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The policeman's beard is half constructed.

What is the difference between: * Distance from stars in the centre of galaxy A to the stars in galaxy B and * Distance from stars at the edge of galaxy A to the stars in galaxy B ? And - what is the distribution of stars in the average galaxy? I think you'll see the answer there.

That all depends on context. Unless you are talking relativity, kicking a ball harder, doesn't give it more mass, it gives it more acceleration. It's (your green bit) more the reverse, the greater the mass, the greater the force needed (to give the same acceleration). You're confusing yourself by picking a parameter to change, without thinking about what that change really means. Only if the units make sense. Force has certain units, mass has certain units and acceleration has certain units. It doesn't matter which set of units you are using (e.g. slugs or kilograms) as long as they are all consistent. It's bad to use metric for one thing and imperial for another; it's even worse to measure Force in tulips and mass in whales per second. The units have to make sense for the formula to make sense. Well, in a way it's true. If the acceleration stays the same, a greater force will kick a ball of greater mass. That doesn't mean the force is creating or changing the mass. If you kick a ball with some force, you will accelerate it by some amount. If you kick the ball with more force, you won't change it's mass - you'll give it more acceleration. In your example you arbitrarily chose to triple the acceleration. For F=ma to still balance, that would mean kicking a less-mass ball with the same force, or a same-mass ball with more force. Edit: and it's not "... the numerical value of force that is proportional to numerical value of mass", it is "... the numerical value of force that is proportional to numerical value of acceleration x mass". You can't just ignore one parameter. If you choose to keep acceleration fixed, a greater force will accelerate a greater mass. Like kicking a heavier ball, harder, to get the same acceleration. The harder kick is not creating more mass - kick a ball of the same mass, with greater force won't give the same acceleration. You can't have your cake and eat it too. No. The example was pay per day, and that's proportional to hourly rate and hours worked per day. It's a direct comparison.

If you kick a ball, you apply a force to it. That accelerates it, and it ends up moving at some speed. The formula F=ma shows the relationship between the force and the acceleration - and the mass of that ball. If you kick the ball with 3 times the original force, it'll get 3 times the original acceleration. Makes sense doesn't it? Kick the ball harder, it'll go faster (and further). If you kick another ball that has 3 times the mass of the first ball, you'll have to kick it with 3 times the original force, to accelerate it as much as the first ball with the first kick. Makes sense doesn't it? Kick a heavier ball, and it'll need a harder kick to accelerate as much as a lighter ball. If you want to kick a ball with 3 times the mass of the first ball, and you want to give it 3 times the acceleration that you gave that first ball with that first kick, you'll need to kick it 9 times as hard as the first kick. (That's the a x m coming into it). Maybe to give you a totally different (or not?) example: Say you work at a shop where you get paid $5.00 an hour and you work 8 hours a day. What do you earn in a day? Days pay = Hourly rate x Hours worked per day Days pay = $5.00 x 8 = $40.00 Now, that's not saying that the concept "day" equals the concept of "dollar". The hourly rate of 5 isn't making the hours you work in a day 40 ... it's still 8. The formula is just relating the values. Someone earning $10 dollars an hour, would only have to work 4 hours to earn as much. Someone earning $2.50 an hour would have to work 16 hours, to earn as much. The unit analysis works out, too. Days pay unit is dollars per day; Hour rate unit is dollars per hour; Hours worked per day = hours per day; so ... dollars per day = dollars per hour x hours per day dollars per day = (dollars x hours) per (hour x day) dollars per day = (dollars x hours) per (hour x day) dollars per day = dollars per day ... the units always need to balance; try that with F=ma.

... and from someone with this in their sig?

If that's all there is to it (relative speed) then yes. That's the whole point of calling the "twins paradox" a "paradox". ... why does one twin/clock age more if A and B have symmetrical views of each other? The answer is that, due to acceleration, in the twins paradox you don't have a simple case involving only relative speed. The situation isn't symmetrical.

Is there an encoding (or encoding of encoding) issue? \\bar ... in the call seems to emit as: \x08ar ... that is, the second backslash is being treated as an escape sequence. (The combinations of double-quotes inside the single-quotes also look weird to me.)

Your expectations are on the science fiction end of the scale.

Is the content of the JSON good? Maybe in the server environment (due to firewalls or whatever - something different to your own PC) you're getting an error message back instead of well formed JSON data? Can you dump/print/save the raw JSON before trying to decode?

I think that's a bit simplistic (though you may not have meant it quite as simply as it reads). This sort of thinking leads to (what seems to me a very common structure) where the data is model "OO-ishly" but then for processing the data objects are just passed to business logic in separate classes. That's actually fine for many uses, but when the business logic is modelled with the data you start to get a lot more of the benefits of OO (polymorphism and all that). For example, it's very common to see a Person object and a Dog object, then see them passed to the EstimateLifeSpan class, which then uses a switch/case statement to see what type of object it was passed, and do the appropriate calculation. It can be nicer for the Person and Dog objects to not only encapsulate their own data, but also the relevant methods (or calculations wrapped in properties), so something wanting the animals life span can just - whichever kind of animal object it has - call the EstimateLifeSpan method on the object itself. No big case/switch statement. (All helped along by interfaces and all that, with common properties and methods coming from some base Animal class ...) I do think the first structure is quite common, and most often the style people can be used to; the second "more OO-ish" structure can be "scary" to them. Anyway, my two cents on the OP (I'm sure I'm not the first), if someone has a Windows PC - just get one of the free "Express" versions of Visual Studio and go with C#. Plenty of help. Plenty of forums. A good free IDE. From there it's no big stretch to go Java or C++ or similar if required.

