416_Physics ProblemsTechnical Physics

416_Physics ProblemsTechnical Physics - 418 P14.18 Fluid...

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418 Fluid Mechanics P14.18 (a) Using the definition of density, we have h m A w == = water water 23 g 5.00 cm g cm cm 2 100 100 20 0 ρ . . ej (b) Sketch (b) at the right represents the situation after the water is added. A volume Ah 22 bg of mercury has been displaced by water in the right tube. The additional volume of mercury now in the left tube is 1 . Since the total volume of mercury has not changed, FIG. P14.18 1 = or h A A h 2 1 2 = (1) At the level of the mercury–water interface in the right tube, we may write the absolute pressure as: PP g h w =+ 0 water The pressure at this same level in the left tube is given by g hh P g h w + =+ 02 0 ρρ Hg water which, using equation (1) above, reduces to Hg water h A A h w 1 1 2 + L N M O Q P = or h h w A A = + water Hg 1 1 2 . Thus, the level of mercury has risen a distance of h = + = 200 13 6 1 0490 10 0 500 .. . . . . gcm cm cm 3 3 af ch above the original level. P14.19 ∆∆ Pg h 0 3 266 10 × . P a : PP P =+ = × = × 00 55 1 013 0 026 6
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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