Newton's law of gravitational attraction
in [Physics]
|
Prev: New Volcanic Activity This Week- Three~~ Total 16 active
Next: Solutions manual to Intermediate Accounting 13e Kieso
From: doug on 19 Oct 2009 17:00 NoEinstein wrote: > On Oct 16, 2:12 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: Kinetic Energy is identical to > MOMENUTM. Except, of course, that this is completely wrong. If they were the same, they would have the same name. And the same units. And the same formulas. So john is zero for three there. It's how hard an object of known weight will impact due to > having some specific velocity. It turns out that KE will increase ONE > WEIGHT UNIT of FORCE for every 32.174 ft. per second increase in > velocity. In the case of falling objects, that velocity increase > occurs every second of the fall. This is just word nonsense and does not rise to the level of wrong. > > Work�like we both agree�is force x distance. Move something twice as > far against some resistance, and in the desired direction, and you do > twice as much WORK. But the distance of fall of falling objects isn't > against any resistance! That is pretty funny. You quote the formula but then demonstrate complete ignorance of what it means. So, no work can be calculated. Wrong again, john. What is > known, however, is that the WEIGHT of the object is uniformly acting > DOWN, for however long the object falls. > > *** The majority of the distance of fall is due to the accruing > coasting velocity each second. This is also pretty funny and has nothing to do with science. What that is like is taking your foot > off of the accelerator when you reach 32.174 ft. / sec. During the > next second (ignoring road friction and wind resistance) the car will > COAST 32.174 feet. The latter distance doesn't increase KE, because > the velocity hasn't changed! If you would actually take the following > Pop Quiz for Science Buffs, It is good for a laugh. you might better understand that distance > of travel doesn't increase KE, nor momentum (the same thing), Wrong again john. in any > way. The equation for momentum is mass x velocity (in 'g' > multiples). 'D' is nowhere to be seen in such. John is trying to increase the density of stupid statements in his posts. � NoEinstein � > > Pop Quiz for Science Buffs! > http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... > > Dropping Einstein Like a Stone > http://groups.google.com/group/sci.physics/browse_thread/thread/989e16c59967db2b?hl=en# > >>On Oct 16, 12:03 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: >> >> >>>On Oct 13, 1:39 pm, jbriggs444 <jbriggs...(a)gmail.com> wrote: >> >>>Dear jbriggs444: Your definition of work is like a time-motion >>>study. My definition is AFTER the work has been done, and its >>>quantity gets evaluated. My formal definition of work is: W = FD x >>>cos. of angle of deviation from the desired direction of motion. >> >>Good. Now, let's look at YOUR definition. >>Notice that term D. >>What do you think D stands for? >> >>It is the DISTANCE the object moves while the force is acting on the >>object. >> >>Now go back and look at what I told you a few days ago about why the >>work increases each second as an object falls, even as the force >>remains the same. You see? You AGREE with me, by invoking that formula >>above. >> >> >> >> >>> No >>>motion in desired direction would = zero work done. Effective KE is >>>AFTER all of the forces have been added or subtracted according to >>>their vectors. The resultant velocity times the mass that is in >>>motion yields the KE (in weight multiples, like pounds). � NoEinstein >>>� >> >>>>On Oct 13, 12:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: >> >>>>>On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: >> >>>>>PD: You are off-your-rocker. Go into a small room and talk to >>>>>yourself. No one with half a brain reads your replies. I only scan >>>>>down to your first foot-in-the-mouth nonsense. � NE � >> >>>>So do I. Usually that results in reading clear to the bottom of a PD >>>>post. >> >>>>>>On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: >> >>>>>>>On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: >> >>>>>>>Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as >>>>>>>KE! >> >>>>>>The contribution to the energy that changes KE is called work. Please >>>>>>consult your 7th grade science book. >> >>>>This one gave me pause. It's too causal for my taste. It implies >>>>that if two opposite forces act on a uniformly moving object that >>>>neither of them do any work since neither of them change KE. >> >>>>One can argue that this definition isn't quite right for non-rigid >>>>bodies and that it requires clarification for rotating bodies. >> >>>>Still, it's not a bad starting point. A simplification suitable for >>>>pedagogical purposes. Especially when the discussion doesn't involve >>>>rigidity or rotation. And if you read it just right, it is equivalent >>>>to what I consider to be a good definition. >> >>>>Me, I define work as the vector dot product of instantaneous point >>>>force times the incremental motion of the target object at the contact >>>>point integrated over the life of the force. If it's not a point >>>>force, you can integrate over the contact area as well. >> >>>>I've had discussions previously with folks who think that you're >>>>supposed to do a path integral of force times incremental movement of >>>>the force along a path. That's wrong. It's force times incremental >>>>movement of the object to which the force is applied. For point >>>>objects it's the same thing. For extended objects it's not.- Hide quoted text - >> >>- Show quoted text - > >
