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Unformatted text preview: PHY 317K Homework 5 Solutions 1. E p = gh , so p = g h . Here h = 10 3 m, = 8 . 5 10 2 kg / m 3 , and of course g = 9 . 8 m / s 2 Then p = (8 . 5 10 2 kg / m 3 )(9 . 8 m / s 2 )(10 3 m) = 8 . 3 Pa. 2. E The pressure at a given depth depends only on that depth. Because the distance from the surface of the fluid to the bottom of the vessel is the same for all four vessels, the pressures at the bottom are the same. 3. B The height of the oil column is 5.0 cm. The water column rises 3.0 cm above the BOTTOM of the oil column, as is clear from the figure showing the situation. (The top of the water column is 2 cm below the top of the oil column, so it is 3 cm above the bottom.) The pressure must be the same at the bottom of the oil column and at the corresponding level in the water column, so: oil g (5 . 0 cm) = water g (3 . 0 cm). Then: oil = 3 5 water = oil = 3 5 (1 . 0 g / cm 3 ) = 0 . 60 g / cm 3 . 4. B Because the object floats in each fluid, the buoyant force provided by each fluid must be equal to the weight of the object. Let V 1 , V 2 , V 3 be the volumes displaced by the object in each of the three fluids. Then because buoyant forces are equal, we have: . 9 V 1 g = V 2 g = 1 . 1 V 3 g , so that 0 . 9 V 1 = V 2 = 1 . 1 V 3 . Therefore, V 3 < V 2 < V 1 5. B The bouyant force provided by water when the object is submerged is 10 N (since 30 10 = 20). In a fluid with half the density of water, the bouyant force will be half this, 5.0 N. Thus, the spring balance will read 30 5 = 25 N. 6. C An incompressible fluid delivers the same pressure to all locations (at the same height). Thus, a force F 1 acting on a piston with area A 1 corresponds to a pressure p = F 1 /A 1 . At the other piston, with area A 2 , the pressure will be the same, so the force F 2 must satisfy: F 1 A 1 = F 2 A 2 . Here, let piston 1 be the larger one, with A 1 = (8 . 0 cm / 2) 2 = 16 cm 2 . Then for piston 2, A 2 = (2 . 0 cm / 2) 2 = cm 2 ....
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 Spring '07
 KOPP
 Physics, Work

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