Why when you look at a clock does the second hand always appear stationary?
396067
You need about 0.1 second to notice the motion of any moving thing, since the movement of she second hand takes about that time so it is not easily to notice it along its movement between to seconds locations
514998
If you are using a rotatable lcd monitor, it probably came with software to change the orientation from landscape (like a regular crt monitor) to portrait. If you are running Windows, check your Start Menu for programs that look like they might have come from your monitor's manufacturer. Also check the SysTray (where the clock is) for icons that look like they might have something to do with a monitor.
638729
The grandfather clock works on a mechanical system of gears, spring, levers, catches and a pendulum for keeping time. Energy is stored in a wind-up spring or a weight driven axis.\n\nThe basic time keeping is achieved by a pendulum that counts time in an accurate fashion within seconds per month. This uses the principle of Periodic motion and may be adjusted to be slightly faster or slowly by shifting the pendulum ball's position slightly.\n\nThe pendulum counter then controls the movements of a system of mechanical levers, gears & wheels that are calculated to drive the second, minute and hour hands in the desired rotation ratios so that they will show the correct time.\n\nUpon the hour, the system will activate a chime via the spring driven hammers that strike tubular bells or other sounding vibrating elements. A series of activation may be composed to produce the desired melody. Some system will also provide the date for the month. In its peak, many other noviety displays such as birds, toys, animals, etc would be used to announce the top of the hour much to the delight of children and adults alike.
234240
Because if it went counterclockwise time would be going in reverse? :)
925766
In a modern clock, a voltage is applied to a crystal such as quartz.  This causes the molecules in the crystal to vibrate at a fixed frequency.  A sensor then counts the vibrations over time and determines how much time has elapsed.
308200
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.
611283
I understand that when a watches (or clocks) hands are at 10:10, it looks the cleanest and most symmetrical and shows the brand name, or logo, which is usually just below the center of the face. Some companies have started using 10:08 or 10:12 as the time in an advert because it will keep the longer minute hand from blocking the numeral.
393981
you get used to the hat being on your head. I get that same feeling with wristwatches...you get so used to it just being there, that it kinda throws you off when you go to check the time on a nonexistent watch.
985574
generally, the doomsday clock is refered to in regards to the minute hand.  Technically there is no second hand.  During the height of the cold war it was as close as two minutes to midnight if I'm not mistaken, but a second hand would just be overly morbid, wouldn't you say?
397782
If the clock is constantly moving at constant velocity, then you're right, the people on earth will say the plane's clock is going slower, whereas the people on the plane will say the earth's clocks are slow.\n\nHowever, when the plane starts and stops in these experiments, it undergoes an acceleration, which differentiates the two frames. That's why the plane's clock is behind when it's compared to the earth clock. Basically it's because the plane changes which inertial frames it is a part of when it accelerates. Your point is similar to the twin paradox, I suggest you check out my source, that should clear things up, you sound like you have a solid enough background to understand it.
940431
This seems really dumb, because clocks do not change speed because of gravity.  Even if they were, their acceleration would be different at different times, not first constant and then zero like in your question.  But if you really need an answer...\n\nThe hands of the clock that are not affected by gravity will meet between 6:30 and 6:35.  To find when, we have to look at the angles of the hands.\n\nAssume t counts, in minutes, from 6:00.\nh(t) = 180 + t/60*15 degrees, or 180 + t/4\nm(t) = t/60*360 degrees, or 6t\n180 + t/4 = 6t\n(6 - 1/4)t = 180\nt = 180/5.75 = 720/23 = about 31.30 min, or 31:18.  This is when the hands of the clock not affected by gravity will meet.\n\nNow for clock A.  The rate of acceleration is xv(t), so we know that v' = xv.  This is an ordinary linear differential equation.  Rewrite it as v' - xv = 0.  The characteristic equation is λ - x = 0.\nThe zero of the characteristic equation occurrs when λ = x, so the basis for the solution is e^(xt).  So,\nv(t) = ce^(xt)\n\nThe initial velocity is 360º/hour = 6º/min, so v(t = 0) = 6 = ce^0 => c = 6.\n\nFor the minute hand, until time t=15, v(t) = 6e^(xt).  Integrate this to find the position and convert from radians to degrees...\nm(t) = (6/x)e^(xt) * (π/180) = (π / 30x) e^(xt)\n\nSolve for t=15:\nv(15) = 6e^(15x)\nm(15) = (π / 30x) e^(15x)\n\nFor all time t>15, v(t) is constant.  Integrate:\nm(t) = 6t e^(15x) + c\n\nKnowing that m(15) = (π / 30x) e^(15x),\n(π / 30x) e^(15x) = 6*15 e^(15x) + c\nc = (π / 30x - 90) e^(15x)\n\nSo for time t>15 min,\nm(t) = (6t + π / 30x - 90) e^(15x) º\n\nFor the hour hand, the initial velocity is 1/4º per minute.  The acceleration is -yv, which gives v' = -yv.  Rewrite as v' + yv = 0.  The characteristic equation is λ + y = 0.  The zero occurrs when λ = -y.  The basis for the solution is e^(-yt).\nv(t) = ce^(-yt)\nv(0) = 1/4 = ce^0 => c = 1/4\n\nSo for the hour hand, v(t) = 1/4 e^(-yt).  The hands will meet before 9:00, so we do not have to worry about the condition when the hour hand stops decelerating.\n\nIntegrate to find the position:\nh(t) = 1/(-4y) e^(-yt) + c\nh(t = 0) = 180º = -1/(4y) e^0 + c\nc = 180 + 1/(4y)\n\nSo,\nh(t) = 180 + [1 - e^(-yt)]/(4y)\nFrom above,\nm(t) = [6t + π/(30x) - 90]e^(15x)\n\nSet them equal and solve for t.  I think YOU can handle that.\n\n\n\n\nAbout the acceleration being constant:  Yes, the acceleration due  to gravity is a constant, however, it is acting upon the arm's centre of mass, which is rotating about the fixed end.  The constant force due to gravity acts straight down.  However, if you break the gravitational force into orthagonal components parallel and normal to the arm, only the normal component generates torque, and torque is what accelerates the arm.  The normal component varies as the sine of the angle at which the arm is pointing, hence the torque will vary and so will the acceleration of the arm.  I took statics; I know what I am talking about.\n\nIs this a homework problem, or did you make it up?
107716
Made in China:\nClocks - The first clock that they devised was for astronomical uses. In the first clock ever, there was a puppet that would hold up a plaque that would tell the time. They also invented giant water clocks, which rang every fifteen minutes.
116565
With the goal that a clock signal will not drift, \nPLL (phase locked loop) logic can be added to keep\nthe frequency constant.  It works by adding "feedback"\n(from the output back to the input) with a tuning\ncircuit to keep the two signals (the original input\nas well as the feedback) in synch.
464155
The hands at the 10:10 or 1:50 position, lead the eye to the makers logo, usually found just above center.\nAlso some interesting info here.\nhttp://www.ubr.com/clocks/faq/1010.html
592435
a watch?\n\nlike a watch with an hour, minute, and second hand, and then another clock with two hands inside of it.\n\ndoes that make any sense? haha
227138
Because the hour hand is easier to read i.e. if the hand is in between the 5 and the 6 it's five something.  The minute hand is bigger because it is more precise.  It's harder to see the space between a minute on a clock.  Therefore, the hand is longer!
1150643
It could be a double reflection.  If the image you see is in the correct orientation, than it is a daouble reflection, or reflected any even number of times.  The object is in front of you which reflects off the left window, into the right window, where you are looking.