modest Posted July 12, 2010 Report Posted July 12, 2010 ...as long as you ignore the physics in the calculation... Newton's law of gravity and Coulomb's law are both laws of physics and they are the correct laws of physics to use given the OP. To correct something else you said, the mass of hydrogen-1 is not 1.0794. That is an average of all naturally occurring isotopes in their natural abundances. ~modest
heterotic Posted July 12, 2010 Report Posted July 12, 2010 Newton's law of gravity and Coulomb's law are both laws of physics and they are the correct laws of physics to use given the OP. To correct something else you said, the mass of hydrogen-1 is not 1.0794. That is an average of all naturally occurring isotopes in their natural abundances. ~modest Ha! First of all, you can't focus physics on two laws (this is getting a bit redundant), ignoring other observations within a system. Again, if he's doing basic physics homework (high school or 100-level at a University), then your proportionality equation would be fine. If he wants to apply it to a real situation, like an atom, sorry, but your equation alone will not help him. Secondly, see post #12. You can't correct me when I've already corrected myself. I said that I should have looked it up, because our calculation of the molar mass of H-1 based on the molecular mass of H-H, which I had forgotten (we neglected H-2 and H-3 for obvious reasons at that level, see figures below AMU's and percentages from the AMDC, calculations of atoms per mole are based on those figures.) Obviously, your degree is not in chemistry either, so I don't really feel you've corrected me at all. H-1 ~ 99.9816% ~ 6.02103e23 atoms per mole of H = 1.00782503207 u H-2 ~ 0.0184% ~ 1.10807e20 atoms per mole of H = 2.0141017778 u H-3 ~ 10^-16% ~ 602214 atoms per mole of H = 3.0160492777 u These masses can't even be exact - the mass of Hydrogen will be different for its different states, based on its Kinetic energy. Which was my point to begin with.
modest Posted July 12, 2010 Report Posted July 12, 2010 Secondly, see post #12. You can't correct me when I've already corrected myself. You need to correct post #10. Use the correct mass for H-1 (1.00783), redo your calculations top to bottom, and you'll find the strength is ~2.3 x 1039 times stronger. This is the same answer ("around 10^39 times stronger") that I gave three years ago. All of your objections in this thread have amounted to nothing but detracting from the OP. ~modest
heterotic Posted July 12, 2010 Report Posted July 12, 2010 You need to correct post #10. Use the correct mass for H-1 (1.00794), redo your calculations top to bottom, and you'll find the strength is ~2.3 x 1039 times stronger. This is the same answer ("around 10^39 times stronger") that I gave three years ago. All of your objections in this thread have amounted to nothing but detracting from the OP. ~modest ... I used 1.00794 u in the calculation, but I'll use the one from the AMDC if you'd like. Not that it matters, since the variance in mass is the entire point behind the radius being a factor. Your time frame is incorrect, please try recalculating the amount of time that has passed since you gave a vague inaccurate response. 10^39 versus 2.27e39 versus 2.3e39 is VERY different and variable. Again, you're not disproving my point at all. If you can't accept that the radius matters in real situations, then you're not much of a scientist. If I'm doing nothing other than "detracting from the OP", why are you exacerbating such a detraction?
modest Posted July 12, 2010 Report Posted July 12, 2010 ... I used 1.00794 u in the calculation You used 1.0794. Use 1.00783 and you will get 2.3 x 10^39 times stronger. ~modest
modest Posted July 12, 2010 Report Posted July 12, 2010 Your time frame is incorrect, please try recalculating the amount of time that has passed since you gave a vague inaccurate response. I intentionally gave an approximated response to encourage the OP to do the calculation themselves. This is common practice on this forum. It is also common practice to close threads which are detracting from the OP. ~modest
heterotic Posted July 12, 2010 Report Posted July 12, 2010 You used 1.0794. Use 1.00783 and you will get 2.3 x 10^39 times stronger. ~modest You meant to say then, that I made a typo, which you didn't notice until now. Either way, you're wrong. The radius is extremely important. It's common practice to observe physics properly when instructing others.
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