From: doug on 19 Oct 2009 17:02 NoEinstein wrote: > On Oct 16, 3:40 pm, Jonah Thomas <jethom...(a)gmail.com> wrote: > >>NoEinstein <noeinst...(a)bellsouth.net> wrote: >> >>>Jonah Thomas <jethom...(a)gmail.com> wrote: >>> >>>>NoEinstein <noeinst...(a)bellsouth.net> wrote: >>>> >>>>>Dear Dougie Boy, the leech: Not a soul in the world cares about >>>>>having your... 'respect'. Mentally, you are a NON entity. � NE � >> >>>>It's an unmoderated newsgroup and you have the right to post as you >>>>see fit. >> >>>>But that was unkind. >> >>>Dear Jonah: Though I am a nice person (to nice people), Dougie Boy, >>>the leech, deserves only the disdain of the readers for his psychotic >>>negativity in the face of new science truths. People like him are the >>>black spots on these groups. >> >>He deserves your pity. But if he threatens your self-esteem then it's >>understandable you would not feel merciful. > > > Dougie Boy threatens only the time I spend keeping him in his place. No, you are here for us to laugh at. And we certainly do laugh at you. You are good to show students what happens when you refuse to study and trip over your ego. You also show the dangers of blind hatred and jealousy in your feelings towards Einstein. > � NE �
From: PD on 19 Oct 2009 16:33 On Oct 19, 2:43 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 16, 2:12 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: Kinetic Energy is identical to > MOMENUTM. No sir. Look in a 7th grade science book. I'm astonished that you, an architect, would make so foolish a mistake. > It's how hard an object of known weight will impact due to > having some specific velocity. Sorry, but no. NEITHER kinetic energy nor momentum is defined this way. Never has been. It's not polite to use someone else's word and just make up a new definition for it. If you need a definition of what momentum is, or what kinetic energy is, I could point you to a 7th grade science book. > It turns out that KE will increase ONE > WEIGHT UNIT of FORCE for every 32.174 ft. per second increase in > velocity. In the case of falling objects, that velocity increase > occurs every second of the fall. > > Worklike we both agreeis force x distance. Move something twice as > far against some resistance No resistance is needed, nor is any resistance included in the definition. Consult your 7th grade science book. Work is force x distance, period. > and in the desired direction, and you do > twice as much WORK. But the distance of fall of falling objects isn't > against any resistance! So, no work can be calculated. What is > known, however, is that the WEIGHT of the object is uniformly acting > DOWN, for however long the object falls. > > *** The majority of the distance of fall is due to the accruing > coasting velocity each second. What that is like is taking your foot > off of the accelerator when you reach 32.174 ft. / sec. During the > next second (ignoring road friction and wind resistance) the car will > COAST 32.174 feet. The latter distance doesn't increase KE, because > the velocity hasn't changed! If you would actually take the following > Pop Quiz for Science Buffs, you might better understand that distance > of travel doesn't increase KE, nor momentum (the same thing), in any > way. The equation for momentum is mass x velocity (in 'g' > multiples). 'D' is nowhere to be seen in such. NoEinstein > > Pop Quiz for Science Buffs!http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... > > Dropping Einstein Like a Stonehttp://groups.google.com/group/sci.physics/browse_thread/thread/989e1... > > > > > On Oct 16, 12:03 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On Oct 13, 1:39 pm, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > > > Dear jbriggs444: Your definition of work is like a time-motion > > > study. My definition is AFTER the work has been done, and its > > > quantity gets evaluated. My formal definition of work is: W = FD x > > > cos. of angle of deviation from the desired direction of motion. > > > Good. Now, let's look at YOUR definition. > > Notice that term D. > > What do you think D stands for? > > > It is the DISTANCE the object moves while the force is acting on the > > object. > > > Now go back and look at what I told you a few days ago about why the > > work increases each second as an object falls, even as the force > > remains the same. You see? You AGREE with me, by invoking that formula > > above. > > > > No > > > motion in desired direction would = zero work done. Effective KE is > > > AFTER all of the forces have been added or subtracted according to > > > their vectors. The resultant velocity times the mass that is in > > > motion yields the KE (in weight multiples, like pounds). NoEinstein > > > > > > > > On Oct 13, 12:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > PD: You are off-your-rocker. Go into a small room and talk to > > > > > yourself. No one with half a brain reads your replies. I only scan > > > > > down to your first foot-in-the-mouth nonsense. NE > > > > > So do I. Usually that results in reading clear to the bottom of a PD > > > > post. > > > > > > > On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as > > > > > > > KE! > > > > > > > The contribution to the energy that changes KE is called work. Please > > > > > > consult your 7th grade science book. > > > > > This