Re: FW: Dinosaur Weights

```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

```