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*To*: Dinogeorge@aol.com*Subject*: Re: FW: Dinosaur Weights*From*: dmorgan <dmorgan@exesolutions.com>*Date*: Sun, 24 Aug 1997 17:47:46 -0700*Cc*: EDELS@classic.msn.com, dinosaur@usc.edu*References*: <970823212335_383939708@emout03.mail.aol.com>*Reply-to*: dmorgan@exesolutions.com*Sender*: owner-dinosaur@usc.edu

Dinogeorge@aol.com wrote: > > 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. I was subscribed to this any many other newsgroups by some maliciuus person. Please give me the information on how to unsubscribe. Thanks, Dan

**References**:**Re: FW: Dinosaur Weights***From:*Dinogeorge@aol.com

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