Physics 6B Practice Midterm Solutions

Physics 6B Practice Midterm Solutions - Physics 6B Practice...

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Physics 6B Practice Midterm #1 Solutions
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1. A block of plastic with a density of 920 kg/m 3 floats at the interface between oil of density 850 kg/m 3 and water of density 1000 kg/m 3 , as shown. Calculate the percentage of the plastic which is submerged in the water. interface oil water plastic The block is floating motionless, so the net force is 0. The total buoyant force includes contributions from the oil and the water, as shown in the free-body diagram. F b,water + F b,oil – weight = 0 The key to this problem is to write these forces in terms of densities: ( ) g V V g V F g V F g V mg weight wate o o oi o b water water water , b plastic ρ = ρ = ρ = ρ = = weight F b, water F b, oil Here V is the total volume of the plastic block. Putting these into our formula, and solving for V water: water oil oil oil oil , ( ) ( ) ( ) V % 47 V V 150 70 V 850 1000 850 920 V V V V V g V g V V g V water water oil water oil plastic water oil plastic water oil water plastic water oil water water = = ρ ρ ρ ρ = ρ ρ = ρ ρ ρ = ρ + ρ
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2. A U-shaped tube contains 10 cm of alcohol (ρ a =800 kg/m 3 ), 12 cm of oil (ρ o =850 kg/m 3 ), and water (ρ w =1000 kg/m 3 ) as shown. Find the distance h between the surface of the alcohol and the surface of the water. alcohol oil water 12 cm 10 cm h = ? 22cm-h The gauge pressure on either side must balance at the oil/water interface. On the left, we can add the partial pressures from the alcohol and the oil. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) cm 12 cm 10 cm 22 h ... ebra lg a ... h cm 22 cm 12 cm 10 g h cm 22 g cm 12 g cm 10 g oil alc water water oil alc water oil alc ρ ρ ρ = ρ = ρ + ρ ρ = ρ + ρ ( )( ) ( )( ) ( )( ) cm 8 . 3 1000 12 850 10 800 22 1000 h water = = ρ
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3. An incompressible fluid is flowing in a horizontal pipe. The radius of the pipe gradually decreases from 10 cm to 5 cm. The velocity of the fluid in the narrow portion will be: a) twice as large as in the wide section b) half as large as in the wide section c) the same as in the wide section d) four times as large as in the wide section Continuity dictates that the flow rate is the same throughout the pipe. We can use the formula Since the cross-section of the pipe is circular, 2 2 1 1 v A v A = 1 2 2 1 2 2 2 2 1 2 1 v r r v v r v r = π = π v 1 v 2 ( ) 1 2 1 2 1 2 2 v 4 v v 2 v cm 5 cm 10 v = = = Answer d) This is a problem that you can do quickly if you recognize that the
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This note was uploaded on 09/09/2011 for the course PHYSICS 6b taught by Professor Staff during the Spring '11 term at UCSB.

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Physics 6B Practice Midterm Solutions - Physics 6B Practice...

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