes place? edit: reading this again did you mean 20g of 1molar solution of copper sulphate? The end question will be the same, which one is used up, but the figures will be different. Have you seen this http://www.youtube.com/watch?v=KmhD8BmEFIo Finally this is a displacement reaction so what displaces what? When reporting the heat of reaction I would say __kJ/mole of ___displaced

You neither answered my points nor provided accessible references for your claims, both of which are against the rules of thsi forum. I see no point discussing this further.

Partly depends on what you include in 'Science'. I would say that there has been an accelerating occurrence of paradigm shifts in human experience that started with the Renaissance. Look at an 'old' movie and see how much cellphones, computers, plastics have changed the world..............

Here is a review of the stuation or a restatement of the problem. A retailer has just bought goods for a price of N200. If he pays right now he will receive a discount of 5% (so how much does he owe right now?) But he has no money to pay with at this time. He expects to get the money to pay by reselling the goods in the future. So long as he pays within two months he will still only owe N200, but will not receive the discount. He can borrow money right now and pay right now and so receive the discount. But he will have to pay 9% per annum (=per year) on the money he borrows. If he borrows the money and pays 2 months interest how much is that interest?

Yes this is true for the purposes of this question and is also normal business practice.