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**Unformatted text preview: **Fluids Andrea Bangert January 26, 2009 Required: Archimedes Principle (40 points) Required: The Venturi Effect (40 points) Optional: Bernoullis Equation (10 points) Optional: Atmospheric Pressure (10 points) Optional: Pressure and Elevation (10 points) Optional: The Weight of Air (10 points) 1 Archimedes Principle 1.1 State Archimedes Principle in words. (5 points.) This is the concept which you need to verify. The buoyant force acts upon an object submerged in a fluid and is directed upwards towards the surface of the fluid. The buoyant force is equal to the weight of the fluid displaced by the object. 1.2 Weigh the aluminum object hanging in air and submerged in wa- ter. (10 points.) The balance scale gives the mass of the object in grams . m objectinair = 67 . 8 g 1 m objectinwater = 43 . 2 g Calculate the difference between the mass of the object in the air and the apparent mass of the object in water. m object = m objectinair- m objectinwater m object = 67 . 8 g- 43 . 2 g = 24 . 6 g The balance scale gives you the mass of the object in grams . If you want to obtain the weight of the object in Newtons , you need to use the equation w = mg where g is the acceleration due to gravity, g = 9 . 8 m/s 2 . Then w objectinwater = m objectinwater g w objectinwater = (43 . 2 g) 9 . 8 m s 2 = (0 . 0432 kg) 9 . 8 m s 2 = 0 . 423N w objectinair = m objectinair g w objectinair = (67 . 8 g) 9 . 8 m s 2 = (0 . 0678 kg) 9 . 8 m s 2 = 0 . 664N The difference between the weight of the object in air and the apparent weight of the object in water is w object = w objectinair- w objectinwater w object = 0 . 664 N- . 423 N = 0 . 241 N This difference between the weight of the object in air and the apparent weight of the object in water is the buoyant force, F B = w object = 0 . 241 N Notice that force is measured in Newtons . 2 1.3 Measure the weight of the empty beaker and the weight of the wa- ter which was displaced by the object. (10 points.) The balance scale reports the mass in grams . m empty beaker = 43 . 0 g m beaker withwater = 70 . 7 g The mass of the water displaced by the object is m displacedwater = m beaker withwater- m empty beaker m displacedwater = 70 . 7 g- 43 . 0 g = 27 . 7 g The weight of the water displaced is thus w displacedwater = m displacedwater g w displacedwater = (27 . 7 g) 9 . 8 m s 2 = (0 . 0277 kg) 9 . 8 m s 2 = 0 . 271 N So the weight of the displaced water is w displacedwater = 0 . 271 N 1.4 Conclusion The difference between the mass of the object in air and the apparent mass...

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