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Warp <war### [at] tag povrayorg> wrote:
> gregjohn <pte### [at] yahoocom> wrote:
> > IN CANOE:
> > displacement = Mball /(Ph2o)
>
> No. In the canoe the ball displaces an amount of water equal to the
> weight of the ball.
>
That's what I said, only more precisely.
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Darren New wrote:
> You are sitting in a canoe, in a swimming pool, holding a cannon ball in
> your lap. You throw the cannonball overboard, and it sinks to the
> bottom. Does the level of water in the pool go up, go down, or stay the
> same?
>
> (I've asked this of probably a dozen or more scuba dive instructors, and
> only one has gotten it right. The reasoning behind the correct answer is
> obvious once you hear it. I don't remember if I got it right when I
> heard it.)
It goes down as measured against the edge of the pool. While in the
canoe, the cannonball displaces its mass in water, but in the water it
displaces only its volume. Since the cannonball is denser than water,
it displaces more while in the canoe than in the water. Since there is
less displaced water, the water level of the pool goes down.
Regards,
John
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In article <477b4045@news.povray.org>, pgf### [at] optusnetcomau says...
> Gail Shaw wrote:
> > "Darren New" <dne### [at] sanrrcom> wrote in message
> > news:477ad72c$1@news.povray.org...
> >> You are sitting in a canoe, in a swimming pool, holding a cannon ball
in
> >> your lap. You throw the cannonball overboard, and it sinks to the
> >> bottom. Does the level of water in the pool go up, go down, or stay th
e
> >> same?
> >
> > I believe it should stay the same.
> >
> >
>
> Incorrect. Hint - The cannon ball sinks to the bottom. Think about
> what was preventing it doing so before.
>
Umm. I don't see how that would be incorrect. While in the boat the
*boat's* hull has to displace an amount of water sufficient to account
for the added weight of the ball... Ok, you have a point. The
displacement from the ball is going to be different "in" the boat than
in the water, since one depends on the *weight* being shifted, while the
other depends entirely on the *size* of the object. The question then
is, will the displacement of the boat, from the cannon balls weight, be
larger or smaller than the displacement of the cannon ball itself?
Without knowing the size of the boat, the actual weight of the ball, and
thus what the displacement will be from that, its not possible to
project if the water will be "more" or "less" displaced by the inclusion
of an entire cannon ball in the pool, versus a few millimeters of boat
sinking slightly more into the water from the weight.
I.e., insufficient information to make a ruling.
--
void main () {
if version = "Vista" {
call slow_by_half();
call DRM_everything();
}
call functional_code();
}
else
call crash_windows();
}
<A HREF='http://www.daz3d.com/index.php?refid=16130551'>Get 3D Models,
3D Content, and 3D Software at DAZ3D!</A>
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Patrick Elliott <sel### [at] rraznet> wrote:
> Without knowing the size of the boat, the actual weight of the ball, and
> thus what the displacement will be from that, its not possible to
> project if the water will be "more" or "less" displaced by the inclusion
> of an entire cannon ball in the pool, versus a few millimeters of boat
> sinking slightly more into the water from the weight.
The size of the boat nor the weight of the ball make any difference on
the original question: The original question didn't ask how much the water
level changes, it only asks if the water raises or lowers. To answer this
question the only information needed is whether the cannonball has higher
density than water or not. Since cannonballs are made of metal, the answer
is yes. Thus its weight displaces more water than its volume, and thus
the answer is that the water level lowers.
--
- Warp
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In article <477d49d8@news.povray.org>, war### [at] tagpovrayorg says...
> Patrick Elliott <sel### [at] rraznet> wrote:
> > Without knowing the size of the boat, the actual weight of the ball, an
d
> > thus what the displacement will be from that, its not possible to
> > project if the water will be "more" or "less" displaced by the inclusio
n
> > of an entire cannon ball in the pool, versus a few millimeters of boat
> > sinking slightly more into the water from the weight.
>
> The size of the boat nor the weight of the ball make any difference on
> the original question: The original question didn't ask how much the wate
r
> level changes, it only asks if the water raises or lowers. To answer this
> question the only information needed is whether the cannonball has higher
> density than water or not. Since cannonballs are made of metal, the answe
r
> is yes. Thus its weight displaces more water than its volume, and thus
> the answer is that the water level lowers.
