I know it is the same (almost 300,000 km/s), but I am talking about the relative speed, and not about its own speed.
129292
Same speed as if the car were at rest - but the wavelength of the light will be slightly shorter - shifted towards the blue end of the spectrum.
383466
i may not know some so here is some things velocity is probably 239.03642817m/s againt the therory. Thier kenetikit energy was probably high i cant figure that out and it was definently inelastic. I dont know why but i know it is definenantley inelastic.
305177
Until now, the velocity of light is still the fastest which is about 350km/sec, and no other means can go faster than this.
503687
You can't travel faster than the speed of light in a vacuum, so it's a moot point.
112183
Your perception of time doesn't change if you travel even close to the speed of light. What changes is the perception that will have others respect of you. If they would be able to look at your watch they will see that it moves slower than theirs, and gets slower the more close to the speed of light you move. If you see their watches you will see that they are moving slower, because the movement is relative. It becomes absolute when a force is applied to it, for example an acceleration (the twins paradox). Time and space is relative to the individual that is moving at a certain speed. For you, a minute will always be a minute, you will perceive it as the same amount of time always.\n\n\n---Addition---\n\nThe equations you are thinking of are the Lorentz Transformation, heres a link\nhttp://www.classicalreality.net/Lorentz%20Transformation.htm\n\nWhen you travel nearly the same speed of light, light keeps on traveling at the speed of light from your point of view. So it is not "hardly moving" but in fact keeps on moving for everybody, independent of their velocities, at 300,000 km/sec (in vacumm). The only explanation to this is that time for the different observers is not comparable, everybody follows its own time, and space also.
539613
You can do a theoretical top speed with physics, but it wouldn't account for the two main killers of speed which are aerodynamic drag and internal friction.  Without knowing those, you'd only be getting a false promise from your calculations.
105597
around 1/6th that of earth so around 1.64m/s
371893
it may be that the speed is so fast that it may have reached the maximum speed.... this could be the maximum limit..\nit's critical by common sense...\ncheck the dictionary, or thesaurus....
1802
As I am travelling with a velocity 90 kph & the other car with 120 kph, I will appear as going back to the other driver but with a slower speed ie\na) 120-90 =30 kph\n\nThe other car wil appear to move ahead of me but with a slower speed as I am also chasing it\nb)120-90 =30 kph\n\nBut as we are talking about velocity we have to consider direction also. Hence my velocity will be -ve i.e. (-30 kph) & other car will be +30 kph
45829
do u mean speed? no, the magnitude is stillt he same but the direction is changed if and only if the mirror has a smooth surface. cause if it is not. there would be many angle of reflection
430419
I'm not sure what you are asking, but if gravity is the action acting on the hands of the clock, the increased speed of x% should be the same as the slowed speed of y%.
471759
NO!!  If people would stop (mis)quoting half-remembered precepts here, and do a spot of THINKING, instead, we might get somewhere.\n\n1.  If a moving spherical ball directly hits an identical free, stationary ball head on in an elastic collision, the moving ball STOPS, and the previously stationary one heads off with the original velocity of the other.  So ALL of the moving ball's K.E. has been transferred to the other ball.  \n\nHere are two ways to see this:\n\n1a.  The device known somewhat indelicately to British schoolboys as "Newton's balls" (but in squeamish America as "The Swinging Wonder") provides an excellent table-top demonstration of this.  Most undergraduate physics instructors must haul this out from time to time.\n\n1b.  You can easily solve it analytically, but a thought experiment will also demonstrate it.  In the originally stationary ball's frame, let the other ball approach with speed v.  Now shift (with an appropriately directed speed of v/2) to the frame in which the two equal balls are simply approaching each other, each with a speed in this frame of that same v/2 (oppositely directed, of course)!  Clearly, from the symmetry of the situation, they'll bounce off each other with the same (oppositely directed) speeds in this frame, still of v/2.  Now go back to the original frame: the previously moving ball is now completely stationary, while the originally stationary one is haring off with speed v.  (You may need to make little sketches of what is happening as viewed successively from each of these frames, to see this; I admit that that's what I did.)\n\n2a.  A similar approach can be used to solve the general case for an arbitrary ratio of masses, m (initially moving)/M (initially stationary).  Hint: have the balls approach one another in their center of mass frame with equal but opposite momenta of mv and  MV.  They'll clearly bounce off one another in such a frame, retaining the same values of their respective (but now oppositely directed) speeds.  Then you can do the other frame-jugglings from there. \n\nHowever, this analytical approach, done accurately, is a tedious mess.  So, to appreciate the general problem more physically, consider just the following:\n\n2b.  The case of m << M.\n\nConsider the "equal but opposite momentum" frame for these two balls, in which MV = mv.  This means that if ball M is VERY MASSIVE, the speed it will have in such a frame, V = (m/M)v; that is, it's VERY SMALL compared to little puny m's high speed.\n\nSo, transforming to the frame in which M itself (the big guy) is initially stationary takes only a speed shift of O(m/M)v.  (In fact identically so.)  This only affects the speed of puny m minutely, so m's initial K.E. is still of order O(m v^2).  (There's no need to do picky constant book-keeping for these purposes.)  \n\nNow after the collision: In the original frame, Big M is now moving (and in fact at 2V if you think about it.)  But, being loose about the constants, M's K.E is only O(MV^2).  \n\nSo, the ratio (m's K.E.)/(M's K.E.) ~ (mv^2)/(MV^2)\n\n= (m/M)(M/m)^2 (since mv = MV)\n\n= M/m.\n\nBINGO!  The little bugger still carries by far the larger amount of the kinetic energy.  The massive ball (i) moves sluggishly, and (ii) because  K.E. involves both mass but also the SQUARE of the speed, the TWO POWERS of (small) speed more than cancel out the effect of its (large) mass.\n\nThis is a well-known fact in particle collision theory, that little particles basically bounce off big ones.  \n\nYou could also say that it's well-known in "space billiards": when spacecraft head close to Jupiter to receive a "gravitational sling-shot effect" that sends them off to the outermost planets.  \n\nIt's more complicated because of 3-D effects there, but analogous frame-shifting methods can be used to show that the spacecraft can pick up considerable speed AND change
364091
both are the same if it is drop at the same height because the gravity speed is 10 ms-2 (9.81ms-2 more accurate)
392495
Don't things fall at 32ft/sec.
422234
nope, there is no way, 56k is as fast as it goes
129126
Neither Observer EXPERIENCES a change. The ONLY way any comparison can be made between the two unmentioned but essential clocks is to bring them together in the same rest frame again: then whichever clock ticktocked the least would be shown to be the one which accelerated and slowed down in their relative travel.
436232
3.0m/s is not an acceleration, it is a speed.\nIs the acceleration constant?