Formula needed for displacement

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On Fri, 05 Aug 2005 20:00:59 GMT, "George E. Cawthon"

Last time I saw logic like that Eric Idle was saying "She's a witch !"
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Isn't George correct? If an item sinks in water it displaces its own volume of water.
Eureka! and all that.
If two objects of different mass displace the same volume of water, you can determine which object is more dense than the other.
Eureka!
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snipped-for-privacy@aol.com wrote:

Oh? are you saying that it isn't true?
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In article

I think he's saying you're made of wood. Or weigh the same as a duck.
;-)
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Dave Balderstone wrote:

Don't have a clue, since I don't know who Eric Idle is or what he was talking about. I guess I just don't care, since he obviously has a screw loose, err, I counted them and it is actually two screws, 5 bolts, and 4 rivets loose, plus the back bumper is dragging on the ground, but it is getting sharp.
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The references are to a scene in the film "Monty Python and the Holy Grail". The script for the scene is here:
<http://www.mwscomp.com/movies/grail/grail-05.htm
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Dave Balderstone wrote:

Aha! I've got the VCR tape but didn't remember who Erick Idle was. Actually I don't remember, or ever knew the names of the guys in the troop. I still don't remember "She's a witch." must not be very memorable, at least to me. The knights that say nicht, the frenchmen on the castle walls, and the rabbit must have been way funnier as I remember them. Particularly the rabbit. Almost as funny as Jimmy Carters attack rabbit.
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"Roy Smith" wrote in message

In a tank of water there is floating a tin tray. On the tray is a glass bottle filled up with water. Someone comes along and upsets the whole arrangement. The glass bottle and the tray are both completely submerged under the water. Does this upsetting of the tray and bottle cause the level of water in the tank, taken at the side of the tank, to go up or to go down or does the level remain unchanged?
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Wow, that's a cool problem. I'm going to vote for "the level goes down", but I'll admit I had to think on it for a while.
When the pan-bottle system is floating, the weight of the water it displaces is exactly equal to the weight of the floating stuff. That much is obvious.
Because it sinks, it must weigh more than the water it displaces while submerged. Looking at it the other way, the weight of the water it displaces while submerged is less than its own weight.
Since it displaces less water while submerged, the level in the tank must have gone down when it sank. At least I think that's the right answer :-)
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"Roy Smith" wrote in message

Bingo! ... you get to collect the cabal dues this month and deduct your 25% handling fee.
;)
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That's older than I am; a problem in density ...displacement of water volume equal to ...etc. The clue is to think about the water in the bottle [mass, volume], and to think about what would happen if the bottle was empty.
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Thanks. I thought I was going crazy. Thanks everyone, I did find the answer.
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Displacement is an older term, not now used, for volume, since it could be measured by liquid displacement.
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Roy Smith wrote:

But if the OP wasn't hosing us when he said he couldn't figure out V, r or h, which is more likely: That he wanted the volume and said displacement? or vice versa?
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Roy Smith wrote:

You sure you don't mean a math book for the third year of high school?
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Only for floating objects. An object that is *immersed* in water displaces its volume, not its weight.
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Doug Miller (alphageek at milmac dot com)
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wrote:

Not sure I agree with this....Automobile engines are referred to as having a DISPLACEMENT of xxx cubic inches (or liters) which is a volumetric measure.
Bruce T
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Lew Hodgett wrote:

Yeah right! Most third graders can't add very well and about 1/2 of them can read the title of the book.
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I think LRod posted the correct answer in post #11. Below I have reposted his thoughts on this and have added some notes in case the calculated numbers like 3.97899 might actually be 4 in true measure.
Well, the area of the circle describing the cylinder is pi*r^2, so to find r we must first determine the diameter by dividing the circumference by pi :
12.5"/pi = 3.97899" (diameter) (*true measure might be 4")
The radius then is 3.97899/2 = 1.989" (true measure might be 2")
Then we find the area of the circle described by the cylinder as pi * r^2
1.989^2 = 3.958 (*true measure might be 2 squared = 4) 3.958 * pi = 12.434 in^2 (4 x 3.1416 = 12.5664)
Every linear inch of the length of that cylinder then is 12.434 in^3, so the cylinder volume is 273.555 in^3
If there is an error in the original measurement and the true diameter of the cylinder is 4", then the volume of the cylinder is 22 x 12.5664 = 276.4608 cu. in.
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On 15 Aug 2005 07:29:09 -0700, " snipped-for-privacy@yahoo.com"
Thank you.
You know, if I had used "4" and "2" someone would have come along and criticized my lack of precision.

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LRod

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