[NatureNS] Correction; Fw: Velocity of light

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Date: Sat, 22 Sep 2012 10:35:27 -0700 (PDT)
From: Paul MacDonald <paulrita2001@yahoo.com>
To: "naturens@chebucto.ns.ca" <naturens@chebucto.ns.ca>
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Yes Pat=0APhysics was always my favorite subject!=0AYou could get a top mar=
k and know nothing.=0AIt took me a long way through school.=0AMade me doubt=
 the value of exams but that is another story.=0ASo long as you could work =
out problems.=0ASolved a lot of tricky one at work too - helped my employme=
nt record too.=0APhysics was great=A0=0AThanks for the reminder=0APaul=0A=
=0A=0A=0A________________________________=0A From: Patrick Kelly <patrick.k=
elly@dal.ca>=0ATo: naturens@chebucto.ns.ca =0ASent: Saturday, September 22,=
 2012 11:28:15 AM=0ASubject: Re: [NatureNS] Correction; Fw: Velocity of lig=
ht=0A =0A=0AI'm not sure I see the confusion arises over the kg being a uni=
t of mass. I just covered this in my last astronomy class and why people us=
e the words mass and weight interchangeably when they are two totally diffe=
rent things. Mass is a measure of how much matter an object contains. You m=
easure mass with a scale. A 1 kg bag of sugar has the same mass here as wel=
l as on the Moon, as on a scale you would still need a 1kg mass to balance =
it.=0A=0AWeight is a force, not a measure of mass. The problem is not helpe=
d when people still use the English unit of "pound" for both weight and mas=
s. The English system unit of mass is the slug, not the pound. Your weight =
is the force of gravity acting on your mass and depends on your mass, the m=
ass of the Earth and the radius of the Earth. if you are standing on the Mo=
on (different mass and radius) you will "weigh" less because the gravitatio=
nal force is less. If you stood on a doctor's scale on the Moon you would f=
ind that your mass is the same as it is on Earth.=0A=0AIn the metric system=
, the unit of force is the Newton (the symbol is N). 1 N is the force that =
is required to accelerate a 1 kg mass at the rate of 1 m/s/s. (i.e. Force =
=3D mass X acceleration).=A0=0A=0AAt the surface of the Earth, the gravitat=
ional force =3D G x m x M / (r x r), where G is the gravitational constant,=
 m is the mass of a small object (like a person), M is the mass of the Eart=
h and r is the radius of the Earth.=0A=0AIf you drop that small object, thi=
s force will also produce an acceleration (because=A0Force =3D mass X accel=
eration). So F =3D m x a. Since the force comes from gravity:=0A=0Am x a =
=3D=A0G x m x M / (r x r)=0A=0ANote that since the small object is on both =
sides of the equation, you can cancel it, and thus you get:=0A=0Aa =3D=A0G =
x M / (r x r)=0A=0AThis acceleration gets a special symbol, g, and has a va=
lue of 9.8 m/s/s (which you can round of to 10 m/s/s). This is the rate at =
which all objects at the surface will fall (in a vacuum). If you go to the =
Moon, since M and r are different the surface gravity there is much lower, =
1.7 m/s/s of about 1/5 that of the Earth.=0A=0AYou can also use the surface=
 gravity as a quick way to calculate your "weight" by multiplying your mass=
 (in kg) by g. A 1 kg mass has a "weight" of 9.8 N and a person such as mys=
elf with a mass of about 100 kg would have a weight of 980 N. Thus a proper=
 metric scale would NOT be graduated in kg, but in Newtons. I expect there =
are two reasons they don't. One is that most people would freak if their "w=
eight" appeared to be 10 times what they thought it was, and secondly, sinc=
e they assume you are using the scale at the Earth's surface, they have "co=
nverted" the force back into "mass". Just don't take it to the Moon and exp=
ect it to work properly!=0A=0APat=0A=0APS: For those who are curious, a mas=
s of 1 slug has a weight at the Earth's surface of 32 pounds.=0A=0A=0A=0A=
=0A=0A=0AThe kg a unit oef mass=A0=0A=0AOn Sep 22, 2012, at 9:58 AM, David =
& Alison Webster wrote:=0A=0AHi Fred & All, =A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=
=A0=A0=A0=A0=A0=A0=A0=A0=A0Sept 22, 2012=0A>=A0=A0Explanations of weight/ma=
ss usually resemble inconsistent confessions obtained by torture but I gues=
s that is ok now.=0A>=0A>=A0=A0It seems logical to me that weight should be=
 treated as a special case of F=3Dma where a in this case is g and F is mea=
sured by exerting an equal and opposite force upward to keep the mass from =
falling.=0A>=0A>=A0=A0One could then determine the magnitude of mass as F/g=
 and then proceed to work out units of momentum, inertia etc from there.=0A=
>=0A>=A0=A0But unfortunately kg has been defined (or perhaps redefined) as =
a unit of mass which leads to everything being a hopeless muddle.=0A>=0A>=
=A0=A0I would try to reform the system of units dealing with weight, mass a=
nd distance but I am tied up this afternoon.=0A>=0A>Yt, Dave Webster, Kentv=
ille=0A>----- Original Message ----- From: "Fred Schueler" <bckcdb@istar.ca=
>=0A>To: <naturens@chebucto.ns.ca>=0A>Sent: Friday, September 21, 2012 9:19=
 PM=0A>Subject: Re: [NatureNS] Correction; Fw: Velocity of light=0A>=0A>=0A=
>=0A>Quoting David & Alison Webster <dwebster@glinx.com>:=0A>>=0A>=0A>>=0A>=
I just dug out my 1st yr Physics (Weber, White & Manning, 1952) and =A0they=
 consistently refer to 'speed of light' as opposed to 'velocity =A0of light=
'. Four other sources (1941, 1948, ~1965 &1962) have velocity.=0A>>>=0A>=0A=
>>=0A>* indeed, it would be ineffective to purge society, or even =A0litera=
ture, of all who misuse speed/velocity or weight/mass. Or, to =A0bring it c=
loser to natural history, all who refer to nonhemipterans as =A0"bugs."=0A>=
>=0A>=0A>>=0A>fred.=0A>>=0A>=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=0A>>=0A>=0A>>=0A>----- Original Message ----- From: "=
David & Alison Webster" <dwebster@glinx.com>=0A>>>=0A>To: <NatureNS@chebuct=
o.ns.ca>=0A>>>=0A>Sent: Friday, September 21, 2012 7:44 PM=0A>>>=0A>Subject=
: Velocity of light=0A>>>=0A>=0A>>>=0A>=0A>>>=0A>Dear All, =A0=A0=A0=A0=A0=
=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0=A0Sept 21, 2012=0A>>>>=
=0A>=A0The recent discussion about transmission of electricity, reading =A0=
of HEAT and reading some of Energy... brings to the forefront a =A0question=
 that has nagged me for decades. Perhaps someone can clarify.=0A>>>>=0A&