Post by: throwingks on April 03, 2006, 09:28:00 AM
http://www.hbci.com/...new/gravity.htm
Is there a such thing as Neutrinos?
http://www.ps.uci.ed...k/neutrino.html
Is your weight absolute 0 at the center of the earth?
Tesla said his theory of gravity applied to all things no matter how small or how big. Eisnteins and Newtons theory doesn't. Will the world ever know?
http://www.netowne.c...logy/important/
If the world exploded, where would all the ghosts go?
Post by: _iffy on April 03, 2006, 11:01:00 AM
Post by: throwingks on April 03, 2006, 11:17:00 AM
I like the discussion in the other thread, I just feel like it needs a better home.
Post by: _iffy on April 03, 2006, 11:35:00 AM
most people thing gravity is mass x acceleration of gravity.
this is fine on earth but...the real formula is...
Gravitational force = (G * m1 * m2) / (d2)
Where G has the value of 6.67 x 10E-8 dyne * cm2/gm2
so any two objects that have mass have a gravitation force drawing them together.
QUOTE
Does gravity push or pull?
Is there a such thing as Neutrinos?
I'm not touching theseIs there a such thing as Neutrinos?
but in the centre of the earth... you would float like vishnu, there would still be gravity but it would be from
all sides and angles so they would be in balance.
I think einstien used the above formula but i'm not sure so i can't really comment on that.
all the goshts would go to that planet all the mormons are going to. You know the one were every girl is pregnant. Always. Forever.
Post by: _iffy on April 03, 2006, 12:08:00 PM
That formula is universal. It doesn't matter what planet your on. F=mg is a shortcut.
Gravitational force = (G * m1 * m2) / (d2)
d= distance between
G=6.67 x 10E-8
m1= mass (kg) of first object
m2 = mass (kg) of second object
You use this formula when looking at planets, stars, and orbits.
eg: Two satalites in orbit have gravity, from the earth, the sun, the moon, even mars, and each other.
You use it to find the gravitation constants. Like the force of gravity on the moon.
That formula is correct puck.
Post by: slightly_damp on April 03, 2006, 01:43:00 PM
Post by: damam on April 03, 2006, 02:49:00 PM
that doesnt explain gravity - it just describes it
the real question is why are these objects attracted to each other in this manner . . .
I believe the ghosts would just go into the ether
Post by: throwingks on April 03, 2006, 03:12:00 PM
QUOTE(puckSR @ Apr 3 2006, 12:31 PM) 
its always used...not just when talking about planets
It isn't always used. It is only used most of the time. For objects extremely small or extremely large the equation does not work. Newtons was more obviously wrong that is when Einstein made a better one. It still encorporates Newton's in a normal range but Newton's is more wrong than Einstein's in extreme ranges. As damam said we don't know why.
Post by: jha'dhur on April 03, 2006, 03:54:00 PM
QUOTE(puckSR @ Apr 3 2006, 12:45 PM)
ummm....actually your wrong iffy....
but at least you tried....
F=mg....everywhere....g just changes depending on where you are....
what you may have meant to say was that most people think g=9.8...that may be true on earth...but everywhere else
...........
You are still confusing gravitatinol field and gravitational force.....
QUOTE(_iffy @ Apr 3 2006, 12:08 PM)
Why are you starting this thread?
You are the tool PUCK is the #1 thread crapper and you removed your post becuase he told you too.
Post by: _iffy on April 03, 2006, 04:22:00 PM
their no such thing as a gravitatinol field.
It's not like sheilds in star trek.
It's a force directed from the centres of two objects.
FYI - i removed my post because it had nothing to do with jesus being a jew, also because of Throwingstrikes
request. Seeing how it was this thread, i felt it was the polite thing to do. But for your knowlegde a nuclear reactor works by putting radioactive material into a pot of water to make it boil. The steam is used to power turbine generators. It's nothing like a particle accelerator.
puckSR are we arguing? i can't tell any more. You just agreed with me, after your earlier post where you dissagreed. **F=mg is a shorcut using a calculator**(sorry for the confusion)
one thing to add with f=mg is it doesn't incorporate distance. Like, why aren't we all sucked into black holes?
Their suppose to have near infinit mass?
F=ma is law but f=mg works.
throwingks You can use that formula anytime. It's just sometimes the result is so negligible whats the point.
Post by: damam on April 03, 2006, 04:51:00 PM
I know enough about gravity to know that i really dont know or understand it at all . . .
The proposed model in my physics classes was that matter created curvetures in space making the shortest distance in 4 dimensions between the two bodies, well much shorter than they appear. . . so when you drop an object your witnessing the trajectory moving in curved space taking a geodesic path which we then interpret as an attractive force. (its been a while since I have even thought about this but I think I got it right).
I could be wrong but i believe while I was learning this, the theory had been cast in considerable doubt
Post by: jha'dhur on April 03, 2006, 04:52:00 PM
QUOTE(_iffy @ Apr 3 2006, 05:29 PM)
^^ your an idiot
their no such thing as a gravitatinol field.
It's not like sheilds in star trek.
It's a force directed from the centres of two objects.
[qoute]
Gravitational Field
The gravitational field at any point is the gravitational force on some test mass placed at that point, divided by the mass of the test mass. The dimensions of the gravitational field are length over time squared, which is the same as acceleration. For a single mass (other than the test mass), Newton's law of gravitation tells us that
[/qoute]
1 of many sources with it definition and impirical equation.
http://www.physics.n...ok/node132.html
QUOTE(_iffy @ Apr 3 2006, 05:29 PM)
But for your knowlegde a nuclear reactor works by putting radioactive material into a pot of water to make it boil. The steam is used to power turbine generators. It's nothing like a particle accelerator.
REALLY!!!!!!!!
Post by: _iffy on April 03, 2006, 04:58:00 PM
puckSR
ok we agree.
Post by: jha'dhur on April 03, 2006, 05:02:00 PM
QUOTE(puckSR @ Apr 3 2006, 05:55 PM)
No....Im not....
If you think I am...then your misunderstanding my post
Wow....that is just blatantly wrong....
Ummm...
I dont even know where to start....
In this case the word "field" refers to the range of the interaction...
like an electromagnetic field...
you almost got the idea when you asked about the whole distance thing earlier.
Ok...here is what you are misunderstanding
F=ma LAW
F(g)=mg LAW...
F(g) is gravitational force
g is acceleration due to gravity
this is always going to be true...as long as Newton's second law is true....
what you are not understanding is that g is not a constant....
if you consider g a constant...i.e. g=9.8 then yes...this is a simplification
once again...Im really just arguing semantics...
Based upon your post dont, YOU do not fully understand the concept of a gravitatinal field.
YOU are explaining how objects ACT in the presence of a gravitational field.
We do not know exactly what causes a gravitational field, LAST I read the gravitons were all the rage.
But, I am sure _iffy will be able to explain it better than I.
Post by: _iffy on April 03, 2006, 05:20:00 PM
maybe i'm just thinking of gravity different.
I don't see fields, i see force acting linearly inbetween centres.
using the Gravitational force = (G * m1 * m2) / (d2) equation you get...
I can see how one might view a feild. An area close by where gravity would become significant.
But what damam's talking about i have no idea.
Post by: _iffy on April 03, 2006, 05:27:00 PM
I should explain that better.
When doing the math for that example, you add up all the vectors and end up with a "net result"
Are you saying "A feild would be all those vectors not added up?
Post by: jha'dhur on April 03, 2006, 05:28:00 PM
QUOTE(_iffy @ Apr 3 2006, 06:27 PM)
But what damam's talking about i have no idea.
You dont have any idea bout anything............
Post by: _iffy on April 03, 2006, 05:31:00 PM
Post by: jha'dhur on April 03, 2006, 05:36:00 PM
QUOTE(_iffy @ Apr 3 2006, 06:38 PM)
^^ do you have something to add to the thread or are you just flaming for flaming's sake?
I am confused.
Which one of you is dumber and which is dumberer.
Post by: slightly_damp on April 04, 2006, 12:16:00 AM
QUOTE(jha'dhur @ Apr 4 2006, 12:35 AM)
You dont have any idea bout anything............
Dude if you're gonna be a troll, at least do it properly for gods sake!
You are giving us other trolls a bad mane !
Post by: puckSR on April 04, 2006, 12:35:00 AM
did you get the idea of the field?
if you didnt I'll try a little bit better of an explanation....
Post by: jha'dhur on April 04, 2006, 06:01:00 AM
QUOTE(puckSR @ Apr 3 2006, 06:54 PM)
jha'dhur....seriously...what is your problem?
Instead of trying to explain any of this to iffy...you just keep insulting him....
I have a reputation for being an asshole to people when they are wrong....
but as long as they are trying...I will at least try to help....
You just blatantly are an asshole....your not trying to help...your not being constructive...
Your just being an asshole....
Thank You
OH the irony...
Coming from a self indulgent, marginal prick like you I take as a complement.
BUT SERIOUSLY, which one of you is DUMBERER.
GOOD money is on PUCK...
_IFFY what about or bet!!!!!!!!!!!!!!!!!
Post by: _iffy on April 04, 2006, 10:38:00 AM
QUOTE(_iffy @ Apr 3 2006, 07:48 PM)
^ i never claimed you could turn lead into gold.
What i said was you could add individual electrons, protons, until you end up with gold.
puckSR
I can visualize a spherical volume surounding a plant. And I can paste a label of a feild to it.
See what's confusing me is you have the equation, and gravity is always "turned on", so why have a feild?
what is this feild.
I can see the the feild being that lower left hand corner of that graph i made. An area where gravity becomes significant.
I've been doing some reading, specifically maxwell's equations, and i came across "gravitational lens"
Light, having no mass, is forced to change it's direction, just like a comet would. The only way that could work is if there was a gravitational feild. The gravitation equation doen't apply.
If this is what your getting at then, ok i need more reading. But i can assure you i can visualize it just fine.
Do you understand what i understand?
Post by: _iffy on April 04, 2006, 10:54:00 AM
i wasn't angry at all. I was being serious. (not flaming)
Post by: _iffy on April 04, 2006, 11:12:00 AM
sorry for the missunderstanding
ok
like an onion where the core is a planet, and every layer, would be a "range of gravitational intensity"...
were does the feild end, cause gravity is infinite?
is there a concensus somewhere?
Like the amount of deviation over a given distance of a traveling object is less than "x" gravity is not applied?
Post by: throwingks on April 04, 2006, 03:24:00 PM
The bigger distance gets the closer you get to 0. Although you can never reach 0 it is pretty much the same thing as being 0.
(M1 * M2 * G) / (D^2) = (GF)
Let D = 1 through infinity
100/1 = 100
100/4 = 25
100/9 = 11.111111111111111111111111111111
100/16 = 6.25
100/25 = 4
100/36 = 2.7777777777777777777777777777778
100/49 = 2.0408163265306122448979591836735
100/64 = 1.5625
100/81 = 1.2345679012345679012345679012346
100/100 = 1
100/121 = 0.8264462809917355371900826446281
etc.
