studiot

You have been told that the moon does rotate, but not like the Earth due to a phenomenon called tidal locking. So to complete the final part of this jigsaw. First why does the Earth (or moon or other body rotate at all?) Well during and after the formation of any substantial body in space other (smaller) bodies crash into it. Whether they coalesce with the body or just bump into it they must hit it off centre which means that the collision exterts a moment about that central axis. If all the collisions are from random directions they will tend to cancel out over time. But if they are mostly from certain directions, as in the early solar system, over time this builds up to generate a sizable rotational speed and there is not much in space to cause friction so the rotation will last a long time after the period of planet building has occurred. So we have a body like the Earth, which is slowing down, rotating about its axis, but having substantial rotational momentum The density of the Earth is 551 5510 compared to the Moon which is only 334 3340 kg/m3. Edit see post7 Coupled with the huge size difference between the two bodies this leads to a huge difference in stored rotational momentum. I will leave you to calculate the difference. As strange noted, so far gravity is not involved, but introduce that into the situation and you enter the interaction between two spinning bodies via the tidal locking mechanism when the Moon became captured by the Earth's gravity.

You can't change the rules half way. A rocket is basically a small payload on top of a large fuel tank. The fuel is often 90% of the mass of the rocket. At any one time the payload plus remaining fuel is being propelled forwards thus the mass of this combination diminishes rapidly with progress. It is true that [math]KE = \frac{1}{2}m{v^2}[/math] But m is a function of v or the other way round. So you need to develop a mathematical statement of this relationship to assess the KE of the part travelling forwards.

Then you will realise that the mass of the rocket is not constant.

Your title is a good one, because I think this is a straightforward misunderstanding. Therein lies your difficulty. Yes the thrust (a force) is constant. And if we take gravity as constant over the flight The same acceleration numerically happens in each second of the flight since force is constant. So each second the rocket gains say 5 m/s (or whatever) velocity. So the velocity gain is additive not multiplicative as you have in your '20 times' But of course if the acceleration is a (a + a)2 = (2a)2 = 4a2 and so on for higher powers. It is the difference between an arithmetic progression and a geometric progression. BTW I also wonder if you actually realise what force is doing the work. It is not the thrust alone.

Exactly and where did this equation come from? I have asked you several times what sign convention you are using and tried to make it fit, but you have not replied. Do you know what a sign convention is? Do you understand that the image is real and what this means? I can only help properly if you answer my questions - they are not tricks they are meant to help.

This is a good question and to answer it you need to look at what is well established in astrophysics, as well as a bit of nuclear physics. Here is a simplified diagram showing the important bits. This plots energy against atomic number and important elements are positioned along the axis. You will note that the energy is a minimum at the element iron (which has a lower atomic number than lead). I have drawn a dashed vertical line at this point. Physical systems tend to the lowest available energy state and the diagram shows that processes to the left of the dashed line that combine elements to form new ones release energy or lower their own energy states. These are of course, fusion processes and are the ones that occur in ordinary stars. The diagram also shows that to the right of the dashed line, to continue to assemble heavier and heavier elements (ie greater atomic number than iron) requires a supply of energy. Sufficient is not available in ordinary stars so elements heavier than iron are formed by another process. Such a process happens in a supernova. Ordinary stars are self sustaining for billions of years, but another important supernova process is the violent explosion they undergo. This ejects material in the explosion. So this provides freshly minted uranium for the inclusion in the formation of new planets. Looking further at the diagram you can see that all elements beyond iron have the potential to decay by fission with the release of energy. However not all elements readily do this. Uranium is the element of choice here and lead is pretty stable and one result of the decay of uranium. A great deal of readable further information is provided here. http://large.stanford.edu/courses/2013/ph241/roberts2/

There was a long discussion about this recently, so it obviously bothers some beginners. http://www.scienceforums.net/topic/59887-help-in-explaining-formula-of-kinetic-energy/page-2?hl=%2Bkinetic+%2Benergy#entry949583 my contribution started at post#39

Well you are getting there but your arithmetic is suspect. I make the chlorobutane 0.89/92.6 = .0096 moles or 9.6 x 10-3 moles. and the bromobutane 1.27/137 = .00927moles or 9.3 x 10-3 moles. Edit also present your calculation as a division not a multiplication please.

What was that a reply to, and what did it mean?

I am struggling to find your point in all that wordage. Especially as you have started two threads on essentially the same subject within half an hour of each other. http://www.scienceforums.net/topic/103942-a-question-of-delayed-choice-application/ Your comment that all points on a light beam are simultaneous to the light beam is essentially correct, but as to the rest What is your point sir?

