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Unformatted text preview: FLUID MECHANICS PROBLEMS II SOLUTIONS 1) B ; The pressure experienced by the diver is the sum of the atmospheric pressure and the pressure exerted by the weight of the water. Hence, ( 29 P Pa kg m m s m P Pa Pa = × + ⇒ = × + × 1013 10 1000 9 8 10 1013 10 9 8 10 3 3 2 5 4 . . . . ⇒ = × P Pa 199 10 5 . . 2) D ; The flow rate is given by the equation ( 29 ( 29 Q A v = , however the units of the possible choices does not correspond with our answer if you simply multiply the respective quantities which suggests that a conversion is necessary. Thus, ( 29 ( 29 ( 29 ( 29 Q cm cm s Q cm s Q Q = ⇒ = ⇔ = ⇒ =  3 30 90 90 1 60 90 1 60 2 3 3 1mL 10 L min min ⇒ = Q L 5.4 min . Observe that the mass of the man was irrelevant information…a sort of red herring to lead you off the right path! 3) B ; Since the closed tube was initially evacuated, any pressure above the mercury liquid would be from vapor pressure (which is essentially zero for mercury). Hence, the pressure at the bottom of the column of the mercury is simply due to the weight of the mercury over the respective area. Mathematically, we need to find the hydrostatic pressure which is given by the formula P g h = ρ ∆ . Thus, ( 29 P kg m m s m P Pa = × ⇒ = × 136 10 9 8 93 100 124 10 4 3 2 5 . . . . 4) C ; The buoyant force is not only defined as the weight of the displaced fluid, but also as the difference between the weight of an object in air and the weight of the same object in a fluid. Hence, F W W B air water = ⇒ =  ⇒ = F kg m s kg m s F N B B 100 1000 9 8 25 1000 9 8 0 735 2 2 . . . . Since ( 29 ( 29 ( 29 F V g B water = ρ , then ( 29 ( 29 0 735 1000 9 8 7 5 10 7 5 10 75 3 2 5 3 5 3 3 . . . . 100 N kg m V m s V m V cm V cm = ⇒ = × ⇒ = × ⇒ = . Hence, the density of the body is simply ρ ρ body body g cm g cm = ⇒ = 100 75 13 3 3 ....
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 Spring '09
 TANNY
 Calculus, Buoyancy, Force

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