But, why isnt the equation:
F = [(G1 * M1) * (G2 * M2)] / (D^2)
Post by: jha'dhur on April 04, 2006, 04:42:00 PM
QUOTE(throwingks @ Apr 4 2006, 04:31 PM)
Gravitational Force = (Gravity * Mass1 * Mass2) / (Distance^2)
The bigger distance gets the closer you get to 0. Although you can never reach 0 it is pretty much the same thing as being 0.
(M1 * M2 * G) / (D^2) = (GF)
Let D = 1 through infinity
100/1 = 100
100/4 = 25
100/9 = 11.111111111111111111111111111111
100/16 = 6.25
100/25 = 4
100/36 = 2.7777777777777777777777777777778
100/49 = 2.0408163265306122448979591836735
100/64 = 1.5625
100/81 = 1.2345679012345679012345679012346
100/100 = 1
100/121 = 0.8264462809917355371900826446281
etc.
But, why isnt the equation:
F = [(G1 * M1) * (G2 * M2)] / (D^2)
Because F and F(g) arent the same.
Your friend keeps wanting to exchange the becuase netwon uses that derivation to describe the force acting on an object within earths gravitatinal field relative to that objects mass.
However once you leave the gravity well of a large object that argument (F=ma) falls short of describing the gravitational forces that you would experience. (i.e. in a space shuttle in between earth and the sun)
Where you are "essentially" weightless in microgravity.
Post by: throwingks on April 04, 2006, 05:54:00 PM
QUOTE(puckSR @ Apr 4 2006, 04:34 PM)
Well....first off...where in the heck are you getting G1 and G2...
G is not Gravity...
G is the gravitational constant....it doesnt ever change.....
Second if you are assuming that G1=G2....then you would have G^2.....which would simply be a redefinition of the constant....
G is simply a constant......to balance the equation and make it work....
IT IS NOT "GRAVITY' OR ANY MEASUREMENT OR IN ANY WAY A VARIABLE
I made it up as a reference to the Gravity from object 1 (G1) and the Gravity from object 2 (G2). No matter how big or small an object is it has a gravitational pull. If G is always constant why dont they just use that number instead of giving it a variable? G of the sun != the G of a marble in the middle of space. That is why I am saying the gravitational pull of both objects should be used in the equation. On Earth we are all relatively close to the origination of the earths gravity (we think). So you can assume its consistency. However, an object with a larger mass, even though it weighs the same as an object with a smaller mass, will have more gravitational force. Do I make any sense?G is not Gravity...
G is the gravitational constant....it doesnt ever change.....
Second if you are assuming that G1=G2....then you would have G^2.....which would simply be a redefinition of the constant....
G is simply a constant......to balance the equation and make it work....
IT IS NOT "GRAVITY' OR ANY MEASUREMENT OR IN ANY WAY A VARIABLE
To use your magnetic example. If a huge magnet pulled on a smaller magnet, it would be a stronger pull than if the same huge magnet pulled on a piece of metal the same size as the smaller magnet.
I don't have the answers, these are not rhetorical questions. I am trying to learn.
P.S. All of this describes gravity and how it interacts. None of it pertains to the original questions of the thread.
Post by: _iffy on April 04, 2006, 06:31:00 PM
QUOTE
However, an object with a larger mass, even though it weighs the same as an object with a smaller mass, will have more gravitational force. Do I make any sense?
Weight = mass x acceleration.If m1>m2 then they can't weigh the same. (unless m2 has a rocket pushing it down on the scale)
Using your magnets as an example, if you put two magnets on a smooth table, bring them close to one another, let go, they will both move and meet in the middle.
Same with gravity. Stars can collide. Planetary orbits decay. It's just gravity is a weaker force than magnetism so it takes alot longer.
QUOTE
P.S. All of this describes gravity and how it interacts. None of it pertains to the original questions of the thread.
It's your thread what do you want to do?
Post by: jha'dhur on April 04, 2006, 06:36:00 PM
QUOTE(throwingks @ Apr 4 2006, 07:01 PM)
I made it up as a reference to the Gravity from object 1 (G1) and the Gravity from object 2 (G2). No matter how big or small an object is it has a gravitational pull. If G is always constant why dont they just use that number instead of giving it a variable? G of the sun != the G of a marble in the middle of space. That is why I am saying the gravitational pull of both objects should be used in the equation. On Earth we are all relatively close to the origination of the earths gravity (we think). So you can assume its consistency. However, an object with a larger mass, even though it weighs the same as an object with a smaller mass, will have more gravitational force. Do I make any sense?
To use your magnetic example. If a huge magnet pulled on a smaller magnet, it would be a stronger pull than if the same huge magnet pulled on a piece of metal the same size as the smaller magnet.
I don't have the answers, these are not rhetorical questions. I am trying to learn.
P.S. All of this describes gravity and how it interacts. None of it pertains to the original questions of the thread.
Dude Puck, is conceptually mistaken.
F=ma has no relevance outside of the influence of a large body.
The equation you quote which contains the universal gravity constant is the singular definition.
F=ma is a derivation of that expression which describes gravity on the planet earth.
Gravity acts along the line of the center of masses it is not multidirectional. It pulls you to earths center it doesnt push also.
P.S. _IFFY I do recall you mentioning a gold factory, dont back away from that foolishness now.
What happened to OUR BET!!!!!!!!!!!!!!!!!!!11
Post by: _iffy on April 04, 2006, 06:39:00 PM
the gold factory was sarcasm.
Post by: jha'dhur on April 04, 2006, 06:43:00 PM
QUOTE(_iffy @ Apr 4 2006, 07:46 PM)
^^ haha
the gold factory was sarcasm.
Post by: _iffy on April 04, 2006, 06:45:00 PM
QUOTE
Gold is bought and sold every day. the value of gold is set by the worlds supply aswell as demand. If a "gold makeing factory" was built, it would flood the market with gold - lowering it's value.
notice the quotation marks...Oh yeah i should tell you, it would cost more to make gold useing a particle accelerator, than the gold is worth
Post by: jha'dhur on April 04, 2006, 06:54:00 PM
QUOTE(_iffy @ Apr 4 2006, 07:52 PM)
^^ the pic not showing up for me
anyways here's my quote
notice the quotation marks...
Oh yeah i should tell you, it would cost more to make gold useing a particle accelerator, than the gold is worth
Since you cant manufacture gold your point IS.
What really let everyone know you were a complete idiot was the atomic structure being influenced by atmospheric pressure BS you posted. We had a good laugh on that one.
WHAT ABOUT OUR BET!!!!!!!!!!!!!!!!!!!!!!
Post by: throwingks on April 04, 2006, 07:03:00 PM
QUOTE(_iffy @ Apr 4 2006, 05:38 PM)
It's your thread what do you want to do?
Those questions arent answerable by anyone I know. I am not attacking Puck, I just remember he making a statement a while ago about something being as obvious as the theory of gravity. I remember thinking "is it really that obvious?" So I started this thread. Also to have a new place so, my other thread went back on track. It worked a little.It is understood that the weight at the center of earth would be 0 if you had a perfectly spherical shape in the exact center, because gravity would be pulling equally in all directions. So one, does have an answer, but no way to prove.
QUOTE(jha'dhur @ Apr 4 2006, 05:43 PM)
The equation you quote which contains the universal gravity constant is the singular definition.
So to make it accurate you would have to calculate twice. Once for each object then get the summation?
Post by: jha'dhur on April 04, 2006, 07:17:00 PM
QUOTE(throwingks @ Apr 4 2006, 08:10 PM)
Those questions arent answerable by anyone I know. I am not attacking Puck, I just remember he making a statement a while ago about something being as obvious as the theory of gravity. I remember thinking "is it really that obvious?" So I started this thread. Also to have a new place so, my other thread went back on track. It worked a little.
It is understood that the weight at the center of earth would be 0 if you had a perfectly spherical shape in the exact center, because gravity would be pulling equally in all directions. So one, does have an answer, but no way to prove.
Considering the earth is being slung around the sun with considerable momentum, Its not like you would be weightless a force would be acting upon you like it is on the the surface of the earth.
In the sense of the classical force balance:
The sum of the forces = M*a. And this really only most precisely applies to objects in freefall.
Consider:
- ) a rocket, accelerating upward,
- ) or you walking down the street
- ) swimming in the ocean.
- ) Living on a space station
F = m*a: isnt a catch all.
P.S. From sea level to crest of Mt Everest g(a) fluctuates less than 1%. Which for all intensive purposes is negligible.
Post by: throwingks on April 04, 2006, 07:22:00 PM
As to the different scenarios, for my references I am using a perfect vacuum. So the only 2 forces would be G1 and G2.
Post by: _iffy on April 04, 2006, 07:23:00 PM
QUOTE(jha'dhur @ Apr 4 2006, 09:01 PM)
I'm a stupid idiot, who likes little boys and i smell like dung and im a stupid idiot my name is jha'dhur
I told you, you could make gold. Do you remamber any of that? Do you remember what you type right after you post?
It wasn't just atmospheric pressure, it was also gravity.
As far as the bet... it wasn't a bet, you offerd a reward for a search. I think you should keep every penny you have. You'll need it later.
ThrowingksI think you need to start a new thread entitled "jha'dhur's thread"
Post by: jha'dhur on April 04, 2006, 07:43:00 PM
QUOTE(puckSR @ Apr 4 2006, 08:31 PM)
F=ma is not a derivation of the "law of gravity"
do you even realize that without F=ma you would never be able to arrive at the law of gravity?
The law of gravity necessitates the Gravitational constant...which could not have been calculated without F=ma
Ok....first off...your thinking about this wrong....G isnt gravity.....
Secondly
When two magnets pull on each other....they attract each other....
you can measure the force of their attraction....but you cannot measure the individual force of each magnet...
The equation from earlier describes the force between the two masses.....
F is the gravitational pull....
Not G......
We have a tendency to name constants in Physics....
and besides...G is a constant....but we are constantly refining it....
Its constant to the physical world...not so much to us....
G=6.67 × 10−11....which isnt a lot of fun to write down....
SO
....G is not gravitational pull...F is...
....You cannot measure the individual force caused by an individual mass....since they are interacting....
I think I can explain one other thing....
the marble vs the sun
Gravity is directly proportional to mass....(well for the cases of this discussion)
thats why both masses are included in the equation....
Post by: jha'dhur on April 04, 2006, 08:52:00 PM
QUOTE(puckSR @ Apr 4 2006, 09:49 PM)
Umm actually it is...
i.e. your rocket example....
^ lets say that your rocket is accelerating at 20 m/s^2
it has a mass of 20,000 kg....
F(net)=20*20,000=400,000 Newtons
F(g)=-9.8*20,000=-196,000 Newtons
F(rocket)=F(net)-F(g)=596,000 Newtons
So the rocket is using a force of 596,000 Newtons to rise at the acceleration of 20 m/s^2
As you put it:
If the rocket is traveling at a velocity of 20 m/s and is not accelerating what is its force?