>
Yeah. Read that later. Its not like I a) spend a lot of time thinking
about these sorts of problems, or b) took physics in high school. Took a
second English course, where I got to read all Sci-Fi stuff instead, so
I still count as a nerd. ;) lol
--
void main () {
if version = "Vista" {
call slow_by_half();
call DRM_everything();
}
call functional_code();
}
else
call crash_windows();
}
<A HREF='http://www.daz3d.com/index.php?refid=16130551'>Get 3D Models,
3D Content, and 3D Software at DAZ3D!</A>
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Kevin Wampler wrote:
> If you've been on the beach, you've probably noticed that the sand dries
> out around you foot when you step on it, why is this? Wouldn't you
> expect a small puddle of water to form around your foot instead?
Here's the solution as I recall it:
The sand on the beach is actually in a more or less maximally compressed
state before you step on it. When you step on the sand, you compress it
downward slightly, and to allow for this the sand under your foot
expands outward somewhat. Since the sand was about as dense as it could
get at the beginning the end result of this downward compression and
lateral expansion is that small gaps open up between the grains and the
sand underneath your foot and its density actually *decreases*. The
water that was in that sand thus doesn't fill the whole empty volume
anymore and drops to below the surface of the sand, causing the top of
the sand to dry.
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Patrick Elliott wrote:
> is, will the displacement of the boat, from the cannon balls weight, be
> larger or smaller than the displacement of the cannon ball itself?
The cannonball sinks. That's why I included that part. :-)
(I used to say "bowling ball", until someone pointed out that most
bowling balls actually float.)
--
Darren New / San Diego, CA, USA (PST)
It's not feature creep if you put it
at the end and adjust the release date.
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"Darren New" <dne### [at] sanrrcom> wrote in message
news:477ad72c$1@news.povray.org...
> You are sitting in a canoe, in a swimming pool, holding a cannon
> ball in your lap. You throw the cannonball overboard, and it
> sinks to the bottom. Does the level of water in the pool go up,
> go down, or stay the same?
>
> (I've asked this of probably a dozen or more scuba dive
> instructors, and only one has gotten it right. The reasoning
> behind the correct answer is obvious once you hear it. I don't
> remember if I got it right when I heard it.)
>
> --
> Darren New / San Diego, CA, USA (PST)
> It's not feature creep if you put it
> at the end and adjust the release date.
Well that's interesting:
If one cannon ball level imperceptibly lowers;
If MANY cannon balls the swiiming pool will
fill tis cavity and eventually overflow;
Graphing the water level versus number of
cannon balls does not result in a linear
image.
Bonus points:
1. How many cannon balls before dip becomes rise?
(assume CB V= .001 swimming pools and
CB density equals 7.87 g/mL)
2. What is the type name of such curves?
3. What kind of person would assume a swimming
pool could be *drained* by dropping cannon balls
into it?
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pan <pan### [at] syixcom> wrote:
> If one cannon ball level imperceptibly lowers;
> If MANY cannon balls the swiiming pool will
> fill tis cavity and eventually overflow;
That can't happen in the problem because the canoe cannot stay afloat
with all those cannonballs.
--
- Warp
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pan wrote:
> "Darren New" <dne### [at] sanrrcom> wrote in message
> news:477ad72c$1@news.povray.org...
>> You are sitting in a canoe, in a swimming pool, holding a cannon
>> ball in your lap. You throw the cannonball overboard, and it
>> sinks to the bottom. Does the level of water in the pool go up,
>> go down, or stay the same?
>>
>> (I've asked this of probably a dozen or more scuba dive
>> instructors, and only one has gotten it right. The reasoning
>> behind the correct answer is obvious once you hear it. I don't
>> remember if I got it right when I heard it.)
>>
>> --
>> Darren New / San Diego, CA, USA (PST)
>> It's not feature creep if you put it
>> at the end and adjust the release date.
>
> Well that's interesting:
>
> If one cannon ball level imperceptibly lowers;
>
> If MANY cannon balls the swiiming pool will
> fill tis cavity and eventually overflow;
>
> Graphing the water level versus number of
> cannon balls does not result in a linear
> image.
>
> Bonus points:
> 1. How many cannon balls before dip becomes rise?
> (assume CB V= .001 swimming pools and
> CB density equals 7.87 g/mL)
> 2. What is the type name of such curves?
> 3. What kind of person would assume a swimming
> pool could be *drained* by dropping cannon balls
> into it?
1: If all the cannon balls are in the canoe to start with, then I don't
even see the canoe staying afloat.
2: Given 1, I would assume this is the 'magic canoe' type of curve. =)
3: The pool could not be completely drained by dropping equal sized
cannon balls into it. If the balls fall in a random close packing form,
the best you could get is about 65% of the volume of water out of the pool.
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