P15_082 - 82. (a) We consider a thin slab of water with...

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Unformatted text preview: 82. (a) We consider a thin slab of water with bottom area A and infinitesimal thickness dh. We apply Newton’s second law to the slab: = (p + dp)A − pA = = = dFnet dp · A − dm · g Adp − ρgAdh dm · a = ρaAdh which gives dp = ρ ( g + a) . dh Integrating over the range (0, h), we get h p= ρ(g + a)dh = ρh(g + a) . 0 (b) We reverse the direction of the acceleration, from that in part (a). This amounts to changing a to −a. Thus, p = ρ ( g − a) . (c) In a free fall, we use the above equation with a = g , which gives p = 0. The internal pressure p in the water totally disappears, because there is no force of interaction among different portions of the water in the bucket to make their acceleration different from g . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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