Post by: lordvader129 on April 04, 2006, 09:04:00 PM
QUOTE(jha'dhur @ Apr 4 2006, 09:59 PM)
As you put it:
If the rocket is traveling at a velocity of 20 m/s and is not accelerating what is its force?
the force applied for an acceleration of 0 is 0
F = MA
F = M*0
F = 0
Post by: Chook on April 04, 2006, 09:05:00 PM
F=ma ,is just an equality
Σ(F)=ma ,is the direction in which an object is acclerating (sum of all forces)
F= Gconst * (M1*M2)(r^2), is the attractive force of two objects due to force of gravity
F= Cconst * (q1*q2)(r^2), is attractive/repulsive force due to electromagnetic force (i got some of the constant's proper letters wrong)
g=Gconst * M/r^2, is a substitution for the gravity constant, but this is only for a point attraction. To calculate the attraction you would have to integrate and approximate the forces.
Lightwaves exist as both energy and mass, it depends on what they are interacting with, its from einstein's E=mc^2
That is why light can move, even higher energy waves contain particles, or excite particles, depending on what they are doing.
Were i am confused is that electromagnetic fields tend to terminate on opposite charges, and do not know how the field lines of gravity look. I have always thought of them as a perfect sphere from a point source.
What we know isn't perfect, and we try to improve.
The internet will decrease the change of getting laid.
We are on the internet
And it is in our genes to want to "procreate"
so we all are the dumberer
Post by: jha'dhur on April 04, 2006, 09:13:00 PM
QUOTE(lordvader129 @ Apr 4 2006, 10:11 PM)
the force applied for an acceleration of 0 is 0
F = MA
F = M*0
F = 0
Hey, I'm Vaders Roommate:
Come on PUCKY you can't possibly beat this.
QUOTE(puckSR @ Apr 4 2006, 10:16 PM)
F(g)=m*g
An object in motion tends to stay in motion....so gravitational force must be equal to the force being exerted by the rocket....
which in this case would be 196,000 Newtons
Now...this isnt a very practical situation because something had to start the rocket moving in the first place, and then the gravitational force decreases with distance...
By the way....that whole object in motion thing....Newton's 1st law
Think of it this way....
if I start a marble rolling...and there is nothing to stop it...including friction....I am not continually applying force...I only applied force when the marble started rolling....which if why F=ma
So which is it??????
2 HEADS are better than ONE.....
LOL
Post by: lordvader129 on April 04, 2006, 09:19:00 PM
let me help you wrap your head around that one
since gravity is pulling down on the rocket, a constant force much be applied to maintain a constant velocity, that force is 196,000 N
however, sicne gravity is pulling down on the rocket that force is canceled by the force of gravity (-196,000 N) resulting in a NET force of 0 N
196,000 + (-196,000) = 0
BTW, you running out of bullshit pics? that last one was a rehash
Post by: puckSR on April 04, 2006, 09:29:00 PM
seriously though....we can completely quit talking about this...
I dont expect you to admit that your wrong....but I do expect you to shut up when you realize that you are wrong.....
If you were arguing this entire F=ma thing from the perspective of Einstein or quantum mechanics...I would entertain it all....but this is just proving your an idiot
Post by: jha'dhur on April 05, 2006, 06:06:00 AM
QUOTE(lordvader129 @ Apr 4 2006, 10:26 PM)
actually both are right
let me help you wrap your head around that one
since gravity is pulling down on the rocket, a constant force much be applied to maintain a constant velocity, that force is 196,000 N
however, sicne gravity is pulling down on the rocket that force is canceled by the force of gravity (-196,000 N) resulting in a NET force of 0 N
196,000 + (-196,000) = 0
BTW, you running out of bullshit pics? that last one was a rehash
Lets test your theory let me drive my car at you at 20 mph. You tell me if the F = 0.
Post by: jha'dhur on April 05, 2006, 06:16:00 AM
QUOTE(puckSR @ Apr 4 2006, 10:36 PM)
It really sucks to get another lesson in physics....doesnt it Xmedia
seriously though....we can completely quit talking about this...
I dont expect you to admit that your wrong....but I do expect you to shut up when you realize that you are wrong.....
If you were arguing this entire F=ma thing from the perspective of Einstein or quantum mechanics...I would entertain it all....but this is just proving your an idiot
Stop lying to people and telling them you are an engineer.
When you were riding on the short bus.
Post by: Chook on April 05, 2006, 08:08:00 AM
QUOTE(jha'dhur @ Apr 5 2006, 07:23 AM)
Stop lying to people and telling them you are an engineer.
When you were riding on the short bus.
Nice Dubs, is that what gets you up to 20 mph?
Puck you are not a EE anymore, cause you pissed off bredeson!
Post by: lordvader129 on April 05, 2006, 09:56:00 AM
QUOTE(jha'dhur @ Apr 5 2006, 07:13 AM)
Lets test your theory let me drive my car at you at 20 mph. You tell me if the F = 0.
as long as you arent accelerating the net force is 0
however, i think you are confusing net force with kinetic energy, those are 2 different things
Post by: damam on April 05, 2006, 10:11:00 AM
matter tells spacetime how to curve (ie. the more matter the greater the curve) creating non-Euclidean geometries, and spacetime tells matter how to move by taking the shortest path possible through 4-dimensions.
this is why singularities can actually merge the dimensions and the approach to event horizon slows down time to a near stand still.
F= Gconst * (M1*M2)(r^2) is a description of gravity as we perceive it.
Post by: _iffy on April 05, 2006, 10:26:00 AM
Thanks
Post by: throwingks on April 05, 2006, 10:35:00 AM
Tesla is a genious. He made fun of Edison for taking so long inventing the lightbulb. He said something like, "If Edison were to look for a needle in a haystack he would look at every piece of hay and discard until he found the needle."
Back on point Tesla's theories on Gravity should hold more "weight". <-- </bad joke> He never officially released them though. He unfortunately died.
link in openning post.
Post by: lordvader129 on April 05, 2006, 10:51:00 AM
QUOTE(_iffy @ Apr 5 2006, 11:33 AM)
^^^ what kind of phisics is that? Subject title? I want to do more reading.
Thanks
the first experiment here might help you get a better understanding of general relativity
http://www.jracademy...olj/exper2.html
the second experiment deals with quantum gravity
Post by: damam on April 05, 2006, 11:02:00 AM
thanks for posting it
tesla-
no doubt a very brite individual. his theory claimed that he had united both theories some how. However, tesla really seemed to have lost it later in life, and i take his claim with a grain of salt. remember his death box LOL?
Post by: lordvader129 on April 05, 2006, 03:35:00 PM
http://www.bringyou....ogetics/p67.htm
Post by: jha'dhur on April 05, 2006, 03:36:00 PM
QUOTE(lordvader129 @ Apr 5 2006, 11:03 AM)
as long as you arent accelerating the net force is 0
however, i think you are confusing net force with kinetic energy, those are 2 different things
That is JUST INCORRECT..
The only reason I havent provided you with the derivation is I wish to read more of your little brothers ramblings.
It is shameful that you believe that if an object like a car strikes you while it is moving at a constant velocity the force of that collision will not throw you in the air.
Lets try this once more:
Imagine a steam catapult pushing an F-16 down the runway of an aircraft carrier at some point the catapult will no longer be accelerating, and will begin to deccellerate.
Is it(catapult) no longer exerting a force on the F-16 just because it is no longer accelerating.
Puck almost spit the word out but his genetics keeps him from stumbling on to the truth.
Two words, which have mathematical expressions for them describe real life physics.
Given the rocket lab at Sandia Natl LAB always has these great videos of rocket sleds flying into walls. I thought the rocket example would be a dead giveaway,
Since the rocket is traveling in a horixontal direction along the track (g) is of no significance.
but it seems I "overstimated" you guys.
P.S. I wouldnt have posted a link to that BS gold story either. The source is as probably sketchy as you morons.
Post by: puckSR on April 05, 2006, 04:49:00 PM
QUOTE
Imagine a steam catapult pushing an F-16 down the runway of an aircraft carrier at some point the catapult will no longer be accelerating, and will begin to deccellerate.
Once again...you are an idiot....you now have to fight wind resistance and friction....
How many times do I have to pound this concept into your head....
an object in motion tends to stay in motion unless acted upon by an outside force....
once the F16 is moving...it will continue to move at the same velocity...unless another force is acting on the F16...
I would absolutely love to see your "Derivation" to explain the "Force" that needs to be maintained to keep an object moving in a gravity free...friction free(which includes friction due to air) enviroment....
I then want to see you explain the voyager space probe....which is still moving...despite the fact that it doesnt have any rocket boosters......and will continue to move even when it is far out in space.....
This is the most ridiculous conversation I have had about physics since some idiot told me that the sun provides the gravitational attraction of the earth....and that if the two "gravity wells" didnt exist...you wouldnt be able to stand on the earth.....
Please, im begging you, I would love to see your explanation of how constant force is necessary to maintain velocity in a friction-free, gravity free enviroment......
Post by: jha'dhur on April 05, 2006, 04:53:00 PM
QUOTE(puckSR @ Apr 5 2006, 05:56 PM)
Once again...you are an idiot....you now have to fight wind resistance and friction....
How many times do I have to pound this concept into your head....
an object in motion tends to stay in motion unless acted upon by an outside force....
once the F16 is moving...it will continue to move at the same velocity...unless another force is acting on the F16...
I would absolutely love to see your "Derivation" to explain the "Force" that needs to be maintained to keep an object moving in a gravity free...friction free(which includes friction due to air) enviroment....
I then want to see you explain the voyager space probe....which is still moving...despite the fact that it doesnt have any rocket boosters......and will continue to move even when it is far out in space.....
This is the most ridiculous conversation I have had about physics since some idiot told me that the sun provides the gravitational attraction of the earth....and that if the two "gravity wells" didnt exist...you wouldnt be able to stand on the earth.....
Please, im begging you, I would love to see your explanation of how constant force is necessary to maintain velocity in a friction-free, gravity free enviroment......
Post by: lordvader129 on April 05, 2006, 04:58:00 PM
QUOTE(jha'dhur @ Apr 5 2006, 04:43 PM)
It is shameful that you believe that if an object like a car strikes you while it is moving at a constant velocity the force of that collision will not throw you in the air.
note the word COLLISION in your sentence there, it implys a decrease in velocity, also know as DECELERATION, or negative ACCELERATION, and we established, acceleration has a net force greater than 0, however a car simply moving at a constant velocity (not speeding up or slowing down, in case you dont know what constant velocity is either) will have 0 net force
now kinetic energy is a different story, this is what is transferred to a person when a car (or anything) hits them, if you want i could calculate the kinetic energy of a car moving at 20 m/s, the formula is KE = 1/2 M * V^2, where M is mass and V is velocity
Post by: jha'dhur on April 05, 2006, 05:10:00 PM
Post by: lordvader129 on April 05, 2006, 05:15:00 PM
QUOTE(jha'dhur @ Apr 5 2006, 06:17 PM)
I just wish to see you and your little brother flounder in your ignorance a little longer.
what exactly are we suposedly ignorant of? i dont think youve even explained that much
Post by: jha'dhur on April 05, 2006, 05:25:00 PM
QUOTE(lordvader129 @ Apr 5 2006, 06:22 PM)
what exactly are we suposedly ignorant of? i dont think youve even explained that much
Dude you and your domestic partner just make stuff up.
