studiot

Yeah, but as John says, what does that have to do with general use in connection with radiosotope tracers?

I suppose it depends what you are measuring, can you provide more detail, rather than just repeating the question? Half lives are really only strictly applicable to first order reactions, such as simple radiactive decay. But say you did a radio-isotope dilution analysis with an isotope of half life 10,000 years. What use would knowing that number be during your afternoon visit to the Chemistry lab? The actual activity would be essentially constant for the duration of the experiment. Relaxation methods are often used in Chemistry to investigate fast chemical reactions, when the relaxation to equilibrium is suddenly disturbed the return to equilibrium is mostly first order.

For any one else interested I make the value of 59.4 the solution to the following equation which needs to be formed and solved. Let X be the number of m2 of panels attached to walls and ceiling. Then (0.2 * 48) + (0.4 * X) + (0.05 * (160-X)) = 38.4 I have rounded the factor of 0.05 I anticipate using a more accurate value would yield the required answer of X = 59.7

How would you obtain violet light using only red, green and blue light? Note you can obtain the subtractive colour violet by shining white light onto a violet coloured object, thereby removing all but the violet light from the white.

You first need to set out whether you are discussing the colour (subtractive colour) of some object such as a painted wall or the light from a light source (additive colour).

No.

The answer of this question is 59.7 m2 , can someone point out my mistake???? Well I would start by checking my arithmetic. This comes to 64m2 not 73.6m2, but then I do get 73.6m2 if I put the correct values of the absorbtion coefficient in for the ceiling so perhaps you copied your work incorrectly? I have underlined that part. However 73.6m2 is what happens if you cover the entire wall and ceiling area with the stated absorbtion coefficient as well as carpeting the floor. Put that into your Sabine equation and you will find that it does not yield the correct reverb time. Do you know how to go about reducing the treated area to get that? Hint your carpet area if fixed so what happens to RT if you take it out the the running? What treated area do you then need to finish the job?

Thank you for finding that thread geordief, I half remembered it, but I seem to have trouble finding old threads again. I note I should have proposed the 'effective radius' of curvature to be of the order of 1 x 1018 metres. This is from a simple geometrical calculation of actual measurement, which can be and has beeen made.

Would I be right in calculating that Eddington's value for the grazing deflection of light past the Sun of 1.6 seconds of arc implies a value of around 9.6 x 1017 metres for the curvature of spacetime in the vicinity of the Sun?

Flourescent lights work by using the phosphor coating on the inside of the tube to emit the wnanted light. Different phosphors give different spectra.

Your questions and opening post are a bit jumbled so before going into detail let us get straight exactly what you are asking. Lorentz and Fitzgerald had some experimental results so they developed mathematical formulae that fitted these results, but they had no idea why in terms of the science of Mechanics of the day. Einstein came along with a theoretical hypothesis, based on new thinking he called "The principle of relativity", which produced the same formulae (and others besides as it included electromagnetic phenomena as well as mechanical ones.) This was in 1905 and the new theory was called Special Relativity. It did not involve curved space. So they then had experimental evidence/verification of this theory. In 1915 Einstein produced a paper with a wider ranging subject matter he called General Relativity. This paper did indeed incorporate curved space. Furthermore Einstein again offered the possibility (he did not conduct the experiments himself) experimental verification, but this was much more difficult and could only be achieved over astronomical distances. This was his first paper on the subject and introduced another new principle of Physics. So when you say 'proove the curvature' are you interested in a) Understanding what is meant by this curvature? b) the theoretical development that leads to the conclusion that curvature exists? This is pretty difficult. c) Or are you interested in the experimental evidence that supports this conclusion (which is easier)?

Yes it is a good video but I agree with Strange that SR is Euclidian and not about curvature. +1 for finding it though.

Can't see what you are disagreeing about, perhaps you are smoking the wrong end of your cigar? Nothing you say there is at variance with what I posted.

