For this problem neglect air resistance and remember that weight = mg. How much force acts on a 0.01 kg coin when it is halfway to its maximum height?
2007-02-01 16:29:44 · 3 answers · asked by Tone_Loc 2 in Science & Mathematics ➔ Physics
I'm assuming that this coin was tossed up into the air, or something similar.
If that is the case, then the downward force of gravity is constantly accelerating the coin at -9.8 meters per second squared. F = ma
F = (.01kg)(-9.8m/s^2) = -.098 N
The negative sign indicates the force is in the downward direction.
2007-02-01 16:36:11 · answer #1 · answered by The Ry-Guy 5 · 0⤊ 0⤋
You must have the inital force first, like the amount of force when it was thrown up. If it was dropping, at what height it was dropping? Since there is no air resistance, the way it was dropped will not matter.
So its only left with the height it drop, or the force it was thrown up. :)
2007-02-01 16:36:22 · answer #2 · answered by larry_lum 1 · 0⤊ 0⤋
The flip table has no angular acceleration (that's transferring at a persevering with angular velocity), so the coin won't have any angular acceleration. The coin is transferring in a circle, although, and so has centripetal acceleration. wr = v v^2 / r = centripetal acceleration (wr)^2 / r = acc. (w^2)r = acc w = 33 a million/3 rev/min = one hundred/3 rev/min one hundred/3 rev/min * 2pi radians/rev * a million min/60sec = 3.40 9 rad/s verify you're making r in meters, no longer centimeters acc = (3.40 9)^2(5/one hundred) acc = (3.40 9)^2(10/one hundred) acc = (3.40 9)^2(18/one hundred)
2016-12-17 07:37:58 · answer #3 · answered by ? 3 · 0⤊ 0⤋