What is the real proof of the accuracy of an experiment,Heinsenberg"s Principle?
552178
If you buy several items, all of which are subject to a sales tax, then you can calculate the total tax by summing the tax on each item. However, for the total tax to be accurate to the penny, you must do all the calculations to an accuracy of tenths of a penny (in other words, to three significant digits), then round the sum to the nearest penny (two significant digits). If you calculated only to the penny, then each measurement might be off by as much as half a penny ($0.005) and the total possible error would be this amount multiplied by the number of items bought. If you bought three items, then the error could be as large as 1.5 cents; if you bought 10 items then your total tax could be off by as much as five cents.
1088429
Any demonstration of this false fact will be flawed, and often uses a hidden division by zero\n\nfor instance:\na and b are real numbers different from zero:\na = b\n\na^2 = ab\na^2 - b^2 = ab - b^2\n(a - b)(a + b) = b(a - b)\n\nhence: \na + b = b,\n2b = b, and:\n2 = 1\n\nHere, i divided by "(a-b)", which is illegal since a -b =0
779448
YES. IT WAS SCIENTIFICALLY PROVED.
76447
p1v1t1=p2v2t2\n\nSee Bernoulli's theorem and Pascals Law
593819
It appears that the constants are calculated at one condition (dry,wet,etc) whereas the effective is closer to reality.  See\n\nhttp://www.pnas.org/cgi/content/full/102/45/16293
857924
where is the proof that he/she/it is real.
874570
NO!!  If people would stop (mis)quoting half-remembered precepts here, and do a spot of THINKING, instead, we might get somewhere.\n\n1.  If a moving spherical ball directly hits an identical free, stationary ball head on in an elastic collision, the moving ball STOPS, and the previously stationary one heads off with the original velocity of the other.  So ALL of the moving ball's K.E. has been transferred to the other ball.  \n\nHere are two ways to see this:\n\n1a.  The device known somewhat indelicately to British schoolboys as "Newton's balls" (but in squeamish America as "The Swinging Wonder") provides an excellent table-top demonstration of this.  Most undergraduate physics instructors must haul this out from time to time.\n\n1b.  You can easily solve it analytically, but a thought experiment will also demonstrate it.  In the originally stationary ball's frame, let the other ball approach with speed v.  Now shift (with an appropriately directed speed of v/2) to the frame in which the two equal balls are simply approaching each other, each with a speed in this frame of that same v/2 (oppositely directed, of course)!  Clearly, from the symmetry of the situation, they'll bounce off each other with the same (oppositely directed) speeds in this frame, still of v/2.  Now go back to the original frame: the previously moving ball is now completely stationary, while the originally stationary one is haring off with speed v.  (You may need to make little sketches of what is happening as viewed successively from each of these frames, to see this; I admit that that's what I did.)\n\n2a.  A similar approach can be used to solve the general case for an arbitrary ratio of masses, m (initially moving)/M (initially stationary).  Hint: have the balls approach one another in their center of mass frame with equal but opposite momenta of mv and  MV.  They'll clearly bounce off one another in such a frame, retaining the same values of their respective (but now oppositely directed) speeds.  Then you can do the other frame-jugglings from there. \n\nHowever, this analytical approach, done accurately, is a tedious mess.  So, to appreciate the general problem more physically, consider just the following:\n\n2b.  The case of m << M.\n\nConsider the "equal but opposite momentum" frame for these two balls, in which MV = mv.  This means that if ball M is VERY MASSIVE, the speed it will have in such a frame, V = (m/M)v; that is, it's VERY SMALL compared to little puny m's high speed.\n\nSo, transforming to the frame in which M itself (the big guy) is initially stationary takes only a speed shift of O(m/M)v.  (In fact identically so.)  This only affects the speed of puny m minutely, so m's initial K.E. is still of order O(m v^2).  (There's no need to do picky constant book-keeping for these purposes.)  \n\nNow after the collision: In the original frame, Big M is now moving (and in fact at 2V if you think about it.)  But, being loose about the constants, M's K.E is only O(MV^2).  \n\nSo, the ratio (m's K.E.)/(M's K.E.) ~ (mv^2)/(MV^2)\n\n= (m/M)(M/m)^2 (since mv = MV)\n\n= M/m.\n\nBINGO!  The little bugger still carries by far the larger amount of the kinetic energy.  The massive ball (i) moves sluggishly, and (ii) because  K.E. involves both mass but also the SQUARE of the speed, the TWO POWERS of (small) speed more than cancel out the effect of its (large) mass.\n\nThis is a well-known fact in particle collision theory, that little particles basically bounce off big ones.  \n\nYou could also say that it's well-known in "space billiards": when spacecraft head close to Jupiter to receive a "gravitational sling-shot effect" that sends them off to the outermost planets.  \n\nIt's more complicated because of 3-D effects there, but analogous frame-shifting methods can be used to show that the spacecraft can pick up considerable speed AND change
149491
This is a "rule of thumb" that is the average of a lot of kinetic data.  To blindly apply it is unwise, but as a zero order estimate, it should work OK.  Detailed measurements must be done on your system of interest to find out true kinetic parameters and their sensitivity.
261371
It's not that they are at two places at the same time, but there is an equal probability of finding the proton in both places. When you get to the level of quantum mechanics, due to Heisenberg's uncertainty principle we rely on probability curves to get an approximation of the physical world, the funny thing is that the areas with the most probability, are those predicted by Bohr without using quantum mechanics.
440513
Yes it is wrong.  It might be a good approximation for lower altitudes, but is not truly correct.  In the same way, Newton was wrong with his equations, but they still make good approximations over reasonable ranges of values.
221116
The doctor who performed the experiment back in the early 1900's found that result once, but he was never able to duplicate the results when he and other doctors tried it again.\n\nSince no one ever duplicated the results, we can assume that the original experiment was flawed in some way (maybe the scale they used was broken).\n\nEven though that 20 oz thing was proven wrong, some people still believe in it. If you look hard enough, you will still find people who believe that the Earth is flat.
560034
It demonstrates the physics of inelastic collisions.\n\nSome of the properties demonstrated are conservation of momentum, the properties of inertia, kinetic energy, potential energy . . .
492268
Better question--how can we prove that that it is fixed? We don't want this to keep happening year after year....
164065
The trick with "proofs" like these are generally an obscured division by zero. One of the lines seems to work, though when you divide to cancel out some factors, the quantity you're dividing by could/does equal zero. It's a trick and doesn't say anything substantial about mathematics.
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
82803
Yes acceleration and force should have a linear correspondence in this experiment (noise excluded). Obviously Force = Mass * Acceleration.\n\nIt is difficult to give much more qualitatively without a better description though.
1017047
I'm agreeing with chopperchopster.You won't loose anytime.As Eistein's Theory of Realitivity states:,in a revamped version \nSpecial relativity considers that observers in inertial reference frames, which are in uniform motion relative to one another, cannot perform any experiment to determine which one of them is "stationary". This is known as the principle of relativity. While this principle was not new to Albert Einstein's work, he found that including electromagnetism in this principle required a new formalism with many surprising consequences. In particular, it required the speed of light in a vacuum to be the same for all these observers, regardless of their motion or the motion of the source of the light.\n\nOne of the strengths of special relativity is that it can be derived from only two premises:\n\n    * The laws of physics are the same in any inertial frame of reference. This means that the laws of physics observed by a hypothetical observer traveling with a relativistic particle must be the same as those observed by an observer who is stationary in the laboratory.\n    * The speed of light in a vacuum is constant (specifically, 299,792,458 meters per second relative to the observer).