9 Solids and Fluids PROBLEM SOLUTIONS 9. The elastic liit is the axiu stress, FA where F is the tension in the wire, that the wire can withstand and still return to its iginal length when released. Thus, if the wire is to experience a tension equal to the weight of the perfer without exceeding the elastic liit, the iniu cross-sectional area is A in D in F g 4 elastic liit elastic liit and the iniu acceptable diaeter is D 5.0 0 Pa 4g 4 70 kg 9.8 s elastic liit in 8. 0. 9. (a) In der to punch a hole in the steel plate, the superhero ust punch out a plug with cross- sectional area, A, equal to that of his fist and cs a height t equal to the thickness of the steel plate. The area Ashear of the face that is sheared as the plug is reoved is the cylindrical surface with radius r and height t as shown in the sketch. Since A cs = r, then r Acs and A cs.00 0 c Ashear r t t.00 c 70.9 c If the ultiate shear strength of steel (i.e., the axiu shear stress it can withstand befe shearing) is,.50 08 Pa.50 08 N the iniu fce required to punch out this plug is N 0 c shear 70.9 c 8.50 0 4 F A stress.77 06 N Page 9.
By Newton s third law, the wall would exert a fce of equal agnitude in the opposite direction on the superhero, who would be thrown backward at a very high recoil speed. 9. Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of the plank, separated by distance h.00 and each with area, A.00 c 5.0 c 0.0 c ove a distance x 5.00 0 parallel to each other. The fce causing this shearing effect in the plank is the weight of the an F by g applied perpendicular to the length of the plank at its outer end. Since the shear odulus S is given S shear stress F A Fh shear strain x h x A we have S 80.0 kg 9.80 s.00 4 5.00 0 0.0 c 0 c.05 07 Pa 9.4 As a liquid, the water occupied soe volue V l. As ice, the water would occupy volue.090v l if it were not copressed and fced to occupy the iginal volue. Consider the pressure change required to squeeze ice back into volue V l. Then, V0.09 Vl and V 0.090V l, so V N 0.090 V P B.00 0.65 0 Pa 600 at V V 0 9 l 8.09 l 9.5 Using Y F L0 A( L) with A d 4 and F g, we get Y.0 0.6 4 90 kg 9.80 s 50.5 08 Pa 9.6 Fro Y F L0 A( L) the tension needed to stretch the wire by 0.0 is F Y A L Y d L L 4 L 0 0 8 0 0 Pa 0. 0 0.0 0 4. 0 N Page 9.
The tension in the wire exerts a fce of agnitude F on the tooth in each direction along the length of the wire as shown in the above sketch. The resultant fce exerted on the tooth has an x-coponent of R F Fcos 0 Fcos 0 0, and a y-coponent of R F Fsin 0º Fsin 0º F N. x x y y Thus, the resultant fce is R N directed down the page in the diagra. 9.7 FroY ( F A)( L L) ( stress)( L L), the axiu copression the feur can withstand is 0 0 stress L 60 0 6 Pa 0.50 0 9 L 4.4 0 4.4 Y 8 0 Pa 9.8 (a) When at rest, the tension in the cable equals the weight of the 800-kg object, 7.84 0 N. Thus, fro Y F L0 A( L), the initial elongation of the cable is 7.84 0 N 5.0 4 0 4.00 0 0 0 Pa F L0 L.45 0 =.5 A Y When the load is accelerating upward, Newton s second law gives F g a y F g a y [] If 800 kg and a.0 s y, the elongation of the cable will be 800 kg9.80.0 s 5.0 4 0 4.00 0 0 0 Pa F L 0 L. 0. A Y Thus, the increase in the elongation has been increase L L initial.0.45 0.75 (c) Fro the definition of the tensile stress, stress F A, the axiu tension the cable can withstand is Page 9.