one gave me pause. It's too causal for my taste. It implies > > > > that if two opposite forces act on a uniformly moving object that > > > > neither of them do any work since neither of them change KE. > > > > > One can argue that this definition isn't quite right for non-rigid > > > > bodies and that it requires clarification for rotating bodies. > > > > > Still, it's not a bad starting point. A simplification suitable for > > > > pedagogical purposes. Especially when the discussion doesn't involve > > > > rigidity or rotation. And if you read it just right, it is equivalent > > > > to what I consider to be a good definition. > > > > > Me, I define work as the vector dot product of instantaneous point > > > > force times the incremental motion of the target object at the contact > > > > point integrated over the life of the force. If it's not a point > > > > force, you can integrate over the contact area as well. > > > > > I've had discussions previously with folks who think that you're > > > > supposed to do a path integral of force times incremental movement of > > > > the force along a path. That's wrong. It's force times incremental > > > > movement of the object to which the force is applied. For point > > > > objects it's the same thing. For extended objects it's not.- Hide quoted text - > > > - Show quoted text -
From: PD on 19 Oct 2009 16:34 On Oct 19, 2:48 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 16, 2:15 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: If your activity on these groups bore > any resemblance to science, then, where are your '+new posts'? Two comments. Activity in science has zero correlation with new posts on an unmoderated newsgroup. Zero. Secondly, I've already pointed out to you that I have quite a number of new posts, and I can't be blamed if you don't know how to use your newsgroup reader properly to find them. > Other > than that one plagiarized post of yours on high energy particle > physics, your "position" on science has never been specified. > NoEinstein > > > > > On Oct 16, 12:06 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On Oct 13, 1:50 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 13, 12:39 pm, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > > > > > On Oct 13, 12:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > PD: You are off-your-rocker. Go into a small room and talk to > > > > > > yourself. No one with half a brain reads your replies. I only scan > > > > > > down to your first foot-in-the-mouth nonsense. NE > > > > > > So do I. Usually that results in reading clear to the bottom of a PD > > > > > post. > > > > > > > > On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as > > > > > > > > KE! > > > > > > > > The contribution to the energy that changes KE is called work.. Please > > > > > > > consult your 7th grade science book. > > > > > > This one gave me pause. It's too causal for my taste. It implies > > > > > that if two opposite forces act on a uniformly moving object that > > > > > neither of them do any work since neither of them change KE. > > > > > Fair point. Net work is what is usually referenced in the Work-Energy > > > > theorem. > > > > > > One can argue that this definition isn't quite right for non-rigid > > > > > bodies and that it requires clarification for rotating bodies. > > > > > Well, as for the latter, I didn't specify *linear* KE, but again the > > > > point is taken. > > > > > > Still, it's not a bad starting point. A simplification suitable for > > > > > pedagogical purposes. Especially when the discussion doesn't involve > > > > > rigidity or rotation. And if you read it just right, it is equivalent > > > > > to what I consider to be a good definition. > > > > > > Me, I define work as the vector dot product of instantaneous point > > > > > force times the incremental motion of the target object at the contact > > > > > point integrated over the life of the force. If it's not a point > > > > > force, you can integrate over the contact area as well. > > > > > Agreed, but NoEinstein won't be able to look that up in his 7th grade > > > > science book. > > > > > > I've had discussions previously with folks who think that you're > > > > > supposed to do a path integral of force times incremental movement of > > > > > the force along a path. That's wrong. It's force times incremental > > > > > movement of the object to which the force is applied. For point > > > > > objects it's the same thing. For extended objects it's not. > > > > > For *linear* KE, I see the distinction.- Hide quoted text - > > > > > - Show quoted text - > > > > Science is what I figured out over a lifetime, not something I needed > > > to 'look up' anywhere... like you do. NE > > > Perhaps you should use a different word, other than science. Because > > your activity bears no resemblance to science and does not meet the > > criteria that is used by scientists do define that word. > > > Just because you decide to putter with pipes and water in your > > backyard does not give you the right to claim what you learn from that > > hobby to be expertise in plumbing. > > > PD- Hide quoted text - > > > - Show quoted text -
From: NoEinstein on 20 Oct 2009 11:53