It arises because when you divide by something it is the same as multiplying by the reciprocal of that something. Since metres per second means metres divided by seconds it also means metres multiplied by the reciprocal of seconds or 1/seconds. [math]metres \div \sec onds = metres \times \frac{1}{{\sec onds}}[/math] So if we divide this again by seconds we get [math]\left[ {metres \times \frac{1}{{\sec onds}}} \right] \div \sec onds = \left[ {metres \times \frac{1}{{\sec onds}}} \right] \times \frac{1}{{\sec onds}}[/math] or [math] = metres \times \frac{1}{{\sec onds}} \times \frac{1}{{\sec onds}} = \frac{{metres}}{{\sec onds{}^2}}[/math]

https://books.google.pl/books?id=V7IR6tFfaAEC&pg=PA65&lpg=PA65&dq=non-differentiable+space&source=bl&ots=sual_23EI-&sig=wNszgl3QdvymNGBICd0KNhfH9nI&hl=pl&sa=X&ved=0ahUKEwj19q2UuKHaAhUQJlAKHcdYAZ4Q6AEIiQEwCQ#v=onepage&q=non-differentiable space&f=false "The theory of scale relativity [14) is an attempt to study the Of giving up the hypothesis Of space—time differentiability. One can show [14] [15] that a continuous but nondifferentiable space-time is necessarily fractal. Here the word fractal [12] is taken in a general meaning, as defining a set, object or space that shows structures at all scales, or on a wide range of scales. More precisely, one can demonstrate [17) that a continuous but nondifferentiable function is explicitly resolution-dependent, and that its length C tends to infinity when the resolution interval tends to zero Well I assume that twaddle is a result of improper translation to English since your original was in some other language, that I can't read. To be fair it does confirm what I said that functions are differentiable, although it starts off with something that is clearly mistranslated. The whole point about functions is that they are a two part mathematical object. And that's what makes them differentiable, because differentiation is a two part process. When I learned calculus we always had to write as justification something like ""differentiating .Y. with respect to X" as justification. This definitely brings out that two- part nature containing both that which is being differentiated and that with which the differentiation is being performed. This is especially important for the sort of mathematics you have already invoked here using cycles and frequency ie the wave equation. That is because you differentiate one side with respect to distance and the other with respect to time. The object being differentiated is of course the wave variable, which is a function of both space and time. Thank you for the link about fractals. I am aware of the nature of fractal sets. Do you understand measure theory and Hausdorf dimension theory? These are needed to handle fractals.

Not sure where you got that graph from but do you understand what you are talking about? [math]y = \sqrt x = \sqrt 0 = 0[/math] is perfectly well defined. But so what? What does this have to do with the differentiability or otherwise of a space? [math]y = \sqrt x [/math] is a function, which is indeed not differentiable at x=0 because it is not defined for x<0 It is functions, not spaces that are differentiable or not. Stating that a space is or is not differentiable has no meaning, which is why I asked what you mean by such a statement.

Will you just answer my question since it concerned your words not any one else's?

Well, if his theory, deserves Nobel Prize, then maybe I also deserve somekind of prize I made an extension to his theory, without even knowing about it... But what I was asking was were you even stating his theory correctly, let alone understanding it? I will ask my question again. What does it mean to say a space is or is not differentiable?

Really? What does this statement mean please?

Are we both reading the same thread? Or are you just being funny?

Look at the two blocks you have labelled 'pure cyan' They appear different. The one on the left appears to be pure cyan plus grey to me. The one on the right appears to be a fair pure cyan The two you have labelled pure yellow are more different and I think, though I may be wrong, that they are pure yellow plus some other non grey light left and pure yellow (right). I apologise, I now realise that they are your original colour blocks, I was too hasty.

What are you doing there? These are not the colour blocks you originally posted. I didn't make clear that the "other light in the beam" ias not necessarily grey. One of the colour models uses the neutral grey to adjust the apparent brightness of the pure colour. I think this is called the Hue Saturation Brightness (HSB) model. A % grey can be added to any pure colour this way. But that is a restriction of what can be added to any pure colour. Adding any other mixture but the grey alters the received appearance of the colour ie changes the colour.

But I am challenging that that is the case. Therefore, and if I am right, Yes?

But it doesn't, as I have tried to explain. You are mixing up situations. The way to see the colour of additive light sources is to shine them on a common area on a white screen or paper. You don't stare into the beam. What you then see in your two different situations looks different because it is different. That is because the illuminated spot you see is giving off two different lights. One is the pure narrow bandwidth called yellow. The other is the agglomerate of all the other light in the beam, which is what I meant when I originally asked how much grey there is in the left hand sample.