144 Chapter 12 States of Matter 2-9 BERNOULLI'S EFFECT. Answers to the Questions

Transcription

1 144 Chapter 1 States f Matter -9 BERNOULLI'S EFFECT Gal Describe h the pressure arund an bject changes hen the fluid is mving. Cntent The pressure in a fluid decreases as its velcity increases. Teaching Tips As ith the last sectin, this sectin can be skipped ithut causing any prblems in the remainder f the bk. Its presence here gives the students (and prfessr) a "pay-ff" fr sme rather hard rk in the beginning f the chapter. There are many eamples f Bernulli's effect that range frm the fl f smke up a chimney t hat the big bad lf really did t the three little pigs' huse. We like t use the air supply frm ur air tracks t suspend a styrfam ball in mid-air. The ball remains in the air stream even if the hse is tilted. Clearly the air is nt keeping the ball up by simply supprting its eight. A simpler (but harder t demnstrate in a large class) demnstratin is t pick up a 5 card by bling thrugh a thread spl. A pin is pushed thrugh the card t keep it aligned ith the hle in the spl. If a tube is attached t the spl hle, yu can hld the spl ut aay frm yur face, making the demnstratin a little clearer. Physics n Yur On This is a simple eperiment f lifting a piece f paper by bling ver it. Physics n Yur On If the student has a reversible vacuum cleaner, it can be used t suspend a ping-png ball in mid-air. ide Encyclpedia 1 #1 Pitt Tube # Flettner Rtr # Curve Balls #4 Flating Ball in Air Jet #5 Suspended Plate in Air Jet #6 Suspended Parallel Cards #1 Bernulli s Principle Cmputer Animatins Active Figure Animatins are available n the Multimedia Manager Instructr s Resurce CD. They are rganized by tetbk chapter, and each animatin cmes ithin a shell that prvides infrmatin n h t use the animatin, eplratin activities, and a shrt quiz. Ansers t the Questins 1. Slid, liquid, gas, and plasma. The average kinetic energy f the mlecules f a liquid is larger because energy as required t melt the slid.. Density is an intrinsic prperty f aluminum and therefre must be the same fr bth samples. 4. They have the same densities because they are bth diamnds. 5. The less dense magnesium ill take up mre space. 6. The mre dense gld ill have mre mass fr the same vlume. 7. The gld atms must be clser tgether than the uranium atms. 8. The ater in the pp epands upn freezing. 9. The crystal structures are nt the same. We kn this because the salt has cubic symmetry hile ice has heagnal symmetry. 10. Mica essentially has a t-dimensinal structure hile salt's structure is three-dimensinal. 11. Diamnd has strng bnds in all directins; graphite has strnger bnds beteen atms in tdimensinal layers. 1. The crystal structure must have strng bnds in t-dimensinal planes but relatively eak bnds beteen these planes. 1. Slid ygen has a ler melting temperature than slid nitrgen.

2 Chapter 1 States f Matter Because the biling pint f liquid ygen is ler than that f liquid nitrgen, e assume that the bnding in liquid nitrgen is strnger. 15. The drp f ater uld assume a spherical shape because surface tensin minimizes the surface area and a sphere has the smallest surface area fr a given vlume. 16. The frces beteen the ater mlecules are strnger than the frces beteen the ater mlecules and the a mlecules. 17. The surface tensin alls the surface f the ater t rise ithut verfling. 18. Adding sap t ater reduces the surface tensin. 19. The ater mlecules are mre strngly attracted t the glass than t each ther. 0. The mercury atms are mre strngly attracted t each ther than t the glass. 1. Gas particles are neutral atms r mlecules, hereas the particles in a plasma are electrns and charged ins.. The an Allen radiatin belts are frmed f ins, that is, the state is a plasma.. The same frce must be applied t supprt yur eight. The reduced surface area in the case f the gravel requires greater pressure. 4. The upard frce must stay the same s a reduced pressure must result in a larger area, r tire print. 5. The vlume f the liquid ater is much less than the vlume f the vapr, hich results in much ler pressure inside the can. The inside pressure n lnger balances atmspheric pressure, creating a net inard frce n each all f the can. 6. The air eerts an equal upard frce n the bttm f the ckie sheet. 7. It is the atmspheric pressure pushing in n the half-spheres that hlds them tgether (nt the sucking frm the inside). Denver has a ler atmspheric pressure s feer hrses uld be needed. 8. The effect f the atmsphere is t prduce a small buyant frce. Reducing the atmspheric pressure ill result in a slight increase in the scale s reading. 9. Because f its altitude Denver uld alays be listed as a l-pressure regin. 0. The altimeter assumes that ler pressure is a result f greater altitude. A l-pressure eather system fls the altimeter int reprting a higher than actual altitude. 1. The pressure in a fluid is independent f surface area. Yur ears uld hurt the same in bth.. The bdy is nt crushed because the pressure inside the bdy increases t cmpensate fr the eternal pressure.. It ill be the same. The pressure in a fluid is independent f surface area. 4. The secnd barrel ill burst hen the ater in the small tube reaches 0 feet. The pressure in a fluid is independent f surface area. 5. Pressure in a fluid depends n the eight f the material abve yu. This uld be greater fr the denser cld ater. 6. Because the pressure is prprtinal t the density f the liquid, the scuba diver needs t g deeper in the less dense, fresh ater. 7. The atmspheric pressure determines the maimum height f the ater clumn; its eight per unit area is equal t the atmspheric pressure. 8. Less than 10 meters; the atmspheric pressure, hich is hat pushes the fluid up the stra, is less in Bzeman. 9. With the pump at the tp, the pressure that pushes is limited t the atmspheric pressure. With the pump at the bttm, yu can apply much higher pressures. 40. Liquid X must have ler density and therefre uld flat n ater. 41. The purple liquid must have a larger density than the clear liquid fr it t sink. 4. Because the gasline flats n tp f the ater, the density f gasline must be less than that f ater. 4. The bat ill flat higher in salt ater because it desn't have t sink as deep t displace its eight.

