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*To*: EDELS@classic.msn.com, dinosaur@usc.edu*Subject*: Re: FW: Dinosaur Weights*From*: Dinogeorge@aol.com*Date*: Sat, 23 Aug 1997 21:23:35 -0400 (EDT)*Reply-to*: Dinogeorge@aol.com*Sender*: owner-dinosaur@usc.edu

In a message dated 97-08-23 20:06:10 EDT, EDELS@classic.msn.com (Allan ) writes: << I guess it is a matter of approximation. We say the density of water is 1 gram to 1 cubic centimeter. I suspect that it is true until you reach the cubic meter size and above. I don't know in which direction (without looking it up somewhere) the discrepancy lies.>> A long time ago, I learned that a kilogram was defined to be the mass of one liter of water = 1000 cubic centimeters. Later I found out that one liter of water is not quite 1000 cubic centimeters (or vice versa: I don't remember either). The difference is minute--a few thousandths of a percent. Which suggests a temperature effect of some kind--measuring the liter of water at one temperature but a cubic centimeter of water at another temperature, thereby slightly altering the density or whatever. But I sure don't know whether this is correct. << This is sort of like "E=mc2" - which is true, but not precise in all instances. The more complete version of the formula is "E=mc2 + m2c4" [in case this doesn't pass through the email correctly - the first formula is E = m c (squared) and the second formula is E = m c (squared) plus m (squared) c (to the fourth)]. The lesser terms tend to fall out of the formulas because they are usually unimportant to the task at hand. On occasion, it does matter. Some scientists like to speak as precisely as possible, so they will remind you that this number is not exact. Just as I wish some paleontologists would say that "This is what I think the dinosaurs were like" vs. "This is what the dinosaurs were like"!!! >> I think the higher-order terms in the equation depend on higher powers of v/c, not just c. So they would not become important except at velocities >very< close to c. Otherwise the higher order terms totally overwhelm mc^2. At velocity zero, of course, e=mc^2 exactly: the rest mass of the object.

**Follow-Ups**:**Re: FW: Dinosaur Weights***From:*dmorgan <dmorgan@exesolutions.com>

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