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Constant Acceleration Equation - physicist please! |
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| Mar11-06, 05:54 AM | #18 |
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Mentor
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Constant Acceleration Equation - physicist please![example: 1 = 1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1) = i*i = -1 - nothing wrong with that?] There is a difference in doing pure number theory and manipulation versus something with a physical quantity. If you wish to do the latter, you shouldn't have done it in the PHYSICS section of the forum. And just so you know, in many areas of physics, h=k=c=1 by "definition". But this isn't a new definition. It is for "simplicity" and people in the field know what is being done so there's no confusion. It is why you see people often quoting the mass of a particle simply in terms of MeV rather than MeV/c^2. We all know how to convert those back to their true units. Moral of the story: please do mathematics in the mathematics forum, especially if you wish to neglect the physical dimension of the quantities you are manipulating. If you wish to carry the discussion in the physics forums, then all the physics rules apply. Zz. |
| Mar11-06, 06:18 AM | #19 |
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agreed
but surely you're not suggesting that SI units are universal constants, carved in stone. in many cases equations are only approximations. (such as mc²) if a unit changes then it's correlation to other units also changes and this effects the values in dimensional analysis. in the example... (assuming that tomorrow the BIPM decided to redefine the meter and second) this is not a problem for the maths forum it's a physics problem I would hope this question gets more than a sweeping dismissal, and lengthy explaination as to why such questions are not acceptable in Physics. |
| Mar11-06, 07:13 AM | #20 |
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And note that you were the one who said that you were more interested in the manipulation of the mathematics. If you notice, you did something that was physically nonsensical, but you justified it simply by the fact that you're not doing anything wrong numerically. That is what prompted me to request that if you wish only to care about the numerical manipulation of numbers with utter disregard to the physical significance of that number, then the physics forum is NOT the place to do it. I'm only going by YOUR intentions. Zz. |
| Mar11-06, 07:14 AM | #21 |
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| Mar11-06, 02:26 PM | #22 |
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I simply converted the result to µm and I did get the correct result (at least) (what's that you say lucifer??) rather than 9.80665 m/s² neat-o huh! ok, don't hurt me so nnnnn!! and you know what? I'm not going to ride my bike past your Synchotron anymore!
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| Mar11-06, 04:00 PM | #23 |
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And note, I said, if you read carefully, that one cannot ADD (not convert) a length in meters to a length in micrometer and expect a meaningful answer. Do you think 3 +2 = 5 if 3 = 3 meter and 2 = 2 microns? Puhleeze! But that is what you are trying to do, and even worse, you multiply things just because they all result in "1". Zz. |
| Mar11-06, 04:25 PM | #24 |
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I do enjoy that you are trying to think outside the box, Gregor, but this is turning to nonsense quite quickly and we are turning in circles.
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| Mar11-06, 04:41 PM | #25 |
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a = 9.80665 m x 0.001 s x 0.001 s = 0.00000980665 m do you see any units here other than meters and seconds? naturally 0.00000980665 m = 9.80665 µm and 0.001 s = 1 ms but these were converted after the operation the point is that the answer is correct a at 1 g = 9.80665 µm / ms / ms and I derived it with the calculation above, which works for any time interval shall we try 0.5 s ? sure! 9.80665 m x 0.5 s x 0.5 s = 2.4516625 m so a = 2.4516625 m / 0.5 s / 0.5 s feel free to tell me this result is incorrect |
| Mar11-06, 04:43 PM | #26 |
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| Mar11-06, 04:49 PM | #27 |
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Gregor, I would suggest that you not multiply numbers at will and consider the definition of acceleration:
Acceleration is equal to the rate of change of velocity with respect to time. Now, no where in the definition do you see anything about units. It says rate of change of velocity (in what ever units you want to call velocity), divided by rate of change in time (in what ever units you want to call time). As long as you are consistent, you will get the same answers. |
| Mar11-06, 04:57 PM | #28 |
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I guess I have to spell it out for you
a = 9.80665 x 0.001 x 0.001 = 0.00000980665 m / 0.001 s/ 0.001 s now do you get it? are you saying 1 g does not = 9.80665 µm / ms / ms ? because if you are - I'm not the one who needs help here |
| Mar11-06, 05:00 PM | #29 |
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Gregor, you are still multiplying numbers at will without paying attention to their meaning.
the units of acceleration are, in SI, [tex] \frac {m}{s^2} [/tex] What you have done, a = 9.80665 m/s x 0.001 s x 0.001 s, with much carelessness, will result in units of [itex] m-s[/itex], which is not what acceleration represents. So that does not equal a. |
| Mar11-06, 05:01 PM | #30 |
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1 g = 2.4516625 m / 0.5 s / 0.5 s
yes or no ? 1 g = 0.00000980665 m / 0.001 s / 0.001 s yes or no ? please do not answer these questions with anything other than a. yes or b. no |
| Mar11-06, 05:01 PM | #31 |
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| Mar11-06, 05:03 PM | #32 |
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Yes, that is correct.
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| Mar11-06, 05:05 PM | #33 |
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a = 9.80665 m/s x 0.001 s x 0.001 s and that is certainly not a true statement because you end up with meters*seconds which is not and never will be a measure of acceleration. Pay attention to your units and don't carelessy move them around. |
| Mar11-06, 05:10 PM | #34 |
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Mentor
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If you want acceleration, you need to divide by s^2, not multiply by s^2 and then pretend you divided by putting in units and numbers that don't match what you actually did! Gregor, the things you are doing here are so obviously wrong that we're having a hard time believing you aren't doing it on purpose. And if you are doing it on purpose, that's a no-no here. |
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