3 146 Chapter 1 States f Matter 44. The laded freighter must displace mre ater befre the buyant frce is equal t its eight. 45. By Netn s secnd la the buyant frce must balance the ship s eight in bth cases and is therefre the same. 46. By Archimedes principle, the buyant frce is equal t the eight f the displaced fluid. This ill be greater in the denser salt ater. 47. The slight reductin in mass uld cause yu t flat higher. The much larger reductin in vlume reduces the buyant frce and causes yu t sink. 48. The scuba diver's density decreases ith mre air in the lungs, resulting in a net upard frce that causes the diver t rise. 49. A blck f lead ill flat in mercury because lead is less dense than mercury. Gld ill sink in mercury. 50. The density f cpper is less than that f silver s the cpper blck flats higher. 51. The vlume f the submarine des nt change s the buyant frce is cnstant. Epelling the ater simply reduces the eight. 5. When she descends, the increased ater pressure causes the vest t cntract thus reducing the buyant frce. Because the eight des nt change, the net frce is n dnard. The equilibrium as unstable. 5. The buyant frce is the same because they bth displace the same vlume. 54. The buyant frce n the aluminum blck is greater because it displaces mre fluid. 55. When a 5-pund blck f ice flats, it displaces 5 punds f ater. When the ice melts int 5 punds f ater, it just fills in the hle. The level des nt change. 56. The ater level ges dn. When the gld bricks ere in the bat, they displaced a vlume f ater equal t the eight f the gld. Because gld is much mre dense than ater, the vlume displaced as much larger than the bricks vlume. Once the bricks are placed in the ater, the vlume displaced is the same as the vlume f the bricks. 57. The fast mving air creates ler pressure abve the dime, hich results in a net upard frce. 58. The reduced pressure due t the Bernulli effect pulls yu sideays. 59. S the balls ill "drp" ver the net and land n the table. 60. The fling ater reduces the air pressure at that spt. 61. The ind bling thrugh the base causes the air pressure t decrease and the higher pressure inside the building frces the drs pen. 6. In additin t the eakened arterial alls that caused the aneurysm, the sling f the bld fl causes the pressure t increase. Ansers t the Eercises M 84 g 8cm M 6.75 g 0. cm.7 g cm ; silver M 180 g 0. g cm 600 cm M 1000 g 10 cm.5 g cm ; smium 8. g cm ; nt gld

4 Chapter 1 States f Matter M D ( ) ( ) 6. M D ( ) ( ) = = 000 kg m 0.4 m = 100 kg = = 11. g cm cm =.9 g M 90 kg = = = 0.09 m D 1000 kg m M 48 g = = = = = D 6g cm 8cm and cm g cm = g = 1000 cm = 1 cm D 0.9 g cm 10. ( ) ( ) M = D = 0.9 g cm 10 cm = 90 g ater ice ice M 90 g = = = 90 cm D 1g cm ater 11. W Mg ( ) ( ) ( ) 1. = = = 1000 kg m 1 m 10 m 10 m s = 100,000 N W 100,000 N P = = = 100,000 Pa ; same as atmspheric pressure. A 1m W 100,000 N h = = = 0.75 m DAg ( 1,600 km m )( 1 m )( 10 m s ) 1 1. h = ( 0 in. ) = 7.5 in. DHg 1.6 g cm 14. h = hhg = ( 0 in. ) 408 in. 4f = = t D 1g cm 15. Imagine a clumn f mercury ith a crss-sectin f 1 in. Then the pressure is W ( 1 in. ) ( 0 in. ) ( 0.5 lb in. ) P = = = 15 lb in. = 15 psi A 1in. = Mg 1000 kg m 150 m 10 m s N m A = A = = = Because this is apprimately 10 atm, atmspheric pressure is nt very imprtant. 16. P Dhg ( )( )( ) 17. M 150 kg 0. m 750 kg m ; flat M 1000 kg kg m ; flat 1.6 m 19. The blck must displace an amunt f ater equal t its eight. W = mg = 0.5 kg 10 m s = 5 N ( ) ( ) 0. The aluminum sinks and s must displace an amunt f ater equal t its vlume. W = mg = = 1g cm 400cm 10m s = 4N ( ) ( ) ( ) 6 1. Find mass hen it flats. It equals mass f ater displaced, hich is 18 g. Find vlume hen it sinks. It equals vlume displaced, hich is 0 cm.