You should supply a link to that BS gold story.
Nevermind.
www.unsubstantiatedbullsh*t.com..
Post by: lordvader129 on April 05, 2006, 05:39:00 PM
QUOTE(jha'dhur @ Apr 5 2006, 06:32 PM)
Dude you and your domestic partner just make stuff up.
You should supply a link to that BS gold story.
Nevermind.
www.unsubstantiatedbullsh*t.com..
well if you know we are making stuff up why cant you provide any "real" facts?
http://en.wikipedia...._an_accelerator
Post by: jha'dhur on April 05, 2006, 05:55:00 PM
QUOTE(lordvader129 @ Apr 5 2006, 06:46 PM)
well if you know we are making stuff up why cant you provide any "real" facts?
http://en.wikipedia...._an_accelerator
You know I went through this with the other dummy.
I can find
Bigfoot
LochNess
and Chupacabra on some bull shucks website that does not really impress anyone let alone myself. What dont you undertand about that. Its not raining someone is pissing on you.
The reason you cant find any information about gold synthesis on anything with any degree of technical "pedigree" is becasue it is BOGUS.
Thats why I stopped responding to you idiots all you understand are google search results and pictures.
So you and your wife(Puck) can keep on stalking me I am going to respond in kind with illustrations that define your intelligent level.
Keep Trying, though I dont know too many wanna be GEEKS. You will get there.
Post by: jha'dhur on April 05, 2006, 07:52:00 PM
QUOTE(lordvader129 @ Apr 5 2006, 06:05 PM)
now kinetic energy is a different story, this is what is transferred to a person when a car (or anything) hits them, if you want i could calculate the kinetic energy of a car moving at 20 m/s, the formula is KE = 1/2 M * V^2, where M is mass and V is velocity
BOGUS.....
A force will act upon the moron(you) that is stroke by the vehicle.
Post by: lordvader129 on April 05, 2006, 09:02:00 PM
QUOTE(jha'dhur @ Apr 5 2006, 08:59 PM)
BOGUS.....
A force will act upon the moron(you) that is stroke by the vehicle.
uh yeah, if you actually read the post you quoted there i said kinetic energy is transferred
and puck, dont forgot the, lets call it "unique," style of quoting, the way he quotes posts that were made hours ago, even though hes posted several times since then, he responds to the same post multiple times, and actually becomes irate at having to repeat himeslef (when hes the one who doesnt realize hes already responded) ask throwkingks about that one, lol
Post by: throwingks on April 05, 2006, 09:42:00 PM
QUOTE(lordvader129 @ Apr 5 2006, 08:09 PM)
...ask throwkingks about that one, lol
You lost me. 
My head hurts.
Post by: Chook on April 05, 2006, 09:43:00 PM
QUOTE(jha'dhur @ Apr 5 2006, 08:59 PM)
BOGUS.....
A force will act upon the moron(you) that is stroke by the vehicle.
use the word struck instead of stroke next time!
Check out hyperphysics
dP/dt = force, is another way of saying it. Any change in momentum will be the result of a net sum of forces not equal to zero.
g - D(v) + T(t) = M(t) * dv/dt is what you wrote
should be
mg - D(v) + dP/dt = M(t) * dv/dt
Hyperphysics on the force of gravity aka weight
Hyperphysics on Thrust
Hyperphyiscs on thrust
Jha'dhur, please go read something like hyperphysics or get a physics book. I already took the FE and know plenty about Newtonian physics. If you can do this with tesla's or einstein's versions of physics, i'd like to see it. I do not know about quantum physics that well, and i'll admit to that.
Post by: lordvader129 on April 05, 2006, 09:47:00 PM
QUOTE(throwingks @ Apr 5 2006, 10:49 PM)
You lost me.
My head hurts.
sorry, it was iffy_ not you
http://forums.xbox-s...dpost&p=3335517
quotes 3 and 4
Post by: throwingks on April 05, 2006, 09:52:00 PM
Gotcha.
Guys, in my opinion I think you are letting him win. Just a theory of mine, I think he is pulling chains on purpose. He is pulling up older threads that have been dormant for a little bit, and nit-picking. Seems obvious to me, but I have been wrong before. I agree with puck it is entertaining, but to only us few. How long can we let him keep other people from posting. I like statecowboys input a lot more, but I have a feeling we won't see him for a long time again.
Who knows who else would have good insight that is too scared to start.
Post by: jha'dhur on April 05, 2006, 09:55:00 PM
QUOTE(puckSR @ Apr 5 2006, 10:22 PM)
The fact that he just misrepresented the derivation for momentum...classic Xmedia
Like when you dont bother to read what anyone is posting...and then haphazardly try to claim that momentum is = mass * acceleration
Equation #1
Equation #2
Funny, acceleration is the derivative of velocity, dv/dt
[a = dv/dt]
m*a = m (dv/dt)
Since F = m*a
F = m(dv/dt)
Integral of F =
The sum of the forces as a function of time is equal to the change in momentum.
Post by: Chook on April 05, 2006, 10:14:00 PM
QUOTE(jha'dhur @ Apr 5 2006, 11:02 PM)
Equation #1
Equation #2
Funny, acceleration is the derivative of velocity, dv/dt
[a = dv/dt]
m*a = m (dv/dt)
Since F = m*a
F = m(dv/dt)
Integral of F =
The sum of the forces as a function of time is equal to the change in momentum.
your reverse does not account for any constant momentium. Its like using a Bdot or Ddot sensor, only detects changes in the field, not the overall field strength.
Post by: jha'dhur on April 05, 2006, 10:28:00 PM
QUOTE(Chook @ Apr 5 2006, 11:21 PM)
your reverse does not account for any constant momentium. Its like using a Bdot or Ddot sensor, only detects changes in the field, not the overall field strength.
QUOTE(Chook @ Apr 5 2006, 11:21 PM)
Its like using a Bdot or Ddot sensor, only detects changes in the field, not the overall field strength.
Nada amigo. Wrong course.
Pull out a physics book, arent you "supposed" to be at a college. I posted a link to a online physic tutorial.
constant momentum (P) implies that the velocity is constant.
You do understand that delta is shorthand [mv(f) - mv(o)/ dt] or the derivative.
Derivative of a constant = 0
Therefore LHS expression = 0(equilibrium) but RHS still has a time dependance.
Post by: jha'dhur on April 05, 2006, 10:39:00 PM
QUOTE(puckSR @ Apr 5 2006, 11:28 PM)
so....you were wrong...glad we cleared that up
I was wrong let me remind you:
QUOTE(puckSR @ Apr 4 2006, 09:49 PM)
^ lets say that your rocket is accelerating at 20 m/s^2
it has a mass of 20,000 kg....
F(net)=20*20,000=400,000 Newtons
F(g)=-9.8*20,000=-196,000 Newtons
F(rocket)=F(net)-F(g)=596,000 Newtons
U 2:
QUOTE(lordvader129 @ Apr 4 2006, 10:11 PM)
the force applied for an acceleration of 0 is 0
F = MA
F = M*0
F = 0
You know VADER the whole question was a set up. And I knew some dummy would say F=0, Didnt expect it to be you, smart money was on _IFFY. But you win the prize amigo. I caught a big one.
HOOK
LINE
SINKER.
Post by: lordvader129 on April 05, 2006, 10:42:00 PM
QUOTE
lets say that your rocket is accelerating at 20 m/s^2
QUOTE
the force applied for an acceleration of 0 is 0
equations using different numbers have different answers, welcome to math jha'dur
Post by: jha'dhur on April 05, 2006, 10:46:00 PM
QUOTE(lordvader129 @ Apr 5 2006, 11:49 PM)
equations using different numbers have different answers, welcome to math jha'dur
At least you spelled my name CORRECTLY
Post by: Chook on April 05, 2006, 11:21:00 PM
QUOTE(jha'dhur @ Apr 5 2006, 11:35 PM)
constant momentum (P) implies that the velocity is constant.
If P was constant, then you could not calculate any net force, as there would not be any.
ie dP/dt=0 if you consider P a constant.
QUOTE(jha'dhur @ Apr 5 2006, 11:35 PM)
You do understand that delta is shorthand [mv(f) - mv(o)/ dt] or the derivative.
would actually be [(mv(f) - mv(o)) dt]
I graduate EE/CS this semester, I hope my math skillz are good enough for you! I noticed you can google some math up, and have been wrong several times on the board. Puck tells me that you work at a reactor, which one?
Post by: jha'dhur on April 06, 2006, 05:54:00 AM
QUOTE(Chook @ Apr 6 2006, 12:28 AM)
If P was constant, then you could not calculate any net force, as there would not be any.
ie dP/dt=0 if you consider P a constant.
would actually be [(mv(f) - mv(o)) dt]
I graduate EE/CS this semester, I hope my math skillz are good enough for you! I noticed you can google some math up, and have been wrong several times on the board. Puck tells me that you work at a reactor, which one?
One more time "puch" I mean chinook
IF dp/dt is constant there is NO CHANGE in the net force, hence equilibrium balance.
Everything is bigger in Tx, even the "MA ROONS."
Tell you what post a link to momentum balance and point out my mistatements.
I will send you a pound of "fools" gold
LOL
Post by: lordvader129 on April 06, 2006, 09:40:00 AM
Post by: PhatIrishBastard on April 06, 2006, 02:14:00 PM
You should let it go. CockSR and Vader. Are only exhibiting there typical behavior they pulled the same highschool crap with my roommate the xmedia.
Its funny how if you dont agree with his foolish notions you are xmedia or bluedeath.
He is going to stalk you everyday replying to every post you make even those intended for others.
Its nothing new, he is a simply a bully, lordvader is his accomplish. You have proved him wrong since you started posting I sat back and read as he jumped your sh*t for the race thread, and now he claims you are the flamer.
Vader doesnt have any credibility whatsoever, he may as well have a vagina as far as I am concerned. You notice how he appeared out of the blue to smear your name.
Same sh*t different dude.
Post by: PhatIrishBastard on April 06, 2006, 03:51:00 PM
QUOTE(puckSR @ Apr 6 2006, 03:19 PM)
this is where jha'dhur gets confused...he believes that F=ma only applies to freefall because that is the only time that there is an acceleration.....
He fails to realize that a is a component of the force...in other words if the force is known and the opposing force is removed...a would be in action.....
F=ma is a law of physics....its isnt a "shortcut"
go ask a doctorate in physics....he will have to explain the same thing
the derivative of a constant = 0
in other words
if P is constant over time...then dP/dt=0This is basic calculus.....
Make up your mind
First.