ax F A stress ax 4.00 0 4. 08 Pa 8.8 04 N Then, Equation [] above gives the ass of the axiu load as F 4 ax ax g a 8.8 0 N 9.8.0 s 6.9 0 kg 9.9 Fro the defining equation f the shear odulus, we find the displaceent, x, as S S A 6.0 0 Pa 4 c h F A h F 5.0 0 0 N 0 4 c x.4 05 0.04 9.0 The shear odulus is given by S shear stress stress shear strain x h Hence, the stress is stress x 0 5.0 S 6.5 0 Pa 7.5 0 Pa h 0 0 9. The tension and cross-sectional area are constant through the entire length of the rod, and the total elongation is the su of that of the aluinu section and that of the copper section. F L L L F L F Lrod LAl LCu AY AY A Y Y 0 Al 0 Cu 0 Al 0 Cu Al Cu Al Cu where A r with r 0.0 c.0 0. Thus, 5.8 0 N..6 Lrod 7.0 0 0 Pa 0 0.0 0 Pa.9 0.9 c Page 9.4
9. The acceleration of the fear has agnitude a k 0 h 80 v h k 600 s 4.4 0 s t 5.0 0 s The copression fce exerted on the ar is F a and the copressional stress on the bone aterial is Stress.0 kg 4.4 0 s.4 c 4 0 c F A 5.5 07 Pa Since the stress is less than the allowed axiu, the ar should survive. 9. The average density of either of the two iginal wlds was M M M 0 V 4R 4R The average density of the cobined wld is so Mtotal M 4 M M V 4 R 9 R R 4 0 M 4R 8 9R M 7 4.74 4.740 9.4 (a) The ass of gold in the coin is # karats total Au total 7.988 0 kg 7. 0 kg 4 4 and the ass of copper is 7.988 0 kg 6.657 0 4 kg total Cu Page 9.5
The volue of the gold present is V 7. 0 kg Au 7 Au.79 0 Au 9. 0 kg and the volue of the copper is V 6.657 0 kg 4 Cu 8 Cu 7.46 0 Cu 8.9 0 kg (c) The average density of the British sovereign coin is 7.988 0 kg total total 4 av.76 0 kg V 7 8 total VAu VCu.79 0 7.46 0 9.5 (a) The total nal fce exerted on the botto acrobat s shoes by the flo equals the total weight of the acro bats in the tower. That is n g total 75.0 68.0 6.0 55.0 kg 9.80 s.55 0 N n n.55 0 N P A 4 total Ashoe 45 c 0 c sole.00 04 Pa (c) If the acrobats are rearranged so different ones are at the botto of the tower, the total weight suppted, and hence the total nal fce n, will be unchanged. However, the total area Atotal Ashoe sole, and hence the pressure, will change unless all the acrobats wear the sae size shoes. 9.6 We shall assue that each chair leg suppts one-fourth of the total weight so the nal fce each leg exerts on the flo is n g 4. The pressure of each leg on the flo is then P 95.0 kg 9.80 s 4 0.500 0 n g 4 leg A leg r.96 06 Pa 9.7 (a) If the particles in the nucleus are closely packed with negligible space between the, the average nuclear Page 9.6
density should be approxiately that of a proton neutron. That is 7 proton proton 7 nucleus 4 0 kg V proton r 5.67 0 kg 4 4 0 The density of iron is Fe 7.86 0 kg kg/ and the densities of other solids and liquids are on the der of 0 kg. Thus, the nuclear density is about 04 ties greater than that of coon solids and liquids, which suggests that atos ust be ostly epty space. Solids and liquids, as well as gases, are ostly epty space. 9.8 Let the weight of the car be W. Then, each tire suppts W 4, and the gauge pressure is P F A W 4 A Thus, W A P 4 4 0.04.0 05 Pa.9 04 N. 9.9 The volue of concrete in a pillar of height h and cross-sectional area A is V Ah, and its weight is Fg Ah 4 5.0 0 N. The pressure at the base of the pillar is then P 5.0 04 N F g Ah h5.0 0 N A A 4 Thus, if the axiu acceptable pressure on the base is, P 7 ax.7 0 Pa, the axiu allowable height is h P.7 0 Pa 5.0 0 N 5.0 0 N 7 ax ax 4 4.4 0 9.0 Assuing the spring obeys Hooke s law, the increase in fce on the piston required to copress the spring an additional aount of a fluid is x is F F F P P A k x. The gauge pressure at depth h beneath the surface P P gh 0 0 0, so we have, gha k x the required depth is k 50 N, A r with r.0 0, with.00 0 kg h k x ga, If, and the fluid is water ( =.00 0 kg/ ), the depth required to copress the spring an additional is 0.750 c 7.50 0 is Page 9.7