5 148 Chapter 1 States f Matter M 18 g 0.9 g cm 0 cm. In ater, the buyant frce is enugh t balance the eight. In gasline, an additinal frce (a nrmal frce frm the bttm f the beaker) is required s the buyant frce is less.. The buyant frce equals the eight f the displaced ater, hich is (1000 kg) (10 m/s ) = 10,000 N. T = W F B = 890 kg 10 m s 10,000 N = 79,00 N ( ) ( ) 4. Wnet Wball Water ( ) ( ) = = 0.00 kg kg 10 m s = 0.5 N W E,b T c,b F buyant N t,b F buyant W E,b Ansers t the Prblems in Prblem Slving M 96 g 0.96 g cm = 96 kg m 1000 cm M 1.58 kg 1 L D = = 790 kg/m 0.798/cm L = = 10 m 6 M M kg = 6 kg m = 0.6 g cm ; flat πr π ( 6 10 m) 0 M M 1.4 ( kg) 17 D = kg m 4 4 = πr = π 16,000 m = ( ) 5. M D ( ) ( ) = = 19. g cm 0.05 cm = g 6. ( ) ( ) M = D = D =.7 g cm 4 cm = 17 g M 1000 g = = = 7.5 cm D 1.6 g cm M 0 g = = =.65 cm D 11. g cm k ( 1kg) ( 9.8m s ) = F mg = = 0.06 m = ( 45 kg) ( 9.8 m s ) 16 N m F mg k = 4410 N m = = 0.1 m =

6 Chapter 1 States f Matter k ( 1.5 kg) ( 9.8 m s ) = F mg m.68 cm = = 400 N m = = ( 9kg) ( 9.8m s ) F mg k = m 5.88 cm = = 1500 N m = = ( 4kg) ( 9.8m s ) W = = = k 100 N m t tal stretch = = m k eff W k = = = 50 N m 0.9 m kn 14. W = k = knn = kn = kn = k here k n is the spring cnstant f the cmbinatin and n is the crrespnding stretch. 15. P h ( ) ( ) ( ) = = = kg m 9.8 m s 1 m N m 16. P h ( ) ( ) ( ) = = = kg m 9.8 m s 1000 m N m Because atmspheric pressure is abut 1% f this, it is nt imprtant. 5 P 5 atm 1atm N m h = = =.0 m 1atm ( 1,600 kg m )( 9.8 m s ) 5 P 5 atm 1atm N m h = = = 41. m 1atm ( 1000 kg m )( 9.8 m s ) 19. P = h = ( 1000 kg m ) ( 9.8 m s ) ( 0 m) = N m ( abve atm) 0. P = h = ( 680 kg m )( 9.8 m s )( 8 m) = N m ( abve atm) 1. P ( ) ( ) ( ) = 1000 kg m 9.8 m s 5 m = 49,000 Pa atm 1 Ttal pressure at 0 m is abut 4 atm. Rising t 5 m is a 1.5 % decrease. Ttal pressure at 5 m is abut 1.5 atm. Rising t 0 m is a % decrease, hich is much mre dangerus.. ( ) ( ) ( ) ( ) ( ) ( ) 5 Pressure at 0 m = 1000 kg m 9.8 m s 0 m + 101,000 Pa = Pa 5 Pressure at 0 m = 1000 kg m 9.8 m s 0 m + 101,000 Pa = Pa 5 P Pa 0 = 0 = 5 ( 4 L) =.01 L P Pa Buyancy reduced by eight f 1 L f ater = 9.8 N prducing a net frce f 9.8 N dnard.. = ( ) F D D g s g ( ) kg kg 1 kg = 19, m s 9.9 N = m m 19,00 kg m

7 150 Chapter 1 States f Matter 7.86 kg kg 4. Fs ( D D) g ( )( ) al m = = m 9.8 m s = 67. N m m W 0.9 N 0.6 N 5 = = = m = 0.6 cm 1000 kg m 9.8 m s ( )( ) W 40 N = = = = m = 4080 cm ( 1000 kg m )( 9.8 m s ) ( al Dater ) ( 9.8 N/kg)( 700 kg/m 1000 kg/m ) 4 ( 700 kg/m )( 6 10 m ) 1.6 kg W 10 N = = = 6 10 m g D = D = = al al al M.7 kg = = =.7 10 m = 700 cm D 1000 kg m M 65 g = = = = = D 65 cm Frac g cm 65 4 M 7 kg = = = = = D.7 10 m Frac kg m 0.1