| Topic: |
Science > Physics |
| User: |
"Don1" |
| Date: |
11 Jul 2005 10:48:39 AM |
| Object: |
weight-density vs density |
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
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| User: "Steve Ralph" |
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| Title: Re: weight-density vs density |
11 Jul 2005 11:30:32 AM |
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"Don1" <dcshead@charter.net> wrote in message
news:1121096919.328698.27290@f14g2000cwb.googlegroups.com...
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
sr
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:36:33 PM |
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Is anyone familiar with Specific Gravity? I'm wondering why Tables show
both weight-density, and density. Isn't the specific gravity of a
substance the same using either one?
Don
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:43:35 PM |
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Is anyone familiar with Specific Gravity? I'm wondering why Tables show
both weight-density, and density. Isn't the specific gravity of a
substance the same using either one?
Don
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| User: "Don1" |
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| Title: Re: weight-density vs density |
12 Jul 2005 08:06:45 AM |
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Steve Ralph wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1121096919.328698.27290@f14g2000cwb.googlegroups.com...
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
sr
It has been found empirically that:
The weight-density of pure water is about 62.42# per cubic foot.
The mass-density of pure water is about (62.42#/32.174' per sec^2)
equals 1.94 slugs per cubic foot.
Don
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| User: "CrankHater" |
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| Title: Re: weight-density vs density |
12 Jul 2005 10:55:54 AM |
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Don1 wrote:
It has been found empirically that:
you are a fool
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| User: "Martin" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:23:18 PM |
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Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
How many ways are of asking the same question again and again before it
starts to look a bit obsessive?
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:41:37 PM |
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Martin wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
How many ways are of asking the same question again and again before it
starts to look a bit obsessive?
To whom?
Don
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:41:55 PM |
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Martin wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
How many ways are of asking the same question again and again before it
starts to look a bit obsessive?
To whom?
Don
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:42:05 PM |
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Martin wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
How many ways are of asking the same question again and again before it
starts to look a bit obsessive?
To whom?
Don
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| User: "Randy Poe" |
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| Title: Re: weight-density vs density |
11 Jul 2005 11:13:14 AM |
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Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
However, the mass and volume are well defined, and so
it is sensible to define the density as M/V, a definition
which is not connected to the particular local gravitational
field.
- Randy
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 11:38:54 AM |
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Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location; but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
Don
However, the mass and volume are well defined, and so
it is sensible to define the density as M/V, a definition
which is not connected to the particular local gravitational
field.
- Randy
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| User: "T Wake" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:22:47 PM |
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"Don1" <dcshead@charter.net> wrote in message
news:1121099934.441997.168590@g47g2000cwa.googlegroups.com...
Of course you must mean that g varies with location; but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
Don
I am glad you don't get bored of being proven wrong over this issue. It is
fun to see you try to re-couch your terms to get your (false) point across.
What is the mass of the sun? What is the mass of Pluto? How are the two
masses related by your "formula."
Is this going to lead to f=wa/g again? I like that one.
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| User: "Martin" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:22:41 PM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location;
Do you have learning difficulties? Are you just thick?
You've been told and told, g is defined, therefore g does not vary with
location.
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 03:49:17 PM |
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Martin wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location;
Do you have learning difficulties? Are you just thick?
You've been told and told, g is defined, therefore g does not vary with
location.
Oh come on; it does so vary with location! That's why weight varies; so
that the ratio of a body's weight divided by g at the support it
"rests" on is a constant: For any body, w/g is a constant; equal to the
product of its bulk and density.
Don
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| User: "Martin" |
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| Title: Re: weight-density vs density |
11 Jul 2005 04:54:50 PM |
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Don1 wrote:
Martin wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location;
Do you have learning difficulties? Are you just thick?
You've been told and told, g is defined, therefore g does not vary with
location.
Oh come on; it does so vary with location!
no matter how much you wish it, no matter how often you say it, it doesn't.
Do you have a learning disability? You've been told often enough, I'm
not so patient with idiots as some others.
This is easier than alt.conspiracy
That's why weight varies; so
that the ratio of a body's weight divided by g at the support it
"rests" on is a constant: For any body, w/g is a constant; equal to the
product of its bulk and density.
you've been informed, therefore you are lying now.
Don
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 05:39:40 PM |
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Martin wrote:
Don1 wrote:
Martin wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location;
Do you have learning difficulties? Are you just thick?
You've been told and told, g is defined, therefore g does not vary with
location.
Oh come on; it does so vary with location!
no matter how much you wish it, no matter how often you say it, it doesn't.
Do you have a learning disability? You've been told often enough, I'm
not so patient with idiots as some others.
This is easier than alt.conspiracy
That's why weight varies; so
that the ratio of a body's weight divided by g at the support it
"rests" on is a constant: For any body, w/g is a constant; equal to the
product of its bulk and density.
you've been informed, therefore you are lying now.
Being informed by the likes of you doesn't count.
Don
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| User: "Martin" |
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| Title: Re: weight-density vs density |
11 Jul 2005 06:07:18 PM |
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Don1 wrote:
Martin wrote:
Don1 wrote:
Martin wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location;
Do you have learning difficulties? Are you just thick?
