2006-10-25 23:26:18 · 7 answers · asked by pavan kumar NC 2 in Science & Mathematics ➔ Physics
Turbulence.
To go straight down there would have to be perfect laminar flow of the liquid over the coin. This almost never happens.
2006-10-25 23:30:57 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Remember that the deeper you are in the water, the greater the pressure. When the coin is inclined slightly to the left, the pressure on the leading edge is greater than the trailing edge. So the pressure below will trip the coin so that it will go up. The trailing edge cannot balance this torque because the pressure on its level is less.
2006-10-26 02:24:52 · answer #2 · answered by Binangkaw 2 · 0⤊ 0⤋
The coin is flat and acts like a wing in air - as quickly as there is the slightest resistance greater on one component or the different the coin's weight pushes the coin sideways, whilst the coin turns to be almost flat, against it extremely is trajectory, it extremely is going to slide back interior the alternative path.
2016-11-25 21:25:17 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Resistance from the water against the coin.
2006-10-25 23:33:33 · answer #4 · answered by claire 5 · 0⤊ 0⤋
A coin isn't a hydro dynamically stable shape. The tiniest forces will always tilt it one way or the other, and it will glide this way and that depending on the tilt.
2006-10-26 00:39:31 · answer #5 · answered by Nomadd 7 · 0⤊ 0⤋
the buoancy(?) in the water pushes upwards on the coin. when it's lying flat, there's a large surface area for the force to act on, when it's vertical (or at least at an angle) the force isn't enough to support the mass of the coin, so it 'slides' through.
2006-10-25 23:30:36 · answer #6 · answered by phedro 4 · 0⤊ 1⤋
hydro-dynamics
2006-10-25 23:28:48 · answer #7 · answered by Anonymous · 0⤊ 0⤋