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: NoEinstein on 12 Oct 2009 20:54 On Oct 12, 5:43 am, "Autymn D. C." <lysde...(a)sbcglobal.net> wrote: > On Oct 10, 8:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Oct 8, 1:04 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > Dear Y.Porat: The failure of scientists to make charged particles > > reach 'c' was due to the presence of ether inside the vacuum tubes. > > The ridiculous notion behind SR resulted from extrapolation of the > > particle experiments. In space travel, the density of ether drops off > > the further away from massive objects you stay. In the 'Swiss Cheese' > > voids between the galaxies, there is little or no ether. Plot your > > intra-universe courses of travel to stay in those voids, and your > > maximum velocity is limitless. Extra-terrestrials, and yours truly, > > know that, because ether is polar, it can be magnetized and made to > > RIP apart so that the spaceships don't ever impact the ether. > > This doesn't contradict SR. SR placed a non-existent speed limit on light and on objects. I've REMOVED that speed limit, conclusively! NE
From: NoEinstein on 12 Oct 2009 20:57 On Oct 12, 4:12 pm, "Autymn D. C." <lysde...(a)sbcglobal.net> wrote: > On Oct 10, 8:04 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > will impact with a KE of 2 of its static weight units. For each > > second of fall its KE will increase one additional weight unit. All > > objects at rest have a 'KE' = their static weight. The CORRECT > > formula is: KE = a/g (m) + v / 32.174 (m). At the end of second one, > > KE = 2. > > That is the formula of nothing. > > > feet per second. FORCE OF IMPACT, i.e., KE is a function of the > > VELOCITY and the static weight of the object. *** Since the velocity > > i.e. -> q.e. > energhy != forse > mvv/2 != ma = mjt > > > of all dropped objects increases uniformly, or LINEARLY with respect > > to time, NOT the distance traveled, KE must, therefore, be increasing > > LINEARLY, too. The latter conforms to the Law of the Conservation of > > Energy. But your errant "distance" notion violates the L. of the C. > > Momentum, cretin. KE IS momentum, Jerk! NE
From: doug on 13 Oct 2009 00:08 NoEinstein wrote: > On Oct 12, 4:12 pm, "Autymn D. C." <lysde...(a)sbcglobal.net> wrote: > >>On Oct 10, 8:04 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: >> >> >>>will impact with a KE of 2 of its static weight units. For each >>>second of fall its KE will increase one additional weight unit. All >>>objects at rest have a 'KE' = their static weight. The CORRECT >>>formula is: KE = a/g (m) + v / 32.174 (m). At the end of second one, >>>KE = 2. >> >>That is the formula of nothing. >> >> >>>feet per second. FORCE OF IMPACT, i.e., KE is a function of the >>>VELOCITY and the static weight of the object. *** Since the velocity >> >>i.e. -> q.e. >>energhy != forse >>mvv/2 != ma = mjt >> >> >>>of all dropped objects increases uniformly, or LINEARLY with respect >>>to time, NOT the distance traveled, KE must, therefore, be increasing >>>LINEARLY, too. The latter conforms to the Law of the Conservation of >>>Energy. But your errant "distance" notion violates the L. of the C. >> >>Momentum, cretin. > > > KE IS momentum, Jerk! � NE � Wrong as usual john.
From: doug on 13 Oct 2009 00:09 NoEinstein wrote: > On Oct 12, 5:43 am, "Autymn D. C." <lysde...(a)sbcglobal.net> wrote: > >>On Oct 10, 8:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: >> >> >>>On Oct 8, 1:04 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: >> >>>Dear Y.Porat: The failure of scientists to make charged particles >>>reach 'c' was due to the presence of ether inside the vacuum tubes. >>>The ridiculous notion behind SR resulted from extrapolation of the >>>particle experiments. In space travel, the density of ether drops off >>>the further away from massive objects you stay. In the 'Swiss Cheese' >>>voids between the galaxies, there is little or no ether. Plot your >>>intra-universe courses of travel to stay in those voids, and your >>>maximum velocity is limitless. Extra-terrestrials, and yours truly, >>>know that, because ether is polar, it can be magnetized and made to >>>RIP apart so that the spaceships don't ever impact the ether. >> >>This doesn't contradict SR. > > > SR placed a non-existent speed limit on light and on objects. I've > REMOVED that speed limit, conclusively! � NE � No, you have demonstrated your complete ignorance of science and have tripped over your ego.
From: doug on 13 Oct 2009 00:10
NoEinstein wrote: > On Oct 10, 9:24 pm, Jerry <Cephalobus_alie...(a)comcast.net> wrote: > >>On Oct 10, 10:04 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: >> >> >>>On Oct 8, 12:21 pm, Jerry <Cephalobus_alie...(a)comcast.net> wrote: >> >>>Dear Jerry ("another" hard-head): You need to take the following Pop >>>Quiz for Science Buffs. If you do�and will learn from the rationale >>>(Though that's doubtful for you...)�you will realize THIS basic fact: >>>A compact object is traveling 32.174 feet per second, as at the end of >>>one second of free-fall, and will have traveled 16.087 feet. Such >>>will impact with a KE of 2 of its static weight units. For each >>>second of fall its KE will increase one additional weight unit. All >>>objects at rest have a 'KE' = their static weight. The CORRECT >>>formula is: KE = a/g (m) + v / 32.174 (m). At the end of second one, >>>KE = 2. >> >>Your formula may be YOUR definition of energy, but it is not >>anyone else's definition. Kinetic energy is not "splat power". >> >>Let us go back to MY questions. >> >>1) If I lift an object 64 feet, is the energy it took me to lift >> it the first foot equal to the energy it took me to lift it >> the second foot, third foot, fourth foot... sixty-third foot, >> sixty-fourth foot? >> >>Yes/No >> >>2) If I drop an object 64 feet, is the kinetic energy acquired >> by the object equal to the energy that it took to lift it 64 >> feet? >> >>Yes/No >> >>3) If I drop an object, does it fall about 16 feet the first >> second, and after two seconds, is the total distance fallen >> approximately 64 feet? >> >>Yes/No >> >>4) Is 16 equal to one-fourth of 64? >> >>Yes/No >> >>5) Is the amount of potential energy converted to kinetic energy >> after one second equal to one-fourth the amount of potential >> energy converted to kinetic energy after two seconds? >> >>Yes/No >> >>I need to know at what point you start saying "No". >> >>Jerry > > > Dear Jerry: Lifting objects causes PE to accrue. The correct formula > (mine) is: PE = w + (h / 16.087 ft.) w. PE is 100% recoverable as > useful KE only if there is a continuous connection, or mechanism, for > a slow energy transfer. Because most objects in free drop have no > such connection, the recoverable KE is always less than the 'stored > up' PE. In actuality, the only thing stored up is a potential > distance of fall. Until such fall there is ZERO PE stored in the > object itself! � NoEinstein � Amazingly, you keep thinking up more stupid things to say. You cannot have been an architect or my respect for that profession is totally gone. |