QUOTE(puckSR @ Apr 4 2006, 08:49 PM)
So the rocket is using a force of 596,000 Newtons to rise at the acceleration of 20 m/s^2
Now you are using his momentum equation to BS the fact that you were WRONG
Lordvader129, it darn sure isnt ZERO.
The mass of the rocket its velocity and the drag it feels all are time dependant.
m(t) mass of the rocket will vary with burn up of fuel.
D(t) the wind direction and speed vary with time.
v(t) the velocity will vary because of feedback from the above two.
P = M(t) * V(t)
so
dP/dt is never equal to ZERO.
When off the launch pad.
Post by: puckSR on April 06, 2006, 04:30:00 PM
QUOTE(jha'dhur @ Apr 4 2006, 07:43 PM)
Dude Puck, is conceptually mistaken.
F=ma has no relevance outside of the influence of a large body.
.....
This was jha'dhurs original claim.....
It is completely fallacious....
If there is a gravitational force...there will be a gravitational acceleration...any time there is a force applied to a mass...there will be an acceleration...as long as the mass is constant
Post by: jha'dhur on April 06, 2006, 04:43:00 PM
QUOTE(puckSR @ Apr 6 2006, 05:37 PM)
This was jha'dhurs original claim.....
It is completely fallacious....
If there is a gravitational force...there will be a gravitational acceleration...any time there is a force applied to a mass...there will be an acceleration...as long as the mass is constant
Losing the momentum balance arguement HUH.
Yeah, I would also attempt to change the subject and deflect attention from my stupidity.
Good Strategy.
Now throw in some semantics. Or some other irrelevant technobabble.
Post by: lordvader129 on April 06, 2006, 05:20:00 PM
QUOTE
The mass of the rocket its velocity and the drag it feels all are time dependant.
m(t) mass of the rocket will vary with burn up of fuel.
D(t) the wind direction and speed vary with time.
v(t) the velocity will vary because of feedback from the above two.
P = M(t) * V(t)
so
dP/dt is never equal to ZERO.
When off the launch pad.
m(t) mass of the rocket will vary with burn up of fuel.
D(t) the wind direction and speed vary with time.
v(t) the velocity will vary because of feedback from the above two.
P = M(t) * V(t)
so
dP/dt is never equal to ZERO.
When off the launch pad.
all true, however this is compeltely irrelevant to the orginal question, which you probably never read (assuming you arent jha'dhur) so why would you feel the need to reply at all?
QUOTE(jha'dhur)
If the rocket is traveling at a velocity of 20 m/s and is not accelerating what is its force?
NOT accelerating, meaning V is constant, so A is 0, so net force is 0
now, if mass is changing, then P is also changing, HOWEVER, since V is still constantm net force is still 0
now naturally as you say, V is never constant due to changing circumstances, but that wasnt jha'dhurs question
QUOTE
not accelerating
Post by: lordvader129 on April 06, 2006, 05:51:00 PM
QUOTE(puckSR @ Apr 6 2006, 06:43 PM)
actually...that is wrong...
you have to integrate it out to get F=d(mv)/dt
if mass is changing, then while v is constant m is the variable with respect to time
therefore there would still be a force....
my figures are based on the assumption that the force exerted by gravity is also changing due to the decreasing mass this would balance the net force equation at 0
but yes, i do realize that simply saying V = constant doesnt automatically make F(net) = 0
Post by: PhatIrishBastard on April 07, 2006, 11:32:00 AM
QUOTE(puckSR @ Apr 6 2006, 08:01 PM)
Ok....so do you finally get it jha'dhur?
Your absolutely correct, they can use the "momentum balance" equation(even though they would use it with respect to varied momentum over time which is force)....
Im glad you think that Physics is a list of equations....from a book...
I hope you now realize that almost every part of classical physics is based around the 3 Laws of Newton.
F=ma being the main equation
I hope you also realize that Newton's second law is an integral part of modern physics, and that F=ma is important for more than just free-fall equations.
If you are still convinced that F=ma is only applicable in free-fall equations, there isnt really much more I can post. IrishBastard already posted the NASA link....
These might be helpful
U of Penn
NASA with some explanation
Yet another link
All of these links will tell you that Newton's second law...commonly expressed as F=ma....is one of the foundations of Newtonian physics(physics without relativity). Your obsession with momentum is not "unwarranted", but you have to realize that any time you talk about "changing momentum with respect to time" your actually talking about dP/dt=F(net)***Newton's Second Law****
You may have meant to say that F=ma is only applicable to free body diagrams which might have gotten some argument from me....but would technically be accurate.
Nice lesson class....hope we can do this again sometime.
Your post is classic COCK SUCKER at his best.
- First you are proven wrong, you and your boyfriend.
- You posted some B.S. calc claiming F= "BS" amount of newtons, or simply zero.
- You keep insisting F = ma, when it is the sum of the forces.
- F = ma is an oversimplification only valid for objects in freefall until they reach terminal velocity.
Now your pride will not let you be a man and admit you were just WRONG!!!!!!!!!!!
P.S. Everything in physical existence can be mathematically examined with a free body diagram
HILARIOUS
NEWTON even uses the momentum development.
Note the text in RED.
ONCE again you have made 30 posts in the advancment of not conceptual understanding but SEMANTICS..........
Post by: jha'dhur on April 07, 2006, 07:33:00 PM
QUOTE(puckSR @ Apr 7 2006, 05:39 PM)
Are you saying that the only objects with constant mass are objects in freefall?
No only a FOOL would utter such a notion
QUOTE(puckSR @ Apr 7 2006, 05:39 PM)
Yeah, except for planetary motion....but im sure you were aware of that...."theory of relativity"
Let me guess the sum of the forces equals m*a
Centriptial Force - Gravitational attraction = mass * acceleration
No acceleration:
Consequently:
centriptal force = gravitational attraction(of Sun)
Therefore Equilibrium (i.e. SUM of forces equal 0)
QUOTE(puckSR @ Apr 7 2006, 05:39 PM)
"terminal velocity" is a characteristic of wind resistance(friction due to air)....if a planet has no atmosphere...such as the moon...there is no such thing as terminal velocity
DUH no shit SHERLOCK.
However a ball dropped from the empire state building will have a drag coefficient based upon its geometric properties, and this along with initial velocity will predict the time to terminal velocity.
When the Fd (drag force) = Fg (gravity) therefore v= constant
DU MASS
P.S. Lordvader321646 please interject some more of your idiocracy into this conversation. PUCK is becoming boring.
Post by: puckSR on April 08, 2006, 09:02:00 PM
QUOTE
The force is proportinal to the momentum, derived from the mass and veloctiy of the plane at the moment of impact. The F = 0 only when P = 0. (i.e. m or v =0)
Post a mathmatical expression to the contrary.
Post a mathmatical expression to the contrary.
Ok...this is the actual statement of Newton's second law....
Nasa site explainng the different forms of Newton's second law
F=dP/dt....
in case you are completely unaware...the derivative of a constant is 0
if the constant is multiplied times a variable...then you derive the variable and multiple it times the constant....
i.e.
d(2*x^2)/dx=4x
d(x)/dx=1
d(4)/dx=0
Explanation of derivative of a constant
f'(x)=d(x)/dx
we use the form d/dx for physics because the denominator tells us what variable we are deriving the equation by...in the case of dP/dt we are deriving with respect to time...
There....i provided an equation....and I explained basic calculus to you...and i did it all with a fever....
Damn...this cold is killing me.....
The fact that you just said F=0 only when m=0 or v=0 clearly proves you have no understanding of Calculus...which makes Newtonian physics basically impossible to comprehend....since Newton also invented Calculus and used it for physics....
So...will you please just shut up now....me and my roommate were starting to believe that perhaps you and I were arguing over a misunderstanding(we thought you were attempting to argue against newtonian physics...or you werent allowing multiple forces to be the same as a singular force)....now it is relatively clear that you are an idiot...
Post by: jha'dhur on April 09, 2006, 02:27:00 AM
QUOTE(puckSR @ Apr 8 2006, 10:09 PM)
Ok...this is the actual statement of Newton's second law....
Nasa site explainng the different forms of Newton's second law
F=dP/dt....
in case you are completely unaware...the derivative of a constant is 0
if the constant is multiplied times a variable...then you derive the variable and multiple it times the constant....
i.e.
d(2*x^2)/dx=4x
d(x)/dx=1
d(4)/dx=0
Explanation of derivative of a constant
f'(x)=d(x)/dx
we use the form d/dx for physics because the denominator tells us what variable we are deriving the equation by...in the case of dP/dt we are deriving with respect to time...
There....i provided an equation....and I explained basic calculus to you...and i did it all with a fever....
Damn...this cold is killing me.....
The fact that you just said F=0 only when m=0 or v=0 clearly proves you have no understanding of Calculus...which makes Newtonian physics basically impossible to comprehend....since Newton also invented Calculus and used it for physics....
So...will you please just shut up now....me and my roommate were starting to believe that perhaps you and I were arguing over a misunderstanding(we thought you were attempting to argue against newtonian physics...or you werent allowing multiple forces to be the same as a singular force)....now it is relatively clear that you are an idiot...
Corky, you are wrong, incorrect, mistaken.................
P.S. Your roommate,slightly damp, and Cockvader, can eat a dick...........
Post by: jha'dhur on April 09, 2006, 08:13:00 AM
QUOTE(puckSR @ Apr 9 2006, 04:02 AM)
YES
Were you wrong when you claimed that F=ma is a simplification for falling bodies only? YES
So please...explain how I am wrong...
And lets see if you can do it without insulting everyone.....
or by actually using mathematics....
I tried to be nice...I tried to explain this all to you....but if your going to continue to be so blatantly wrong.....Im going to call BULLSHIT
Im waiting......
Fnet = ma
A ball dopped fom the top of a tall building will only experience the gravitational force and if you neglect drag, (the ball never reaches terminal velocity). The resulting free body diagrham looks like
F = ma
This is a simplification because drag does exsist, and since this describes only a small class of situations/problems it is pretty much simplistic.
In the "real" world we have "propulsion" this is were momentum derivation comes into play. This derivation is used to examine the vast majority of physical.
P.S. I asked you to present a mathematical expression that describes the time dependant velocity of an object through the air. That doesnt rely on the momentum treatment.
Not some 1st grade derivative that you still didnt present correctly.
Post by: jha'dhur on April 11, 2006, 06:11:00 PM
QUOTE(puckSR @ Apr 9 2006, 01:13 PM)
Drag only exists under "certain situations"....so your drag model is simplistic....
F=ma still applies...except you are dealing with "equilibrium"...as you so eloquently put it...
However...your forgetting that F=ma is also used to describe the forces on stationary objects...
it is also used to describe any force where mass is constant...and if mass isnt constant...you simply use the integral of the above equation....
How many times do I have to tell you this....
