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This started to bother me today. In the first picture, there's a cup wheres a certain amount of water in it. A ball made of aluminium hasnt been dropped in the cup yet.
In the second picture, the ball has been dropped in.
The ball has not sinked, a string is holding it.
Now please tell me, how did the pressure on the botton of the cup created BY THE WATER change?
I mean, the water level just rose. So the pressure should also.
But there seems to be another solution to the problem, and I just can't figure it out.
The equation for liquid pressure is p= RO*g*h.
Thanks in advance,
______________________________________
[PLAIN]http://www.cornerpocketpool.com/images/catalog/category57.jpg [Broken]
http://www.billiardlover.com/mcdermott-pool-cue.html [Broken]
In the second picture, the ball has been dropped in.
The ball has not sinked, a string is holding it.
Now please tell me, how did the pressure on the botton of the cup created BY THE WATER change?
I mean, the water level just rose. So the pressure should also.
But there seems to be another solution to the problem, and I just can't figure it out.
The equation for liquid pressure is p= RO*g*h.
Thanks in advance,
______________________________________
[PLAIN]http://www.cornerpocketpool.com/images/catalog/category57.jpg [Broken]
http://www.billiardlover.com/mcdermott-pool-cue.html [Broken]
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