Are chlorobutane and bromobutane liquids? If so can you not work out how many moles there are in 0.1 mL of pure substance. It will be different in each case. http://www.odinity.com/nucleophilic-substitution/

Two questions for you to think about/investigate. 1) Will the iodide replace/be replaced by both the chloride and the bromide? 2) How much of each reagent are you starting with?

I did say to watch out for the sign convention, yours is not the easiest. I also said to post your working - it makes it a lot easier to help that way. As it is all I can do is list the equations you should have, which you have not done. So Equation 1 do + di = 320 Equation 2 di/do = -1* (98/2) = 49 Note I have put the minus sign to a 1 here because neither the object nor the image distance is negative. The -1 tells you that the image is inverted, but you should also know (the equations don't tell you) that the image is real and therefore the image distance is positive. or di = 49do or do = di/49 Substituting first for do and then for di yields do = 6.4 di = 313.6 (These figures correctly add up to 320) Note that the quantity (do + di) is called the 'throw' of the lens and is often given the symbol T. Can you now use your third equation to calculate the focal length? and show your working please.

The OP actually did state one. However, only two equations were stated and three are available from the question information. Working it through I find that this does not work out to particularly simple numbers, but my figures place the object on the correct side of the focus. Solution strategy 1) Form two equations in the two unknowns, di and do, and solve. 2) Substitute these values into the lens equation you have stated and solve for f. Take care with the sign convention you are using with the negative sign in the magnification equation What is its significance? Post your working.

So did you find any congruent triangles, and if not, why not?

Look for congruent triangles.

I really think you should practise drawing half way decent diagrams. Yours are , frankly, not good enough to work from. Here is a better one. Note that I have labelled all 14 angles, even though I don't yet know if I am going to use them. Now first question what is your strategy for proving that PQDR is cyclic? What property of the angles of a cyclic quad do you know? If you do not know the answer to this ask as it is the key to the proof. Once this is answered you can assemble the necessary information. There are 5 triangles, 3 quadrilaterals and 4 points where some of the angles add up to 180. These will give you lots of very simple equations between some of the angles. But not enough. You should get used to using all the information provided in a question. This analysis has not yet used the fact that P, Q and R are midpoints. What do you know about lines joining midpoints of triangle sides? Applying this will yield enough further equations to reduce the number of unknown angles since many can be shown to be equal with this extra information. I don't see that the negative mark in post#3 is either productive or warranted so I have added a +1.

It is common to measure in wavelengths rather than (micro or nano)metres. That is why you have the factor of one half. Yes delta-d (normally just delta) would be twice the thickness if viewed square on so B coincided with A. However think about it. What would you see in this case? The whole point is that there is another partial reflection at B (which I did not draw) for oblique incidence. And this will in turn generate another partial reflection and so on. So you will see a succession of (weakening) fringes. By the way, why was this in the homework section? It seems like a genuine physics question to me, seeking information, not requiring you to perform a calculation or answer a question. You would probably get more response there.

Itoero Great subject +1 It is difficult to 'compete' with the supermarkets on price for most crops. It can be economic for fruit, soft fruit in particular. However supermarket produce is a) not properly ripened b) of a variety that is easy to grow and harvest rather than tasty or nutritious. Home gardeners can remedy both these faults. So our the main crop in our garden is the bramley apple. We are down to our last tray and a half from last autumn ( though there is some in the freezer). Typically the fresh and stored apples last from sometime in August to the end of the following March. We also grow rhubarb, runner beans and soft fruit (raspberries and strawberries).

I helped a Doctor get and setup a macbook for an online Masters in Clinical Pharmacology recently. It was a real pig to get running and the University IT department were very unhelpful about it because it is a Mac. Linux or other systems would be even worse. So, unless you are willing to become your own expert you will need to keep up with a Microsoft machine to interact with the powers that be. Sorry.

Delta d is the optical path difference between successive rays, ie the difference between travelling twice through the film and reflecting off the top surface. I will add a quick sketch.

+1 for the weblink. A fascinating site. I know Self as a talented electronics designer and former writer of Wireless World Articles, but I didn't know about his mechanical dimension.

Don't you put the longer posts at the bottom to keep the top of the fence level? +1

Well yes I agree I should have specified a straight run of fence, as one which forms a closed loop will have equal numbers. O did not consider or include other fence constructions. Well you could try ordering "45 of something" from Jewsons and see if you get "The Jewson Lot" See also my note below to Delta I don't think I ever said or implied that you could compare two amounts of any units indiscriminately. In fact it is worth considering the reverse question:_ "Can two quantities with the same dimensions always be compared numerically?" Well what setting on my torque - wrench do I require to exert a force of deliver 10 Joules? Edited After all they have the same dimensions.

Clear (properly) your internet cache.