University Physics with Modern Physics with Mastering Physics (11th Edition)

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14.75: a) Consider the fluid in the horizontal part of the tube. This fluid, with mass , Al ρ is subject to a net force due to the pressure difference between the ends of the tube, which is the difference between the gauge pressures at the bottoms of the ends of the tubes. This difference is ), ( R L y y ρg - and the net force on the horizontal part of the fluid is , ) ( R L Ala A y y g = - or . ) ( R L l g a y y = - b) Again consider the fluid in the horizontal part of the tube. As in part (a), the fluid is accelerating; the center of mass has a radial acceleration of magnitude , 2 2 rad l a ϖ = and so the difference in heights between the columns is . 2 ) )( 2 ( 2 2 2 g l g l l = Anticipating Problem, 14.77, an equivalent way to do part (b) is to break the fluid in
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Unformatted text preview: the horizontal part of the tube into elements of thickness dr ; the pressure difference between the sides of this piece is dr r dp ) ( 2 = (see Problem 14.78), and integrating from , 2 gives to 2 2 l p l r r ρϖ = ∆ = = giving the same result. c) At any point, Newton’s second law gives pAdla dpA = from which the area A cancels out. Therefore the cross-sectional area does not affect the result, even if it varies. Integrating the above result from 0 to l gives pal p = ∆ between the ends. This is related to the height of the columns through y pg p ∆ = ∆ from which p cancels out....
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This document was uploaded on 02/05/2008.

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