F=m*a=dP/dt
the derivation of momentum is force.....
the derivation of momentum is also the derivation of mass*velocity....and the derivative of velocity is acceleration
"propulsion"-a driving or propelling force....it is still a force you numbskulll
Look I dont know how else to explain this to you.....momentum is unimportant most of the time.....
momentum requires no energy input and it is not dependent on time...if a mass has momentum...it will continue to have momentum FOREVER...until an outside force acts upon it...it is unimportant because we normally care about when that momentumed mass interacts with something else...this is known as force...or change in momentum...not momentum
?????
Time dependent velocity? Do you mean acceleration?
ok
F=ma or F=d(mv)/dt or F= d(d(mx))/d^2t
What the heck is the momentum treatment? Are you just making up words again?
So did you at least figure out how this statement is false?
Or do i need to explain things again
First off, that presentation of calculus is completely correct...
Secondly, I had to use a simple explanation...your apparently too dumb to understand complex math
Mr. "derivative of a constant is not 0"
Your really not getting this.....
Go back to the baseball....
When i throw a baseball in space...the Force of the throw is equal to the mass * the acceleration
Once the ball is moving at a constant velocity...it has a constant momentum which is equal to mass times velocity...it will continue to move with this momentum infinitely....unless...
An outside force acts on the ball...when the outside force acts on the ball...the velocity will change...
this change in velocity is known as acceleration....this action on the ball is known as a force, and can be detected by the change in velocity times the mass or accleration times mass.
This is the simplest way I can explan classical physics to you...
Your pretty dense, but this site explains the relationship between force and momentum once again
You are nothing but a poser.
At least Vader had the common sense to STFU because he didnt know what he was talking about.
Post by: jha'dhur on April 11, 2006, 08:09:00 PM
QUOTE(puckSR @ Apr 4 2006, 07:43 PM)
Umm actually it is...
i.e. your rocket example....
^ lets say that your rocket is accelerating at 20 m/s^2
it has a mass of 20,000 kg....
F(net)=20*20,000=400,000 Newtons
F(g)=-9.8*20,000=-196,000 Newtons
F(rocket)=F(net)-F(g)=596,000 Newtons
So the rocket is using a force of 596,000 Newtons to rise at the acceleration of 20 m/s^2
Funny one week ago you had no idea what momentum (P) was no you are an expert on the topic.
QUOTE
The sum of the forces = 0, when the mass = 0, or v = 0.
Post by: PhatIrishBastard on April 13, 2006, 10:57:00 AM
QUOTE(puckSR @ Apr 11 2006, 08:34 PM)
Analog:
You ask us how much gasoline a 747 is using to travel from LA to New York...
We say none...
You say none, it uses jet fuel....
Well duh....but you asked us how much gasoline(force).....you didnt ask us how much jet fuel(momentum)
Another one of your pee brain analogies, that shuffles semantics. You are one CLICHE after another.
QUOTE(puckSR @ Apr 11 2006, 08:34 PM)
Do you even understand that momentum and force are two different things?
Force is "change" in momentum....
just like acceleration is "change" in velocity....
Ok honestly....if you cant understand the derivation F=d(mv)/dt...and if you cannot understand that momentum and force are different.....this conversation is over...
Yet another website explaining the relationship between force and momentum
More semantics.....
The two terms can be equated, and are in the equation you post. Force and temperature are in no way equatable. (i.e. no expression to explicitly relate the two) Newtons Second law not its common derivation defining gravity does this.
QUOTE(puckSR @ Apr 11 2006, 08:34 PM)
this conversation is over...
Yet another website explaining the relationship between force and momentum
I dont see why jha insists on arguing with a JA like yourself you posted that bogus answer because you didnt know now you are trying to steal his answer like you came up with it.
THANKS!! COCK SENIOR!!!!!!!!!
Jha'dhur I believe you lost the BET.
Post by: jha'dhur on April 13, 2006, 04:31:00 PM
QUOTE(PhatIrishBastard @ Apr 13 2006, 12:04 PM)
Another one of your pee brain analogies, that shuffles semantics. You are one CLICHE after another.
More semantics.....
The two terms can be equated, and are in the equation you post. Force and temperature are in no way equatable. (i.e. no expression to explicitly relate the two) Newtons Second law not its common derivation defining gravity does this.
I dont see why jha insists on arguing with a JA like yourself you posted that bogus answer because you didnt know now you are trying to steal his answer like you came up with it.
THANKS!! COCK SENIOR!!!!!!!!!
Jha'dhur I believe you lost the BET.
Yeah, yeah, yeah, you can read PuckSr like a book. Like that is a difficult. Since he thinks that I am you, mediax, and blueballs cant we call it even.
How about double or nothing. You have been playing pucksitter for some time now. I am new.
Besides Vader jumped in a a lot later than you said he would. I got him to change his answer a couple of times and just plain ran Vader off. That should count for something.
But, you were right. He does make ish up, and play word games as well as word meanings. I wonder is he a republican.
At least he has been introduced to the concept of momentum, all he does is keep repeating F(g)=mg
claiming that its equal to F(net) = ma. Did you read the "1000" atoms malarky. HILARIOUS
I wish I could meet him he probably would be fun to get drunk and make fun of him while he is making up one of those "DU MASS" analogies.
Post by: jha'dhur on April 13, 2006, 06:19:00 PM
QUOTE(puckSR @ Apr 13 2006, 07:02 PM)
LOL
LOL....
So...you introduced me to the concept of momentum?
LOL
Ok...I have a question for you jha'dhur...since you really love momentum....
a rocket is travelling through outerspace.(not near any planets)....it has a mass of 100,000 kg. It is travelling at a constant velocity of 100 m/s....
how much momentum does the rocket have?
How much force is being exerted on the rocket?
How much fuel is the rocket using?
There is no gravitational force acting on the rocket(nor any other forces for that matter)
It looks like we may have to take this one step at a time.....
DUDE for like the umptenth billionth time the momentum is only a constant when the rocket is on the ground.
As the rocket (i.e. the big fuel tank) which is like 80% fuel 20% cargo takes off its mass & velocity are constantly changing.
Momentum = [mf(t +1) - (Fuel Burn Rate * Time)] * v(t+1) - v(t)
Now you apply intiial(Boundary) conditions to determine simplify
Like m(t) intial mass = real mass at launch time or whenever
Fuel burn Rate (kg/s) self explanatory
v(t) = 0 at t=0 (on the ground) or actual measured velocity at t=whatever.
v(t+1) this is the GEM, because it is dependant upon Thrust, drag, and the magnitude/direction of v(t)
Thrust is a function of burn rate.
Drag is a function of (v) in outer space lets say it is 0
Weight m (t)*g(t) Mass in changing will fuel burnup.
g(t) although microgravity in space it is small but not zero and the direction and magnitude will change with the relative distance and direction of a gravity source.
So LETS Gather the sum of the forces:
+/-W(t) *(g)t + Thrust(t) - Drag(v) = mass * accelleration
The above is Newtons 2nd law not the simplification, for gravity acceleration on earth IF the rocket is accelerating.
IF the rocket is not accelerating.
+/-W(t) *(g)t + Thrust(t) - Drag(v) = d(m*v)/dt
MOMENTUM BALANCE(i.e the equation you use 99% of time)
Remeber:
The rate of change in the mass = (Fuel Burn Rate * Time)
Mass time relationship:
M = m(t+1) - m(t)
M(t+1) = mass intitial - Burn * dt
M(t) = m(t)
Change in Position
(v) = v(t+1) - v(t)
The derivation for the Thrust (t) and Drag (t) is a little more invloved but can be looked up since they are generic .
+/-W(t) *(g)t + Thrust(t) - Drag(v) = d(m*v)/dt
This equation can be integrated to predict the velocity over time.
Or you can take the derivative to determine if the craft is accelerating.
F(g) = M*g(a) will not provide the desired answer, since you already posted that BOGES answer.
P.S.
It is clearly not zero.
P.S.S.
This is my last post in this thread since your stubborness already cost me a 12 pack...
Post by: puckSR on April 13, 2006, 07:16:00 PM
QUOTE
Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.
Translation:
QUOTE
An object at rest or traveling in uniform motion will remain at rest or traveling in uniform motion unless acted upon by a net force.
Now this is important because our rocket is travelling at a constant(uniform) velocity(motion). Now according to Newton's Law(not a simplification) the velocity will remain constant as long as all forces acting on the object add up to a net force of 0.
That means that an object, once moving at a certain velocity, will continue to have the same velocity(and thus momentum). This means that your rocket wouldnt be using any fuel....if you need proof of this examine the Voyager probe in deep space....
Now...according to Newton, the rocket would require a net force of 0 to keep moving at the same velocity. If the net force was not 0 then velocity would not be constant....(if mass was not constant...i.e. the rocket was burning fuel...then the thrust from the rocket would change the "net force" acting on the rocket....and unless mass was increasing it would be increasing the velocity of the rocket).
Do you finally get it jha'dhur?
What...are you a janitor at a nuclear facility?
My God your dense....
This is basic stuff....simple stuff...
BTW...momentum conservation is the idea that momentum doesnt change....only force changes momentum.
Your a nice guy...good political ideas....
but your clearly way over your head in this discussion
Post by: jha'dhur on April 13, 2006, 07:37:00 PM
QUOTE(puckSR @ Apr 13 2006, 09:23 PM)
Now...according to Newton, the rocket would require a net force of 0 to keep moving at the same velocity. If the net force was not 0 then velocity would not be constant....(if mass was not constant...i.e. the rocket was burning fuel...then the thrust from the rocket would change the "net force" acting on the rocket....and unless mass was increasing it would be increasing the velocity of the rocket).
LMAO..........
The velocity is increasing and decreasing.
Why?
Because of negative feedback from the drag.
the plane speeds up, the madnitude of the drag increases
as a result the
plane slows down, the magnitude of the drag decreases
A Rocket that has a solid chemical rocket booster that burns, there is no first gear or third gear it is a chemical reaction akin to a moderated explosion. Simply put Thrust is on or off.
There may or may not be a throttle, but drag is the dominating effect in atmosphere.
I will concede space may be different but I dont believe drag to be absolute zero in space as a result of cosmic debris.
Post by: jha'dhur on April 13, 2006, 07:57:00 PM
QUOTE(puckSR @ Apr 13 2006, 09:50 PM)
Your missing the point....cosmic debris...microgravity...and all of that are examples of forces acting against the rocket....
That is the reality of the problem, if you wish to keep morphing the scope of the problem to weasel your way into legtimacy JUST SAY SO.
QUOTE(puckSR @ Apr 13 2006, 09:50 PM)
Imagine a space that has absolutely nothing in it.....
In this space...once the rocket is moving...it requires no force to keep it moving...
it moves at a constant velocity....
Can I also imagine that you are an assasaurus Rex, and I am Wolverine.
QUOTE(puckSR @ Apr 13 2006, 09:50 PM)
When you are trying to learn something...it is best to start with the simplest idea...and build your way up...
I have a firm grasp on the concept I have solved this problem in undergrad math simulations.
First off, gravity exist in every micrometer of the universe(as far as humans are concerned)
Secondly, the speed of griavity in vacuum (space) is changing as a result of space debris.
