qz10sol_7193f10 - 2 . a) If the alligator were to swallow...

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TA: Tomoyuki Nakayama Monday, November 15, 2010 PHY 2048: Physic 1, Discussion Section7193 Quiz 10 (Homework Set #12) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ Lurking alligators. An alligator waits for prey by floating with only the top of its head exposed, so that the prey cannot easily see it. One way it can adjust the extent of sinking is by controlling the size of its lungs. Another way may be by swallowing stones (gastrolithes) that then reside in the stomach. The figure below right shows a highly simplified model (a ''rhombohedron gater'') of mass 120 kg that roams with its head partially exposed. The top head surface has area 0.400 m
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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.

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