Talk:X87 edit

What does most of that stuff have to do with improving the X87 article? -- AnonMoos (talk) 15:01, 6 September 2012 (UTC)Reply

Your recent edits edit

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Your recent edits edit

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Your recent edits edit

  Hello. In case you didn't know, when you add content to talk pages and Wikipedia pages that have open discussion, you should sign your posts by typing four tildes ( ~~~~ ) at the end of your comment. You could also click on the signature button   or   located above the edit window. This will automatically insert a signature with your username or IP address and the time you posted the comment. This information is useful because other editors will be able to tell who said what, and when they said it. Thank you. --SineBot (talk) 13:40, 4 October 2012 (UTC)Reply

Your recent edits edit

  Hello. In case you didn't know, when you add content to talk pages and Wikipedia pages that have open discussion, you should sign your posts by typing four tildes ( ~~~~ ) at the end of your comment. You could also click on the signature button   or   located above the edit window. This will automatically insert a signature with your username or IP address and the time you posted the comment. This information is useful because other editors will be able to tell who said what, and when they said it. Thank you. --SineBot (talk) 19:40, 9 October 2012 (UTC)Reply

Talk pages edit

  Please refrain from using talk pages such as Talk:Kinetic energy for general discussion of the topic or other unrelated topics. They are for discussion related to improving the article in specific ways, based on reliable sources and the project policies and guidelines; they are not for use as a forum or chat room. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. See here for more information. Thank you. - DVdm (talk) 21:20, 14 June 2022 (UTC)Reply

Kinetic energy formula is wrong only in specific case/example edit

I read about kinetic energy realativity in Talk:Kinetic energy Archive_1 and understood, that there no easy possible way of proving that kinetic energy formula is wrong... But there is one example where you can safe 1/4 of energy with rocket, which consist of two parts.
Let's say there in cosmos we want sent not moving rocket, which consist of two parts: of part A and part B. Each rocket part is rocket and have mass 1 kg. Part A carring itself and part B and accelerates from 0 m/s to 10 m/s. For this need energy   Now part B separates from part A (both parts have speed 10 m/s) and reaches 20 m/s. But part B from part A reference frame, have speed 0 m/s, so for part B to reach 20 m/s need energy   For part A to accelerate from 0 to 20 m/s need energy   Total energy needed for part A to accelerate itself to 20 m/s and part B to 10 m/s is 200+50=250 (J). And for part B to accelerate from 10 m/s to 20 m/s need energy 50 (J). So total need for both rocket parts 250+50=300 (J) of energy.
Now if accelerate separatly 1 kg mass rocket A to 20 m/s and 1 kg mass rocket B to 20 m/s need energy for each   And total needed energy for rocket A and rocket B to accelerate to 20 m/s is 200+200=400 (J). So we saving 400-300=100 (J) depending in what manner we accelerating rockets (and in both cases rockets reach 20 m/s and have same ~1 kg mass). Say part A of first rocket flying to Moon and part B of first rocket flying to Mars. And rocket A flying to Moon and rocket B flying to Mars. Assuming Moon and Mars is almost in same direction. We saving 1/4 of energy and reaching we same speeds for all parts and rockets. 1/4 energy saving is maximum for two parts rocket. For more parts rocket there, I think, possible to save little bit more than 1/4 energy (but much energy you will not save). Paraboloid01
Update 1. I think I get it. From point P0, when First rocket (consisting of part A and part B) speed in cosmos was 0 m/s, need calculate, what will be First rocket parts A and B final speed. And this speed for part A and part B is 20 m/s. So for part B need 150 (J) to accelerate from 10 m/s to 20 m/s, when part B separates from part A. Let's say, that in First rocket Part B sitting 1 cm Gnom I (inside) with weight 10 grams. Gnom I (sitting in first rocket part B) claiming, that there would be enough 50 (J) of fuel in first Rocket part B to reach speed from 10 m/s to 20 m/s (when at 10 m/s part B splits from part A) and he will be in time for dinner on Gnoms planet "MARS" (this planet "Mars" have same 0 m/s speed like point P0). Another Gnom O (outside) is at point P0 and with smartphone comunicating with Gnom M (on "Mars"). Gnom O claiming, that for Gnom I there will be not enough energy 50 (J) in first rocket part B to accelerate first rocket part B to 20 m/s, and that Gnom I will not be in time for dinner on planet "Mars" (because according to Gnom O and Gnom M for Gnom I need 150 (J) energy [to accelerate from 10 m/s to 20 m/s], but not 50 (J) of energy like thinking Gnom I (Gnom I thinking, that he relative to first rocket part A don't have initial speed 10 m/s, but have initial speed 0 m/s)). Who will be right, or Gnom I or Gnom O and Gnom M? Do for Gnom I inside Rocket part B will be enough energy (50 J) to accelerate from 10 m/s to 20 m/s [relative to point P0 and "Mars"]? - Paraboloid01 (talk)
Note: Please be aware of the fact that Wikipedia is not a chat forum. As long as you don't pollute the articles and article talk pages with this, there's probably no problem, but I have removed the article talk page message that you left pointing to this. You might try an open forum. - DVdm (talk) 16:45, 19 June 2022 (UTC)Reply

Talk page revisited, again edit

  Please stop. If you continue to use talk pages for inappropriate discussion, as you did at Talk:Kinetic energy, you may be blocked from editing. - DVdm (talk) 15:00, 22 February 2023 (UTC)Reply

Is fuel mass changing in different reference frames? edit

Like described in this post for [m=2 kg] toy car need 800 J from ground reference frame and 400 J from transparent aircraft reference frame. Say that car for acceleration using gasoline. In toy car fuel tank is 40 grams of gasoline fuel. Say that this 40 grams of fuel for toy car enough only to accelerate from 0 m/s to 20 m/s on earth. So observer on earth see that in transparent aircraft car accelerating from 10 m/s to 30 m/s and he knows, that for this need 800 J of fuel. So he believs that 40 grams of fuel will not be enough to accelerate car [from 10 m/s to 30 m/s] in 10 m/s flying transparent aircraft. And observer from earth surface strongly believes, that for that (to accelerate car from 10 m/s to 30 m/s) in toy car fuel tank must be 80 grams (equal to 800 J useful energy) of gasoline fuel. So how many grams of gasoline fuel will be used to accelerate toy car to 20 m/s from aircraft RF and to 30 m/s from ground (earth surface) RF? 80 grams or 40 grams? Paraboloid01 (talk) 17:26, 22 February 2023 (UTC)Reply

You might try a forum such as Physics Forums or Google groups for such questions. Wikipedia is not the place for open discussion. - DVdm (talk) 18:18, 22 February 2023 (UTC)Reply