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lecture18 - Exam III Physics 101 Lecture 18 Pg 1 Pressure...

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Physics 101: Lecture 18, Pg 1 Exam III
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Physics 101: Lecture 18, Pg 2 Pressure is force exerted by molecules “bouncing” off container P = F/A Gravity/weight effects pressure P = P 0 + ρ gd Buoyant force is “weight” of displaced fluid. F = ρ g V Today include moving fluids! A 1 v 1 = A 2 v 2 P 1 + ρ gy 1 + ½ ρ v 1 2 = P 2 + ρ gy 2 + ½ ρ v 2 2
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Physics 101: Lecture 18, Pg 3 Determine force of fluid on immersed cube Draw FBD » F B = F 2 – F 1 » = P 2 A – P 1 A » = (P 2 – P 1 )A » = ρ g d A » = ρ g V Buoyant force is weight of displaced fluid!
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Physics 101: Lecture 18, Pg 4 A cube of plastic 4.0 cm on a side with density = 0.8 g/cm 3 is floating in the water. When a 9 gram coin is placed on the block, how much does it sink below the water surface? mg F b Mg Σ F = m a F b – Mg – mg = 0 ρ g V disp = (M+m) g V disp = (M+m) / ρ h A = (M+m) / ρ h = (M + m)/ ( ρ A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm M = ρ plastic V cube = 4x4x4x0.8 = 51.2 g h
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Physics 101: Lecture 18, Pg 5 Buoyant Force (F B ) weight of fluid displaced F B = ρ fluid Vol displaced g F g = mg = ρ object Vol object g object sinks if ρ object > ρ fluid object floats if ρ object < ρ fluid If object floats… F B = F g Therefore: ρ fluid g Vol displ. = ρ object g Vol object Therefore: Vol displ. /Vol object = ρ object / ρ fluid
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Physics 101: Lecture 18, Pg 6 Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the
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