studiot

Well yes it is true that you have to start somewhere to develop a theory or whatever. This is exactly what you will do when you leave college and work in the outside world. There you often have to decide for yourself what level of accuracy, what parameters want to calculate, what level of sophistication your calculation (remembering that in general the more sophisticated the more it costs) ans finally if your answer or result is sufficient. But in school or college you are looking to achieve a predetermined result, set by the curriculum, so you don't have the luxury of trying an approximation and refining it. The nearest I can think of in a college setting would be to check if a formula you think is correct by substituting some simple easy to calculate numbers to see if it yields a known result. Which all still comes back to Don't add things they haven't asked for.

Not always. Just look at some traffic lights. The colour you see is the colour radiated by the source.

You would also need basic Physics to first degree level to pursue what swansont and ajb mean by 'mechanics', So my advice still holds good in this case.

If you are up to the job there are good career prospects if you take Physics as your first degree, specialising in options that will allow you to follow this with postgrad in Mechanical Engineering. Taking things is the other order is not recommended. If you eventually don't like Mech Eng you are a bit stuck with other fields to move on to. But it is relatively easy to move on from Physics to many other technical disciplines. I know medical doctors, surveyors and undertakers with first degress in Physics. Work from the general to the particular. If all else fails you can go with your first option and you will always be able to predict the future.

Yes of course you are. But you don't need to make that assumption, any more than you dont need to make other assumptions like The balloon is not rising through the air and cooling and so shrinking. The expansion of the balloon is not sufficient to burst it. and so on. Each of these possibilities can be calculated by more complicated physics, leading to more complicated maths, but we ignore them. We ignore them because the original question has not mentioned them so we don't either. We concentrate on what the question does mention. If you look, all these assumptions contain the word 'not', they are about what the question is not. Try to avoid negative assumptions about what questions are not. They don't help solve the real question itself.

I am going to look at this question of yours from some different perspectives. What is the context of the question? What does the question want you to find? What is an assumption? Why are any assumptions needed and under what circumstances do you need to make them? Are there any other sources of information? What sort of questions are out there and what is their purpose? How do you become good at extracting information and supplying what’s missing? OK so last two perspectives first (the easy ones together). You become good by lots of practice. It also helps if you can enjoy it. There are several ways to get this practice. Many people enjoy doing puzzles, brainteasers and the like. You can find these in magazines, books of puzzles, online and quiz programs. Older maths books aimed at the 12-15 age range often had many word based questions for pupils to practice with along with advice on how to do them. For example: 54 pots of jam were tested and it was found that some were 1oz overweight and the remainder 1/2oz underweight. The total weight was correct however. How many jars were underweight? A tank X contains 1000 gallons of water and water flows out at a rate of 5 gallons per minute. Another tank Y is empty and water is flowing in at 10 gallons per minute. (1) How much water is in each tank after the water has been running for t minutes? (2) How much more water is there now in tank X than in tank Y? (answer in terms of t) (3) Find this amount when t= 10; t=60; t=100. What does the last answer mean? The second question shows an important point that often part (1) of such a question calculates something you will need in the second or later parts. For the rest of the perspectives I will use your example. So context Had this been from a physics book then physics considerations might have been important. Such as temperature, pressure, Boyles or Charles law, the elasticity of the balloon. But this was from a mathematics book in the mathematics section so mathematics is important and such physics considerations are unimportant and can be discarded. But what mathematical considerations? Well first we extract the information given. This information is only the information given. Nothing else. No assumptions, guesses, inspirations or whatever. The balloon is a sphere. Then we supply relevant (mathematical) information that we know. This information is not an assumption. It is a hard fact or formula that we have learned/been taught. In this case the volume of a sphere is [math]V = \frac{{4\pi {r^3}}}{3}[/math] Do we need to assume that the balloon does not change shape? Not really since the volume of any boxy or blocky object is still proportional to the cube of it’s ‘radius’. What is the context? Well calculus and the books want us to differentiate an equation connecting volume with radius. So do we need any assumptions about holes in the balloon? Well no because holes are not included in the standard formula connecting volume and radius. So long as we extract the facts given and add facts that the book expects us to know as part of our course, we should not need to make any assumptions.

Lizzie L put a lot of effort into this thread. http://www.scienceforums.net/topic/82666-can-a-white-piece-of-paper-reflect-light/

At this time of year, we occasionally see the green ones in the Wiki picture. There were a couple of these orange ones. Thanks acme, I will see if any better information comes along.