On Oct 19, 5:00 pm, doug <x...(a)xx.com> wrote: > NoEinstein wrote: > > On Oct 16, 2:12 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD, the Parasite Dunce: Kinetic Energy is identical to > > MOMENUTM. > > Except, of course, that this is completely wrong. If they > were the same, they would have the same name. And the same > units. And the same formulas. So john is zero for three there. > > It's how hard an object of known weight will impact due to > > > having some specific velocity. It turns out that KE will increase ONE > > WEIGHT UNIT of FORCE for every 32.174 ft. per second increase in > > velocity. In the case of falling objects, that velocity increase > > occurs every second of the fall. > > This is just word nonsense and does not rise to the level of wrong. > > > > > Worklike we both agreeis force x distance. Move something twice as > > far against some resistance, and in the desired direction, and you do > > twice as much WORK. But the distance of fall of falling objects isn't > > against any resistance! > > That is pretty funny. You quote the formula but then demonstrate > complete ignorance of what it means. > > So, no work can be calculated. > > Wrong again, john. > > What is > > > known, however, is that the WEIGHT of the object is uniformly acting > > DOWN, for however long the object falls. > > > *** The majority of the distance of fall is due to the accruing > > coasting velocity each second. > > This is also pretty funny and has nothing to do with science. > > What that is like is taking your foot > > > off of the accelerator when you reach 32.174 ft. / sec. During the > > next second (ignoring road friction and wind resistance) the car will > > COAST 32.174 feet. The latter distance doesn't increase KE, because > > the velocity hasn't changed! If you would actually take the following > > Pop Quiz for Science Buffs, > > It is good for a laugh. > > you might better understand that distance > > > of travel doesn't increase KE, nor momentum (the same thing), > > Wrong again john. > > in any > > > way. The equation for momentum is mass x velocity (in 'g' > > multiples). 'D' is nowhere to be seen in such. > > John is trying to increase the density of stupid statements > in his posts. > > NoEinstein > > > > > > > Pop Quiz for Science Buffs! > >http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... > > > Dropping Einstein Like a Stone > >http://groups.google.com/group/sci.physics/browse_thread/thread/989e1... > > >>On Oct 16, 12:03 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > >>>On Oct 13, 1:39 pm, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > >>>Dear jbriggs444: Your definition of work is like a time-motion > >>>study. My definition is AFTER the work has been done, and its > >>>quantity gets evaluated. My formal definition of work is: W = FD x > >>>cos. of angle of deviation from the desired direction of motion. > > >>Good. Now, let's look at YOUR definition. > >>Notice that term D. > >>What do you think D stands for? > > >>It is the DISTANCE the object moves while the force is acting on the > >>object. > > >>Now go back and look at what I told you a few days ago about why the > >>work increases each second as an object falls, even as the force > >>remains the same. You see? You AGREE with me, by invoking that formula > >>above. > > >>> No > >>>motion in desired direction would = zero work done. Effective KE is > >>>AFTER all of the forces have been added or subtracted according to > >>>their vectors. The resultant velocity times the mass that is in > >>>motion yields the KE (in weight multiples, like pounds). NoEinstein > >>> > > >>>>On Oct 13, 12:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > >>>>>On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > >>>>>PD: You are off-your-rocker. Go into a small room and talk to > >>>>>yourself. No one with half a brain reads your replies. I only scan > >>>>>down to your first foot-in-the-mouth nonsense. NE > > >>>>So do I. Usually that results in reading clear to the bottom of a PD > >>>>post. > > >>>>>>On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > >>>>>>>On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > > >>>>>>>Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as > >>>>>>>KE! > > >>>>>>The contribution to the energy that changes KE is called work. Please > >>>>>>consult your 7th grade science book. > > >>>>This one gave me pause. It's too causal for my taste. It implies > >>>>that if two opposite forces act on a uniformly moving object that > >>>>neither of them do any work since neither of them change KE. > > >>>>One can argue that this definition isn't quite right for non-rigid > >>>>bodies and that it requires clarification for rotating bodies. > > >>>>Still, it's not a bad starting point. A simplification suitable for > >>>>pedagogical purposes. Especially when the discussion doesn't involve > >>>>rigidity or rotation. And if you read it just right, it is equivalent > >>>>to what I consider to be a good definition. > > >>>>Me, I define work as the vector dot product of instantaneous point > >>>>force times the incremental motion of the target object at the contact > >>>>point integrated over the life of the force. If it's not a point > >>>>force, you can integrate over the contact area as well. > > >>>>I've had discussions previously with folks who think that you're > >>>>supposed to do a path integral of force times incremental movement of > >>>>the force along a path. That's wrong. It's force times incremental > >>>>movement of the object to which the force is applied. For point > >>>>objects it's the same thing. For extended objects it's not.- Hide quoted text - > > >>- Show quoted text -- Hide quoted text - > > - Show quoted text - .... Dougie Boy's reply is duly disregarded. NE |