I see Christmas, as it is now, as more about culture than religion. e.g. I'm an Atheist, but my Family and I still "celebrate" christmas; we're happy to have the time off work, we buys presents for each other, we have a big meal. It's just a long-standing tradition, where I am, to celebrate christmas. That different people attach different levels of speciifc religious meaning to it, is to me a different issue. Do pacifists complain about (local version of day acknowledging wars or veterans)? Do anarchists complain about presidents day?

It's not "just" an illusion, it's been experimentally verified. e.g http://en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment What I'm about to write is "wrong", but it's the way I think about this stuff, to help me comprehend it. Again - it's "wrong"; just an analogy. (See above for better technical explanations). Basically: the twin who leaves and comes back has, by changing frames (by accelerating), travelled further in space. Because they travelled further in space, they travelled less in time.

I'll leave someone else to answer the OP (the full derivation is beyond me to explain), but to this I'd say - not understanding something does not make that thing "wrong". The energy-mass equivalence is derived from known principles and has been tested. These things are not just plucked out of the air and agreed to by scientists because they like each others shoes.

Didn't you already do this topic? http://www.scienceforums.net/topic/73539-general-relativity

Any clock at rest with respect to you, will seem to measure the time the same as any other clock at rest with you. Any clock moving relative to you (whether it flies past you on a rocket, or you and your clocks fly past it on a rocket) will seem to be going slow. That gives your own clocks a "special frame" according to your logic; just "maximum" instead of "minimum". Given that's built-in to relativity, what's it going to prove? That is, any measurement you make in your own frame is going to be "special" in its own way. But what makes that "preferred" in the "relativity sense", and what are you trying to show by it?

No. If light routinely passed through matter (we all know about windows) then we'd see nothing. Light wouldn't bounce* off you, to the eyes of someone looking at you. Light wouldn't register in their retinas, it'd pass through. I still have no idea what you are trying to say. (* "bounce" is of course a simplification.)

Unless you are standing in the dark, light is bouncing off you in all directions already. If you were standing in the middle of a circle of people, they'd all be able to see you. So why is it some kind of problem that light bouncing off you, then off a mirror, can be seen by multiple people or from multiple angles? What exactly is it that you don't think makes sense?

In the OP, the equation is shown (in text, outside of the image) as: 36 / 6(2+2+2) = ?? Would people who see it as: 36 / (6 * (2+2+2)) = ?? Have seen it differently if it were written as: 36/6 (2+2+2) = ?? Me, I agree with the basic multiplication=division, addition=subtraction comments in post #13 from ajb; so definitely (sans modifers such as parenthesis) take these from left to right. So ... 36. Who got 32? ;-)

... and the beauty of squaring in this stuff is that it doesn't matter if the difference between the two x's works out as positive or negative (and the same for the difference between the y's). All you need is the difference, to get the sides of the triangle. That is, you could look at it as being from (1,1) to (4,5) or (4,5) to (1,1) - and get the same answer (in terms of distance, if not direction). That is (1-4)^2 = (-3)^2 = 9 Also (4-1)^2 = 3^2 = 9 And (1-5)^2 = (-4)^2 = 16 Also (5-1)^2 = 4^2 = 16 Or - if you look at a rectangle with sides of length 4 and 3, it doesn't matter which diagonal you look at - both diagonals are the same length. And finally: the 3-4-5 triangle is a common thing, as it's so "nice" that the lengths "square and square root" to simple integers. That triangle is used in building as a simple way to find a right angle. Say you need to go out 90 degrees from a point on a wall. Get another point 3 metres along the wall. Measure 5 metres from that point and 4 metres from the first point - and you easily make a right-angle triangle. (I find with math it helps to look at something from multiple directions, and see that they all make sense together.)

It's almost like you have not read post #1 of this thread. Which is odd, as you wrote it. Yes, it's true that the specific make-up of the material in the inner layers of Earth is still a subject under investigation; but how does that relate to your topic? How does that invalidate any of the mainstream knowledge of gravity that's been presented in this thread? And further, how do you think your ideas (which you seem to be avoiding directly stating) explain the "what is gravity" question any better than mainstream science?

... but your comments on "density" and "inner core" seem to imply you think current knowledge is wrong. If what you seem to be implying is true, we'd be measuring different effects, so you need to explain how your ideas could be accurate, while we still get the results we do from our experiments.

Why 4? Why not 8? I object that brunch is excluded.

No that's wrong. The effects of mass and gravity are well understood, (e.g. http://en.wikipedia.org/wiki/Cavendish_experiment ) so the mass of the Earth can be determined by the interactions of Earth/Sun and Earth/Moon. Since the volume of Earth can be measured, that gives us density. http://www.universetoday.com/47217/earths-mass/ http://www.universetoday.com/26771/density-of-the-earth/ Edit: snap!

Say you are standing on scales, on the "floor" of the accelerated box. It shows your weight, as a result of your mass and the acceleration of the box. If the box is accelerating at 9.8 m/s/s, it will show the same weight as if you were at home in your bathroom. If it continues to accelerate at the same rate, and you measure yourself after some time - when your speed (relative to something left behind where you started from) is hugely greater - you'd still see the same weight on the scales. In the frame of the box, nothing's changed, in that constant acceleration. Your relativistic mass is in relation to something else. Accelerate at 9.8 m/s/s for one minute then smack into something (that's at rest relative to where you started from), or accelerate at 9.8 m/s/s for one day then smack into that same thing - then you'll see the difference.

You used the word "obvious" twice in that post.