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Unformatted text preview: 2 . a) If the alligator were to swallow stones with a total mass of 1.50% of its body mass, what would be the change in buoyancy on the alligator? The alligator of mass M is in equilibrium before it swallows stones. The balance of forces equation yields B – Mg = 0 ⇒ B = Mg The alligator reaches equilibrium again after swallowing stones of mass Δ M = 0.015M. Thus we have B’ – (M + Δ M)g = 0 B’ = (M + Δ M)g The change in buoyancy is Δ B = B’ – B = Δ Mg = 0.015Mg = 17.6 N b) How far would it sink? According to Archimedes’ principle, the buoyancy on an object in a fluid is equal to the weight of the fluid displaced by the object. Therefore, Δ B = ρ W Δ Vg = ρ W (A Δ h)g Δ h = Δ B/( ρ W Ag) = 4.49 × 10-3 m = 4.49 mm...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
- Summer '08