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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 2 3 g 5.00 cm g cm cm 2 100 1 00 20 0 ρ . . e j (b) Sketch (b) at the right represents the situation after the water is added. A volume A h 2 2 b g of mercury has been displaced by water in the right tube. The additional volume of mercury now in the left tube is A h 1 . Since the total volume of mercury has not changed, FIG. P14.18 A h A h 2 2 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: P P gh w = + 0 ρ water The pressure at this same level in the left tube is given by P P g h h P gh w = + + = + 0 2 0 ρ ρ Hg water b g which, using equation (1) above, reduces to ρ ρ Hg water h A A h w 1 1 2 + L N M O Q P = or h h w A A = + ρ ρ water Hg 1 1 2 e j . Thus, the level of mercury has risen a distance of h = + = 1 00 20 0 13 6 1 0 490 10 0 5 00 . . . . . . g cm cm g cm cm 3 3 e j a f e jc h above the original level. P14.19 P g h 0 3 2 66 10 = = − × ρ . Pa : P P P = + =
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