P.S.
Even beter, I will imagine YOU are correct and stop wasting my time with you MR. DU MASS.
I just wasted 20 minutes oy my life arguing with a elfing ROCK with LIPS.
Post by: puckSR on April 13, 2006, 11:27:00 PM
QUOTE
the speed of griavity in vacuum (space) is changing as a result of space debris.
You do realize that gravity doesnt have "speed" right?
Post by: PhatIrishBastard on April 14, 2006, 10:03:00 AM
QUOTE(puckSR @ Apr 13 2006, 09:15 PM)
remember F(g)=mg...so....g=F(g)*m1....this will help you determine the change in velocity of the rocket due to gravity.
How does you man say it ...
BOGUS
F(g)=mg
NO......... NO....... NO.........
You use the gravitational force equation. with the UNIVERSAL gravity constant..
I must say you are consistent.
A consistent MORON.............
Post by: jha'dhur on April 14, 2006, 01:37:00 PM
QUOTE(PhatIrishBastard @ Apr 14 2006, 12:10 PM)
How does you man say it ...
BOGUS
F(g)=mg
NO......... NO....... NO.........
You use the gravitational force equation. with the UNIVERSAL gravity constant..
I must say you are consistent.
A consistent MORON.............
Puck what dont you understand about the above.
Your last post is BOGUS regarding a few of the finer points.
F(g) = MA(g) is only valid within the gravity of a heavy body.
This equation (basic 8th grade physics) says the only force acting on the object is gravity.
I.E. it is in FREEFALL, neglecting gravity. (Which isnt very practical)
Because when Newton came up with this discovery from dropping balls and measuring the time from inital hegiht to final height. Everyone that partook in this experiment recorded a different time.
Why, because it is an oversimplification (i.e. useless) Sometimes valid in outer space, but a sidenote in the real world.
YOU DONT GET IT AND NEVER WILL........
Post by: jha'dhur on April 14, 2006, 05:57:00 PM
QUOTE(puckSR @ Apr 14 2006, 06:41 PM)
This is my last attempt.....
THANK YOU..............
Post by: puckSR on April 14, 2006, 06:25:00 PM
jha'dhur are you familiar with a force diagram?
Hmmm...let me show you one...
from this website
This force diagram is modeling an object sitting on a surface
Notice the arrows.....
The labels are important
The arrow labeled F(grav) is the force due to gravity
The arrow labeled F(norm) is the force resisting gravity
The arrow labeled F(frict) is the force due to friction
The arrow labeled F(app) is the force that is acting on the object(pulling on it)
Notice that the direction of the force is important....
Now...
If the object is being pulled at a certain speed...i.e. 1m/s
then that means that F(frict)=F(app)....
If F(frict) != F(app) then the object would be accelerating
This acceleration could be detemined by finding F(net,horiz)=F(app)-F(frict)....
F(net, horiz)=ma...so if we knew the mass of the object we could determine how quickly it would be accelerating.
Post by: CattyKid on April 14, 2006, 11:19:00 PM
Puck, I second, or third, or fourth, or whatever what you are saying.
Post by: jha'dhur on April 15, 2006, 08:57:00 AM
QUOTE(PhysicsProf @ Apr 14 2006, 08:16 PM)
jha'dhur, as you can tell by my screenname, I am a physics professor. And, you are an idiot.
Post by: Chook on April 15, 2006, 04:26:00 PM
QUOTE(puckSR @ Apr 5 2006, 11:28 PM)
right so......
F= dp/dt
= d(mv)/dt
so....you were wrong...glad we cleared that up
Post by: jha'dhur on April 15, 2006, 05:32:00 PM
QUOTE(Chook @ Apr 15 2006, 06:33 PM)
The reason Nasa uses those equations is that it can be simplified becuase of the mass lost directly from the object. Most objects of constant mass can use the F=ma equations. Also yeah hear, get some webspace and stop hotlinking pictures. Also where do you work at?
The reason individuals concerned with physics use this equation is:
#1) What object/vehicle/animal/anything that you know of on gods green earth or the heavens has a constant acceleration or an perpetual increasing equation.
(i.e. the 2nd law is always valid)
The reason I even introduced the whole rocket thing because when I first approached the problem as a freshman I went straight to Newtons 2nd law, which is wrong.
IT WAS A SET UP, I dont have an ego to be served, and I usuallly dont ask questions for which I dont know answer.
#2) It is the only what that you can mathematically resolve the forces on a body(with a = 0)
*****************************************************************************
F(g) = mg
Says the only force acting on the object is gravity. (i.e. freefall)
g = 9.8 m/sec^2 translates to an object in freefall on earth, or the ball drop equation.
QUOTE(Chook @ Apr 15 2006, 06:33 PM)
Also yeah hear, get some webspace and stop hotlinking pictures. Also where do you work at?
When you cease to be a moron, PUCK/CHOOK/NUTTY PROFFESSOR
This is my last post to you cack drips
Post by: PhatIrishBastard on April 17, 2006, 10:51:00 AM
Is not the same expression as
F = ma(g) a derivation of Newtons 2nd Law that describes a specific class of problems.
F = mg states the only force acting on this object is gravity.
What it boils down to is you are playing you semantics game once again.
It is clear that an object with a constant velocity (i.e. a = 0) will exert a force on another object it strikes. You and your boyfriend (Vader) can yell F = 0, until you are blue in the face but I bet neither one of you would place your dense craniums in front of a vehicle moving towards you to test that theory.
Thanks CORKY, you have provided all of us with some great laughs.
Post by: lordvader129 on April 17, 2006, 12:00:00 PM
QUOTE
It is clear that an object with a constant velocity (i.e. a = 0) will exert a force on another object it strikes.
its also clear that A will no longer = 0 (for either object) if a collision occurs, the point is jha'dur never asked for the force of a collision, just net force for an object with constant velocity......0
Post by: jha'dhur on April 17, 2006, 08:40:00 PM
QUOTE(puckSR @ Apr 17 2006, 08:41 PM)
A lot of kids read this forum...teenagers in high school taking physics...
I dont want them to be "challenged" because they read that F=ma is not accurate...
or that momentum is the same as force
or that the derivative of a constant isnt 0
or that F(g)=mg models motion...which it doesnt dumbass
LMAO
Post by: jha'dhur on April 17, 2006, 09:04:00 PM
QUOTE(puckSR @ Apr 17 2006, 11:00 PM)
Awwwww....jha'dhur found another picture
Post by: jha'dhur on April 18, 2006, 06:35:00 PM
QUOTE(puckSR @ Apr 18 2006, 04:18 AM)
a picture is worth a thousand words....
and when your lack of knowledge leaves you speechless....
I guess you think you can make up for it with a picture....
LOL....
So tell us again how it requires constant force to travel at constant velocity...
LOL....
Wait, wait, wait...i know....drag....LMAO.....
What was your major in college...business?
It sure as hell wasnt nuclear physics.....LOL
or anything that required an intricate understanding of mathematics...
Of course I might be wrong....you may just have a problem realizing that laws are based on simple situations...
the complex situations involve several laws acting at once....
that whole idea may just be beyond you...
Post by: Chook on April 21, 2006, 12:41:00 AM
QUOTE(jha'dhur @ Apr 15 2006, 06:39 PM)
This is my last post to you cack drips
You are a real idiot, and what do you have against Texans? Puck still thinks that there is hope for you, but i think that you are a real nutjob and have no hope of a descent future. None of these examples on "god's green earth" follow any of your examples also. These are all simplifications because we do not include minor contributers to the system to avoid confusing people like you, but I am sorry for we failed.
Post by: PhatIrishBastard on April 21, 2006, 03:29:00 PM
QUOTE(Chook @ Apr 21 2006, 12:48 AM)
You are a real idiot, and what do you have against Texans? Puck still thinks that there is hope for you, but i think that you are a real nutjob and have no hope of a descent future. None of these examples on "god's green earth" follow any of your examples also. These are all simplifications because we do not include minor contributers to the system to avoid confusing people like you, but I am sorry for we failed.
Actually I am the one that doesnt care for you hillbillies.
Post by: throwingks on May 06, 2006, 07:13:00 AM
QUOTE(jha'dhur @ May 5 2006, 10:06 PM)
Question:
If I throw a baseball from center fiels to 2nd bases, and it hits you in the jaw, Will the ball have enough force to break your jaw.
The possible answers to that question illustrate why you should not take COCK to seriously.
I think that you would agree the ball would strike you with some force, although it is not accelerating or traveling at a constant velocity.
There are many individuals that are masters of their own opinions and many reside here, which is probably one of the reasons I dont visit as often.
Good Luck
The common sense answer is yes. There is a force exerted on your face. The confusion lies in the original question. There is no longer force needed to keep the ball in motion as it is travelling through the air. As soon as it hits your face the force used to start the motion of the ball is then transferred to your face and it is no longer with the ball. That is why the ball stops moving.
If I throw a baseball from center fiels to 2nd bases, and it hits you in the jaw, Will the ball have enough force to break your jaw.
The possible answers to that question illustrate why you should not take COCK to seriously.
I think that you would agree the ball would strike you with some force, although it is not accelerating or traveling at a constant velocity.
There are many individuals that are masters of their own opinions and many reside here, which is probably one of the reasons I dont visit as often.
Good Luck
Post by: jha'dhur on May 06, 2006, 07:55:00 AM
QUOTE(throwingks @ May 6 2006, 09:20 AM)
The common sense answer is yes. There is a force exerted on your face. The confusion lies in the original question. There is no longer force needed to keep the ball in motion as it is travelling through the air. As soon as it hits your face the force used to start the motion of the ball is then transferred to your face and it is no longer with the ball. That is why the ball stops moving.
COMMON sense, WTF is that, it is not allowed here..
LMAO
Post by: puckSR on May 06, 2006, 08:06:00 AM
You remember your entire comment about drag......
drag is a force exerted on the ball as it travels through the air right?
Ok....so does the ball slow down, stay the same speed, or go faster....
It slows down as it travels through the air.....
why?
Because there is a force due to drag acting on the ball...
however...there is no force "pushing" the ball......so
the ball slowly "accelerates in reverse" or decelerates.....
When it hits your face...it stops......and your face bears the brunt of that force...
Lets pretend you were in outerspace...when the ball hit your face...you would actually start moving....and you wouldnt be able to stop yourself....
Show me an equation that calculates the force a ball travelling through the air "has".....
You can calculate how much force the ball exerts on your face...but as everyone has pointed out to you...the ball is changing speed....(acceleration)....
Find an equation that tells us how much force the ball has without hitting something.....
go ahead....ill give you the rest of your natural life.....
the ball has momentum...but momentum isnt force......
You can also calculate the drag force and gravitational force on the ball.....
but those are trivial forces...not nearly enough to "break your jaw"....so where does that force come from?
The ball must "have force".????
The only thing that "has force" is a damn Jedi....
Everything and everyone else exerts force.....or applies force....or resists force....
