kean (msk955) – Homework 09 – yao – (56725) 1This print-out should have 24 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsA 44.1 kg wo man balances on one heel of apair o f hi gh-heeled shoes.If the heel is circular with radius 0.26 cm,what pressure does she exert on the floor?The acceleration of gravity is 9 .8 m/ s2.Correct answer: 2.03502 × 107N/m2.Explanation:Let : m = 44.1 kg ,g = 9.8 m/s2, andr = 0.26 cm .The pressure is equal to the force she exertson the floor (which is equal to her weight)divided by the area of the heel:P =m gπ r2=(44.1 kg) (9.8 m/s2)π (0.26 cm)2·100 cm1 m2=2.03502 × 107N/m2.keywords:002 10.0 pointsWhat must be the contact area between a suc-tion cup (completely exhausted) and a ceil ingin order to support the weight of a 34.1 kg stu-dent? The acceleration of gravity is 9.8 m/s2.Correct answer: 0.00329891 m2.Explanation:Let : m = 34.1 kg ,g = 9.8 m/s2, andPatm= 1.013 × 105Pa .Since the suction cup is exhausted, the forceon it given by the atmospheric pressure is F =PatmA, where A is the area of the suction cup.In order to support the student, this forcemust be equal in magnitude to the student’sweight, soA =m gPatm=(34.1 kg) (9.8 m/s2)1.013 × 105Pa= 0.00329891 m2.003 10.0 pointsDetermine the absolute pressure at the bot-tom of a l ake that is 20.7 m deep. The accel-eration of gravity is 9.8 m/s2and atmosphericpressure is 1.01 × 105Pa .Correct answer: 3.0386 × 105Pa.Explanation:Let : g = 9.8 m/s2,Patm= 1.01 × 105Pa ,h = 20.7 m , andρw= 1000 kg/m3.The pressure at the bottom of the lake isequal to the pressure at the surface plus thepressure given by the mass of water above, soP = Patm+ ρwg h= 1.01 × 105Pa+ (1 000 kg/m3) (9.8 m/s2) (20.7 m)=3.0386 × 105Pa .004 10.0 pointsA collapsible plastic bag contains a glucosesolution.hGlucosesolutionkean (msk955) – Homework 09 – yao – (56725) 2If the average gauge pressure in the vein is13800 Pa, what must be the minimum heightof the bag in order to infuse glucose into thevein? Assume that the specific gr avity of thesolution is 1.01. The acceleration of gravity is9.8 m/s2.Correct answer: 1.39422 m.Explanation:Let : Pgauge= 13800 Pa andρ = 1.01 ρwater= 1010 kg/m3.The gauge pressure of the fluid at the levelof the needle must equal the gauge pressurein the vein:Pgauge= ρ g hh =Pgaugeρ g=13800 Pa(1010 kg /m3) (9.8 m/s2)=1.39422 m .005 10.0 pointsA simple U-tube that is op en at both ends ispartially filled with a heavy liquid of density1000 kg/m3. A liquid of density 533 kg/m3isthen poured into one arm of the tube, formi nga col umn 11 cm in height, as shown.h11 cmlight liquid533 kg/m3heavy liquid1000 kg/m3What is the difference in the heights of thetwo li quid surfaces?Correct answer: 5.137 cm.Explanation:Let : ℓ = 11 cm ,ρℓ= 533 kg/m3, andρh= 1000 kg/m3.Because the liquid in the U-t ube is stat ic,the pressure exerted by the heavy liquid col-umn of height ℓ − h in the left branch of thetube must balance the pressure exerted bythe liquid of height h poured into the rightbranch, soP0+ (ℓ − h) ρhg = P0+ ℓ ρℓg .h = ℓ1 −ρℓρh= (11 cm)1 −533 kg/m31000 kg/m3=5.137 cm .006 10.0 pointsOne method of measuring the density of a liq-uid is illustrated in the figure. One side of t heU-tube is in the liq ui d being tested, and theother side is in water of density 1000 kg/m3.The air is partiall y removed at the upperpart of the tube and the valve is closed. Theheight of the water above its pool surface is0.8 m . The height of the li quid above its poolsurface is 0.48 m . The di fference in the heightsof the pool surfaces is 0 .17 m .⊗Valvetest liquidwa ter0.48 m0.8 m0.17 mkean (msk955) – Homework 09 – yao – (56725) 3Find the density of the li quid on the left.Correct answer: 1666.67 kg/m3.Explanation:Let : ρw= 1000 kg/m3,hw= 0.8 m ,h = 0.48 m , and∆h = 0.17 m .The pressure at the upper surface of eachliquid is given byP = Patm− ρwg hw= Patm− ρ g h.Therefore,ρ =hwhρw=(0.8 m)(1000 kg/ m3)0.48 m=1666.67 kg/m3.007 10.0 pointsThe small piston of a hydraulic lift has across-sectional area of 1.2 cm2and the largepiston has an area of 60 cm2, as in the figurebelow.F1.2 cm2area60 cm2What force F must be applied to the smallpiston to maintain the load of 55 kN at aconstant elevation?Correct answer: 1100 N.Explanation:Let : A1= 1.2 cm2,A2= 60 cm2, andW = 55 kN .According to Pascal’s law, t he pressure ex-erted o n A1must be equal to the one exertedon A2. The pressure P1=FA1must be equalto the pressure P2=WA2due t o the load.FA1=WA2,F =A1A2W =(1.2 cm2)(60 cm2)(55000 N)=1100 N .008 10.0 pointsSuppose that a volleybal l A floats on the wa-ter, and a bowling ball B, being denser thanwa ter, is completely submerged in water. As-sume they have t he same volume.ABWhich feels a greater buoyant force?1. bowling ball B correct2. volleyball A3. Unable to determine4. They feel the same buoyant force.Explanation:The bowling ball feels a greater buoyantforce because it displaces more water.009 (part 1 of 2) 10.0 pointsA beaker of mass 1.2 kg containing 2.3 kg ofwa ter rests on a scale. A 3.9 kg block of ametallic alloy of density 3900 kg/m3is sus-pended from a spring scale and is submergedkean (msk955) – Homework 09 – yao – (56725) 4in the water of density 1000 k g/m3as shownin the figure.3.9 kgWhat does the hanging scale read? Theacceleration of gravity is 9.8 m/s2.Correct answer: 28.42 N.Explanation:Let : mb= 1.2 kg ,mw= 2.3 kg ,ma= 3.9 kg ,ρa= 3900 kg/m3,ρw= 1000 kg/m3, andg = 9.8 m/s2,The buoyant force on the metallic alloy isthe weight of the water displaced by the alloy.The volume of the al loy is given by V =maρaand B = ρwV g = ρwmag1ρa, so the forceapplied to the upper scale isF = W − B= mag − magρwρa= (3.9 kg) (9.8 m/s2)1 −1000 kg/m33900 kg/m3=28.42 N .010 (part 2 of 2) 10.0 pointsWhat does the lower scale read?Correct answer: 44.1 N.Explanation:The total weight is supported by bothscales, soF + F2= (mb+ mw+ ma) gF2= (mb+ mw+ ma) g − F= (1.2 kg + 2.3 k g + 3.9 kg) (9.8 m/s2)− 28.42 N=44.1 N .011 10.0 pointsAn iceberg i s floating on sea water such thatonly 10% of its volume is above the waterlevel.What is the density of sea water expressedin terms of the density ρiof ice?1.19ρi2.1 −910ρi3.910ρi4.1 −109ρi5.110ρi6. ρi7.109ρicorrectExplanation:Let : f = 10% = 0.1 .Denote the volume of the wa ter displaced bythe ice