You've been told and told, g is defined, therefore g does not vary with
location.
Oh come on; it does so vary with location!
no matter how much you wish it, no matter how often you say it, it doesn't.
Do you have a learning disability? You've been told often enough, I'm
not so patient with idiots as some others.
This is easier than alt.conspiracy
That's why weight varies; so
that the ratio of a body's weight divided by g at the support it
"rests" on is a constant: For any body, w/g is a constant; equal to the
product of its bulk and density.
you've been informed, therefore you are lying now.
Being informed by the likes of you doesn't count.
no wonder there is a call for a moderated group :)
Ignore facts, just shows your mentality. You've been informed, you are
being ignorant rather than accepting being informed.
Don
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| User: "Randy Poe" |
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| Title: Re: weight-density vs density |
11 Jul 2005 04:05:25 PM |
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Don1 wrote:
Martin wrote:
You've been told and told, g is defined, therefore g does not vary with
location.
Oh come on; it does so vary with location!
No, it does not.
That's why weight varies; so
that the ratio of a body's weight divided by g at the support it
"rests" on is a constant
The defined constant g is equal to 9.80665 m/sec^2.
On most places on earth, the acceleration of gravity is
not equal to 9.80665 m/sec^2.
g and "acceleration due to gravity" are different
concepts. g is a standard, predefined acceleration which
approximates but does not necessarily equal the local
gravitational acceleration.
If the local acceleration is 9.9 m/sec^2, then this does
not change the fact that g is 9.80665 m/sec^2. What it
means is that at that location, the local acceleration
is 1.0095 g.
- Randy
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| User: "" |
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| Title: Re: weight-density vs density |
11 Jul 2005 04:06:22 PM |
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Martin wrote:
You've been told and told, g is defined, therefore g does not vary with
location.
Do you have a reference for a defined value for "g" as opposed to a
defined value for "standard g or g_n"?
Like,
say...http://physics.nist.gov/cgi-bin/cuu/Value?gn|search_for=gravity
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| User: "Randy Poe" |
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| Title: Re: weight-density vs density |
11 Jul 2005 01:15:07 PM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location;
No, I don't mean that. I mean that the acceleration of
a body varies with location and is not in general equal
to the defined constant which has been given the name g.
Do you comprehend the term "defined constant"?
but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
"g" at that location is not equal to "acceleration at
that location", and at any rate the acceleration at
any location depends on height (and therefore is
not constant from top to bottom of any object).
- Randy
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 05:34:27 PM |
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Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location;
No, I don't mean that. I mean that the acceleration of
a body varies with location and is not in general equal
to the defined constant which has been given the name g.
Do you comprehend the term "defined constant"?
No.
but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
"g" at that location is not equal to "acceleration at
that location", and at any rate the acceleration at
any location depends on height (and therefore is
not constant from top to bottom of any object).
Oh, you mean it varies with location. Well why didn't you say so(;^?
Don
- Randy
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| User: "" |
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| Title: Re: weight-density vs density |
11 Jul 2005 04:28:40 PM |
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What does this use of "g" to replace "g_n" do to formulas like
"Pressure=rho g h", "Potential Energy = m g h", and especially
"Period = 2 pi (L/g)^0.5
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| User: "" |
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| Title: Re: weight-density vs density |
11 Jul 2005 11:58:57 AM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location; but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
Don
Let's see... "g varies with location", but over the various locations
that make up the volume of any real object, it doesn't vary? Hmmm...
So if I take a one foot cube and move it in 11.99 inch steps from the
pole to the equator, there's no change in g from the pole to the
equator?
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:23:32 PM |
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wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location; but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
Don
Let's see... "g varies with location", but over the various locations
that make up the volume of any real object, it doesn't vary? Hmmm...
Hmmm; you wouldn't be attributing that nonsense to me would you?
Don
So if I take a one foot cube and move it in 11.99 inch steps from the
pole to the equator, there's no change in g from the pole to the
equator?
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 12:43:55 PM |
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wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location; but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
Don
Let's see... "g varies with location", but over the various locations
that make up the volume of any real object, it doesn't vary? Hmmm...
Hmmm; you wouldn't be attributing that nonsense to me would you?
Don
So if I take a one foot cube and move it in 11.99 inch steps from the
pole to the equator, there's no change in g from the pole to the
equator?
.
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| User: "Don1" |
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| Title: Re: weight-density vs density |
11 Jul 2005 11:40:19 AM |
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Randy Poe wrote:
Don1 wrote:
Isn't the density of a mass of matter just its weight-density, divided
by the acceleration (g) at which it will free fall?
No.
"Weight-density" is ill-defined since the gravitational
field will vary over the extent of any object. Furthermore,
at most places on earth the acceleration of free fall
will not be g.
Of course you must mean that g varies with location; but at any
particular location, the mass of any particular body of matter is equal
to the weight it exerts on any support, divided by g at that location.
Don
However, the mass and volume are well defined, and so
it is sensible to define the density as M/V, a definition
which is not connected to the particular local gravitational
field.
- Randy
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