qz9sol_3705s11

# qz9sol_3705s11 - V B = m B ρ B = 8.33 × 10-3 m 3 Since...

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TA: Tomoyuki Nakayama Monday, April 4, 2011 PHY 2048: Physic 1, Discussion Section 3705 Quiz 9 (Homework Set #11) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ A block of wood has a mass of 5.00 kg and a density of 600 kg/m 3 . It is to be loaded with a piece of metal (8.00 × 10 4 kg/m 3 ) so that it will float in water with 0.850 of its volume submerged. a) What mass of metal is needed if the metal is attached to the top of the wood? The volume of the block is
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Unformatted text preview: V B = m B / ρ B = 8.33 × 10-3 m 3 Since the metal is above the surface of water, buoyancy acts only on the block. Newton’s 2 nd law applied to the metal-block system leads t9 B B – W B – W M = 0 ⇒ ρ W (0.85V B )g – ρ B V B g – m M g = 0 m M = (0.85 ρ W – ρ B )V B = 2.08 kg b) What mass of the metal is needed if the metal is attached to the bottom of the wood? Now the metal is under the surface of water, buoyancy acts on the metal too. Newton’s 2 nd law yields the volume of the metal piece. B B + B M – W B – W M = 0 ρ W (0.85V B )g + ρ W V M g – ρ B V B g – ρ W V M g = 0 V M = (0.85 ρ W – ρ B )V B /( ρ M – ρ W ) = 2.636 × 10-5 m 3 The mass of the metal is m M = V M ρ M = 2.11 kg...
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