Can anyone identify this moth or insect please. About half inch long, bright orange, dart shaped two prominent round black spots, Found on a lavender bush.

You need to develop a hypothesis to test. For instance that the measured effect is independent of concentration, normally distributed over some range of concentration or whatever. That is what John's analysis of variance will do for you. Better advice will come if you can state your hypothesis or hypotheses. Better still, next time, formulate your hypotheses before carrying out the measurements.

Well this is a continuous distribution of the probability density function p(x) for x>0 (it is zero for x<0) [math]p(x) = \lambda {e^{ - \lambda x}}[/math] :x > 0:lambda > 0 It has one parameter, lambda which is positive. The mean is given by [math]Mean = \frac{1}{\lambda }[/math] and the variance by [math]Variance = \frac{1}{{{\lambda ^2}}}[/math] As you can see, the probability falls with increasing x so you might use this distribution to model say your chances of living x years beyond 100 years of age.

University of Hawaii? Magnum PI will never look the same again.

This is homework then? Binding energy is the energy that holds the nucleus together, hence the question is about atoms. It is the release of binding energy that allows lighter atoms to undergo nuclear fusion and heavier ones to undergo nuclear fission. There is a maximum around nickel - iron the the periodic table, which divides prospective fission or fusion. Read this and come back if you still cannot complete your homework. http://en.wikipedia.org/wiki/Nuclear_binding_energy

In the past I have helped preserve research confidentiality by use of the private messaging system. However I am not a cosmologist, and ajb knows a great deal more about tensor gravity models than I do so I am not the right person to help further here, since the last time I did that with tensors was in fracture mechanics.

Thank you John, that would make the problem easier. So that would make this first part you did correct. based on the above subsets

Well have you even attempted what I asked? It was a very simple request. Yet your response was to introduce a universal set, that was not correct, and not called for by that part of the question. P, Q and R are all sets of numbers. But they are not sets of the same types of number. In particular, as you have written them in Post#1, P and R are sets of real numbers Q is a set of integers. Back to my question The intersection of Q with R refers to a set that contains only numbers that are in both Q and R. Q contains all the odd integers, excluding 0, which is neither odd nor even. R contains a restricted interval of real numbers, which includes all the integers in that interval. So the intersection of Q with R boils down to all the integers, excluding 0, that are in the interval containing R. That is what I mean by stating in words. Can you now state this in symbols?

Why should I, when you won't cooperate with what I have to say? Your error is in the quote, look again at your first post.

You have gone against what I asked and again gone wrong. What I asked was designed specifically to help.

Individual elements (or parts of one) could be a (coordinate) vector. Unlike vectors, it is not required that all elements represent the same physical quantities, as in my stress tensor example. This means that you must have enough equations to handle different relationships.

If I take a £1 coin out of my right pocket the change in the number of coins is my right pocket is negative. That is I have one less coin in my right pocket. If I place that coin into my left pocket the change is positive. That is I have one more coin in my left pocket. The total coins I have remains the same. The situation with KE and PE is identical. This is because the sign is associated with the change, not the coin or energy itself. Conservation laws of coins (I wish) and energy require that one man's loss is another's gain.

Note also the specification when two material bodies interact. Then yes there is always a second opposing force. Note that Newton also included words to this effect. This extra specification is necessary because he needed to exclude the situation such as gravity in the space around objects. The force of gravity only comes into operation if a second object appears, but we cannot say there is no gravity around an object just because there is no second object there.

Exactly. There is an applied force (often called 'the action') This stimulates a second opposing force, often called the reaction. As I noted bfore they are exactly equal in magnitude, but opposing. Recognise Somebody's Law? Note also that this conclusion you have drawn, does not follow. That is it contradicts your first statement above and of course itself since it talks of creating a second force.

Yes of course and perhaps the drawer is locked or painted shut or something inside is sticking up and catching or....................... But none of these matter since they all apply the same force - just enough to resist your pull no more and no less. The whole point of a model is to extract only the important facts and discard all the rest, even though they are facts. The important facts are The drawer is a box You are applying a force by pulling Something is applying a foce by resisting. One drawer, two forces, no other facts matter.

Did you not say the drawer was stuck? So what is holding it back?