lab07 - Chapter 13 Fluids at Rest 13.2 Archimedes'...

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T 0.154newton = or T m Al g BF - = The tension in the supporting cord plus the buoyant force must equal the weight of the block if it is to be in equilibrium (see figure above). That is T + BF = mg, from which BF 0.091newton = or BF V ρ w g = The block displaces a volume V of water when submerged, so the buoyant force on it is (b) V 9.26cm 3 = or V m Al ρ Al = Because ρ = m/V , we have (a) Solution ρ w 1000 kg m 3 = ρ Al 2700 kg m 3 = Densities: m Al 25 gm = Mass of aluminum: System Parameters The mass of a block of aluminum is m Al = 25 gm. (a) What is its volume? (b) What will be the tension in a string that suspends the block, when the block is totally submerged in water? The density of aluminum is ρ Al = 2700 kg/m 3 . Schaum's Problem 13.14 Archimedes' Principle 13.2 Chapter 13 Fluids at Rest
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m w 73 gm = Density of water: ρ w 1000 kg m 3 = Solution The figure above shows the situation when the object is in water. From the figure, BF + T = mg , so BF m alloy g m w g - = or BF 0.13newton = . But
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lab07 - Chapter 13 Fluids at Rest 13.2 Archimedes'...

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