Post by: jha'dhur on May 06, 2006, 08:55:00 AM
QUOTE(puckSR @ May 6 2006, 10:13 AM)
Find an equation that tells us how much force the ball has without hitting something.....
go ahead....ill give you the rest of your natural life.....
Newtons 2nd law
m dv/dt = F(m){center fielder arm strength} - F(d){Drag Force} - F(g){Mass * gravity}
Post by: jha'dhur on May 06, 2006, 03:52:00 PM
QUOTE(puckSR @ May 6 2006, 03:03 PM)
Hmmm F(m)?
Center Fielders arm strength?
Do you mean to say the force at which the center fielder throws the ball?
If we really want to argue semantics...I will allow that Drag and Gravity are acting against the center fielder's throw....
However....once the ball leaves the center fielder's hand(remember my condition about hitting or touching anything)...the center fielder quits putting force on the ball....once it leaves his hand...it is travelling at its maximum velocity...
Why did the center fielder exert force on the ball?
Because he was accelerating the ball....
Ok....so lets break this down
step by step
Step 1:
Center fielder picks up ball and throws it...
The center fielder exerts a force on the ball...thats why the constant mass ball accelerates.
The center fielder's body is subject to drag, gravity, and other frictions......
Step 2:
Ball is travelling through the air...
It is decelerating because of the force due to drag on it as it travels
No force, however, is "pushing" the ball....
Step 3:
Ball collides with other player....
They player exerts a force against the ball to cause it to stop....
The ball exerts an equal force against the player when it is stopping
The force exerted by both is equal to mass time the deceleration.
This is why the force is stronger if the player "stabs" forward at the ball and weaker if he lets his glove swing back while he catches the ball.....
While the ball travels through the air during step 2....
It has momentum "pushing" it forward....but no actual force "pushing" it...
Momentum...in all honesty is not "pushing" the ball...
The ball is simply moving instead of not moving......
Newton's 1st Law....an object in motion tends to stay in motion....
The problem you seem to have jha'dhur is this:
The ball has force upon it that initially moves it
The ball exerts force when it strikes the player at the other end of the field
Why doesn't the ball have force while it is travelling through the air?(This is your argument correct?)
The basic misunderstanding seems to be your definition of force....
Force describes an interaction....it isnt an inherent property.
The ball can have velocity, momentum, energy, etc.....because those are all properties of the ball
The ball doesnt have force....force is simply a description of its interaction with other objects
Force due to drag is interaction with the air
Force due to gravity is interaction with another mass's gravitational field
Force due to the "outfielder" is an interaction between his arm and the ball...
I seriously hope you understand this time....Im honestly beginning to believe that your either too stupid or too stubborn to understand....either that or your just joking around....
You wrote all this bullshit, and in the end doesnt prove a thing , or even make a coherent point.
The force imparted on the ball by the player can be measured "intially" by the mass of ball and its acceleration of the ball at the release from thrower.
F(arm) = m(ball) a(thrower)
You arent a ENGINEER, ohh let me guess you must attend some BOGUS state school down in Lubbock taught by immigrants.
Explains alot.
Post by: jha'dhur on May 07, 2006, 08:59:00 AM
QUOTE(puckSR @ May 6 2006, 09:18 PM)
The ball doesnt "have" force...
it will exert force once it reaches the other player....
That is amusing I must say, periodic force HUH? Has some initial force, no intemediate force, yet has some final impact force. Intersting theory [cough] (Bullshit) [/cough]
Could you prove that mathematically, So that we all may be blessed by your superior intellect.
LMAO!!!!!!!!!!!!!!!!!!!!!!
Post by: _iffy on May 07, 2006, 11:13:00 AM
Post by: jha'dhur on May 07, 2006, 11:28:00 AM
QUOTE(_iffy @ May 7 2006, 01:20 PM)
Jha... your not confusing force with inertia are you? Inertia acts like a force, but it isn't real.
BTW - both the ball and person have an FN
Inertia is the force that causes you to continue to move forward, when you turn the wheel of your car hard left.
That is "very" real.
QUOTE
No....the ball doesnt "have" force.....
Force is an action.....
the ball has energy
Force is an action.....
the ball has energy
Dumbass
Energy is a force * displacemnt
Hence (ft-lbf) or N - m
What a MA ROON!!!!
Post by: _iffy on May 07, 2006, 11:30:00 AM
QUOTE(jha'dhur @ May 7 2006, 01:35 PM)
Inertia is the force that causes you to continue to move forward, when you turn the wheel of your car hard left.
That is "very" real.
That is "very" real.
That's momentum. Inertia is different. It's not real
Post by: lordvader129 on May 07, 2006, 11:38:00 AM
QUOTE
2. An 2-kg object is moving horizontally with a speed of 4 m/s. How much net force is required to keep the object moving at this speed and in this direction?
QUOTE
Answer: 0 N
An object in motion will maintain its state of motion. The presence of an unbalanced force changes the velocity of the object.
An object in motion will maintain its state of motion. The presence of an unbalanced force changes the velocity of the object.
Post by: jha'dhur on May 07, 2006, 11:41:00 AM
QUOTE(_iffy @ May 7 2006, 01:37 PM)
That's momentum. Inertia is different. It's not real
QUOTE
inertia (n.) The tendency of a body to resist acceleration; the tendency of a body at rest to remain at rest or of a body in straight line motion to stay in motion in a straight line unless acted on by an outside force.
QUOTE
momentum (n.) A measure of the motion of a body equal to the product of its mass and velocity. Also called linear momentum
So, I guess if I was to post the mathematical expression for inertia then what would you say?
Iffy, you are fun to chat with but you have very little credebility when it comes to real world facts.
1) Accelerator Gold Bullsh*t
2) Hiroshima Nagaski
3) Inertia Isnt real
4) Counting atoms
P.S. I did some research it is possible to transmutate platinum atoms to stable gold on a very, very ,very small (atom per atom) scale but nowhere near even a gram amount, plus this is only possible in nuclear reactor not accelerator due to precursors need born from fission events.
QUOTE(lordvader129 @ May 7 2006, 01:45 PM)
Given, the fact you helped me out yesterday and I declared a truce as result I will not point out the flaw in your logic.
Post by: _iffy on May 07, 2006, 11:58:00 AM
QUOTE
inertia (n.) The tendency of a body to resist acceleration; the tendency of a body at rest to remain at rest or of a body in straight line motion to stay in motion in a straight line unless acted on by an outside force.
It's not real. It's not a force. It's a way to describe acts of motion.QUOTE(jha'dhur @ May 7 2006, 01:48 PM)
So, I guess if I was to post the mathematical expression for inertia then what would you say?
QUOTE(jha'dhur @ May 7 2006, 01:48 PM)
Iffy, you are fun to chat with but you have very little credebility when it comes to real world facts.
1) Accelerator Gold Bullsh*t
2) Hiroshima Nagaski
3) Inertia Isnt real
4) Counting atoms
1) i gave you a link from a credible source that proved this
2)wasn't me
3)it isn't a force. It is a way to decribe whats happening. It isn't real.
4)wasn't me. (unless your refering to a particle accelerator adding individual electrons, in which case - proven already)
QUOTE(jha'dhur @ May 7 2006, 01:48 PM)
P.S. I did some research it is possible to transmutate platinum atoms to stable gold on a very, very ,very small (atom per atom) scale but nowhere near even a gram amount, plus this is only possible in nuclear reactor not accelerator due to precursors need born from fission events.
QUOTE(jha'dhur @ May 7 2006, 01:48 PM)
Iffy, you are fun to chat with
Post by: jha'dhur on May 07, 2006, 12:20:00 PM
QUOTE(_iffy @ May 7 2006, 02:05 PM)
It's not real. It's not a force. It's a way to describe acts of motion.
I would say the inertia isn't a force.
1) i gave you a link from a credible source that proved this
A "credible" non-technical website with such crap on it as loch ness and bigfoot. If you were to scroll down in the BOGUS article you will see what everyone on that forum had to say of the "spammer" that wrote your "credible" source
QUOTE(_iffy @ May 7 2006, 02:05 PM)
3)it isn't a force. It is a way to decribe whats happening. It isn't real.
According to you, right.
QUOTE(_iffy @ May 7 2006, 02:05 PM)
Cooooool. Doesn't say you can't make gold in a particle accelerator though.
The particular isotope of platinum(that can be transmutated to gold) is only created from fission(i.e nuclear reactor)
Inertia at work.
Looks pretty real to me, ask Princess Diana if inertia is a "force"
Post by: _iffy on May 07, 2006, 08:18:00 PM
QUOTE(jha'dhur @ May 7 2006, 02:27 PM)
A "credible" non-technical website with such crap on it as loch ness and bigfoot. If you were to scroll down in the BOGUS article you will see what everyone on that forum had to say of the "spammer" that wrote your "credible" source
Post by: Chook on May 08, 2006, 01:16:00 AM
Post by: PhatIrishBastard on May 10, 2006, 03:19:00 PM
QUOTE(Chook @ May 8 2006, 01:23 AM)
You need to give up this arguement with some idiot non-com guy. If he is just doing "research" ie google, and not understanding or willing to take the fact that he may be wrong. You won't get to him, just give it up for now and do things more important.
You and your imaginary friend are amusing. I didnt know that had PC's at the group home. Isnt that against the RENT a Center terms.
QUOTE(puckSR @ May 7 2006, 02:42 PM)
Of course...dont mention the voyager probe to him.....
the Voyager probe has been moving for 20 years without any propellant force....and is continuing to move rather steadily in deep space
This is really sad jha'dhur....
if you would like to know why I think your xmedia?
Xmedia did the same thing...he argued physics with me for a month.....
Dummy, Voyager is in SPACE.
Microgravity, microdrag, same force balances valid just different magnitude of contributors.
You sure are FUCKING obssesed with Xmedia........
But Puck, it does read like your xmedia ramblings, just repeating same old bolgna over and over.
YOU ARE MISTAKEN
Post by: puckSR on May 10, 2006, 06:48:00 PM
QUOTE
YOU ARE MISTAKEN
For the record....what am i mistaken about?
constant velocity constant masses have no force?
F=ma is used for more than modeling free-fall?
the derivative of a constant is 0?
Please....what am I wrong about???
Voyager has had forces acting upon it during its travel....the point of mentioning voyager is to explain that micro-gravity is not equivalent to drag. Gravity can serve both to accelerate and decelerate an object....depending on the relative position to the object.
I am trying to make the point to jha'dhur about force....
Force accelerates objects....force doesnt "keep objects moving" unless a great deal of force is acting on the object in resistance.
You already mentioned "free body diagrams" so I know you know that F=ma is used for more than modeling freefall. You could contribute to the conversation and tell jha'dhur this fact....but instead your arguing with me and you dont even know what we are arguing about.
Im arguing with jha'dhur on very specific points...
He has shown in inadequate knowledge of mathematics and physics...
You keep trying to point out small problems with my analogies...which i wouldnt mind if I wasnt arguing with a certified idiot on the other side.