HW_8.solution - P1112 Fall 2008 Solution to Homework #8 1....

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P1112 Fall 2008 Solution to Homework #8 1. A storage tank 12.0 m deep is filled with water. The top of the tank is open to air. What is the pressure at the bottom of the tank? P= ρ gh=(1.0*10 3 kg/m 3 ) (9.8N/kg) (12.0m) =117 600 Pascal Good to keep in mind that 10m of water is about 1 atmosphere. If the bottom of the tank is not open to air (say sealed to another object), then one should add the air pressure (of 1 atmosphere) to P. 2. A container of sea water is placed on a scale. You then suspend a 15.0 kg solid gold statue in the water. Does the reading on the scale change, and if yes by how much? Note: the density of gold is 19.3 x 10 3 kg/m 3 and the density of sea water is 1 x 10 3 kg/m 3 The scale reads the normal force (upward) that it exerts on the container of water. When the statue (mass M and volume V) is totally submerged in the water, the container of water exerts an upward buoyant force F b = ρ water gV on the statue, and therefore, by Newton's third law, the statue exerts a downward force on the container of water of the same magnitude. Since the net force on the container must remain zero, when F b is applied on it downward, the normal force of the scale (and hence the scale reading) increases by F b . Thus the increase in the scale reading (V = M/ ρ gold ) is : F b = ρ water gV = Mg ρ water / ρ gold =7.62 N That is, 0.78 kg more on the scale. If the statue is suspended by a rope, the tension of the rope decreases by F b when the statue is totally submerged in the water. 3.
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HW_8.solution - P1112 Fall 2008 Solution to Homework #8 1....

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