P15_032 - 32. (a) Since the lead is not displacing any...

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32. (a) Since the lead is not displacing any water (of density ρ w ), the lead’s volume is not contributing to the buoyant force F b . If the immersed volume of wood is V i ,then F b = ρ w V i g =0 . 90 ρ w V wood g =0 . 90 ρ w g µ m wood ρ wood , which, when floating, equals the weights of the wood and lead: F b =0 . 90 ρ w g µ m wood ρ wood =( m wood + m lead ) g. Thus, m lead =0 . 90 ρ w µ m wood ρ wood m wood = (0 . 90)(1000 kg / m 3 )(3 . 67 kg) 600 kg / m 3 3 . 67 kg = 1 . 84 kg 1 . 8kg . (b) In this case, the volume V
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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