Timeline for How can kinetic energy be proportional to the square of velocity, when velocity is relative?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2018 at 17:30 | review | Close votes | |||
| Sep 22, 2018 at 3:05 | |||||
| Sep 16, 2018 at 17:14 | comment | added | knzhou | Possible duplicate of Why is the work done by a rocket engine greater at higher speeds? | |
| Jan 30, 2015 at 15:12 | history | protected | Qmechanic♦ | ||
| Mar 23, 2014 at 17:38 | answer | added | Mellester | timeline score: 1 | |
| Sep 11, 2013 at 11:18 | comment | added | babou | The question, presented as a paradox, is very good. There is of course a fallacy, which has to do with proper consideration of momentum conservation. Not all the energy is spent on accelerating the ship (that is the first key), actually it goes mostly to the ejected reaction mass. Nervertheless you are right that it does accelerate each time by the same amount (assuming same reaction mass and negligible mass change), in any inertial frame, which the bicycles will not do. You can try to find out why, or look at this answer. | |
| Sep 9, 2013 at 16:28 | answer | added | babou | timeline score: 18 | |
| Dec 7, 2012 at 14:16 | answer | added | Bogdan Alexandru | timeline score: -2 | |
| Nov 23, 2012 at 18:13 | comment | added | kηives | @JoseJavierGarcia It is a scalar under rotations; but just try a quick boost $v\rightarrow v'=v+a$ and you will see it is entirely frame dependent. Alternatively, if me and you are standing in a room at rest w.r.t each other, you see my KE as zero. Now start walking. What's my kinetic energy in your frame? | |
| S Nov 23, 2012 at 14:09 | history | suggested | Kitchi | CC BY-SA 3.0 |
improved formatting of latex code
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| Nov 23, 2012 at 12:20 | answer | added | Kitchi | timeline score: 1 | |
| Nov 23, 2012 at 12:07 | review | Suggested edits | |||
| S Nov 23, 2012 at 14:09 | |||||
| Nov 23, 2012 at 11:12 | comment | added | Jose Javier Garcia | the kinetic energy is $ E_{k} = \frac{1}{2}m \vec v \vec v $ so is an SCALAR , an scalar does not depend on the reference ssytem | |
| S Nov 23, 2012 at 8:32 | history | suggested | Curious | CC BY-SA 3.0 |
Equations are written in latex.some corrections are done.
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| Nov 23, 2012 at 4:28 | review | Suggested edits | |||
| S Nov 23, 2012 at 8:32 | |||||
| Nov 23, 2012 at 1:16 | answer | added | Steven Ringo | timeline score: 2 | |
| Nov 23, 2012 at 1:00 | answer | added | Manishearth | timeline score: 3 | |
| Nov 23, 2012 at 0:56 | answer | added | Alfred Centauri | timeline score: 17 | |
| Nov 23, 2012 at 0:50 | review | Close votes | |||
| Nov 28, 2012 at 3:02 | |||||
| Nov 23, 2012 at 0:40 | review | First posts | |||
| Nov 23, 2012 at 4:28 | |||||
| Nov 23, 2012 at 0:33 | comment | added | Qmechanic♦ | Possible duplicate: physics.stackexchange.com/q/535/2451 | |
| Nov 23, 2012 at 0:30 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added link; retagged;
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| Nov 23, 2012 at 0:22 | history | asked | geelen | CC BY-SA 3.0 |