h 50 N 7.50 0.00 0 kg 9.80 s.0 0. 9. (a) P P gh 5 0 0. 0 Pa.00 0 kg 9.80 s 7.5.7 0 Pa The inward fce the water will exert on the window is 5 5.0 0 4 F PA P r.7 0 Pa.57 0 N 9. The gauge pressure in a fluid at a level distance h below where Pgauge 0 is Pgauge when easured in the downward direction. The difference in gauge pressures at two levels is gauge gauge gh with h being positive ( P ) ( P ) g ( h) (P gauge ) = (P gauge ) + g (h) with h being positive if, in going fro level to level, one is oving lower in the fluid. (a) In oving fro the heart to the head, one is oving higher in the blood colun so h 0 and we find P gauge Pgauge g h. 0 Pa 060 kg 9.80 s 0.500 head heart P gauge 8. 0 Pa 8. kpa head In going fro heart to feet, one oves deeper in the blood colun, so h 0 and P gauge Pgauge g h. 0 Pa 060 kg 9.80 s.0 feet heart P gauge 6.8 0 Pa 6.8 kpa feet 9. The density of the solution is.0.0 0 kg. The gauge pressure of the fluid at the level water of the needle ust equal the gauge pressure in the vein, so P gauge gh. 0 Pa, and Page 9.8
h P gauge. 0 Pa g.0 0 kg 9.80 s 0. 9.4 (a) Fro the definition of bulk odulus, B P V V 0, the change in volue of the.00 of seawater will be P.00. 0 8 Pa.0 0 5 Pa V V 0 0.05 8 B 0 water 0.0 0 Pa The quantity of seawater that had volue V 0.00 at the surface has a ass of 00 kg. Thus, the density of this water at the ocean flo is 00 kg.09 0 kg V V V.00 0.05 8 0 (c) Considering the sall fractional change in volue (about 5%) and enous change in pressure generated, we conclude that it is a good approxiation to think of water as incopressible. 9.5 We first find the absolute pressure at the interface between oil and water. P P gh 0 oil oil.0 05 Pa 700 kg 9.80 s 0.00.0 05 Pa This is the pressure at the top of the water. To find the absolute pressure at the botto, we use, P = P + water gh water, P 5 5.0 0 Pa 0 kg 9.80 s 0.00.05 0 Pa 9.6 If we assue a vacuu exists inside the tube above the wine colun, the pressure at the base of the tube (that is, at the level of the wine in the open container) is P ato = 0 + gh = gh. Thus, P 5 ato.0 0 Pa h g 984 kg 9.80 s 0.5 Page 9.9
Soe alcohol and water will evapate, degrading the vacuu above the colun. 9.7 Pascal s principle, F A F A, Fpedal AMaster Fbrake Abrake cylinder cylinder, gives A brake cylinder 6.4 c brake Fpedal 44 N 56 N A aster cylinder.8 c F This is the nal fce exerted on the brake shoe. The frictional fce is f n 0.50 56 N 78 N k and the tque is f r dru 78 N 0.4 7 N. 9.8 First, use Pascal s principle, F /A = F /A, to find the fce piston will exert on the handle when a 500-lb fce pushes downward on piston. A d 4 d F F F F A d 4 d 0.5 in.5 in 500 lb 4 lb Now, consider an axis perpendicular to the page, passing through the left end of the jack handle. = 0 yields F 4 lb.0 in in 0, F. lb 9.9 When held underwater, the ball will have three fces acting on it: a downward gravitational fce, g; an upward buoyant fce, B = water V = 4 water r /; and an applied fce, F. If the ball is to be in equilibriu, we have (taking upward as positive) F F B g 0, y 4r 4r F g B g water g water g giving Page 9.0
F 4 0.50 0.540 kg.00 0 kg 9.80 s 74.9 N so the required applied fce is. F 74.9 N directed downward 9.0 (a) To float, the buoyant fce acting on the person ust equal the weight of that person, the weight of the water displaced by the person ust equal the person s own weight. Thus, B g gv gv sea suberged body total V suberged V total body sea After inhaling, V suberged V total 945 kg 0.768 76.8% 0 kg leaving..% above surface. After exhaling, V suberged V total 00 kg 0.89 8.9% 0 kg leaving.7.% above surface. In general, sinkers would be expected to be thinner with heavier bones, whereas floaters would have lighter bones and e fat. 9. The boat sinks until the weight of the additional water displaced equals the weight of the truck. Thus, wtruck water V g kg 0 4.00 6.00 4.00 0 9.80 s w truck 9.4 0 N 9.4 kn Page 9.
9. (a) Since the syste is in equilibriu, F B w w 0. y r (c) B wgvsuberged wg d A 05 kg 9.80 s 0.04 0 4.00 964 N (d) Fro B w w r 0, w B w B g 964 N 6.0 kg 9.80 s 56 N r s (e) w g 56 N r r foa V r t A 9.80 s 0.090 4.00 0 kg (f) Bax wgvr wg t A 05 kg 9.80 s 0.090 0 4.00.6 0 N (g) The axiu weight of survivs the raft can suppt is wax ax g Bax wr so B w r g 9.80 s ax.6 0 N 56 N ax kg 9. (a) While the syste floats, When steel 0.0 kg, B wtotal wblock w. steel Vsuberged V 5.4 0 w g Vsuberged b 4 b giving gv g. b steel b wvb 0.0 kg V b steel steel w.00 0 kg 408 kg V 5.4 0 4 b If the total weight of the block + steel syste is reduced, by having steel 0.0 kg, a saller buoyant fce is needed to allow the syste to float in equilibriu. Thus, the block will displace a saller volue of water and will be only partially suberged in the water. The block is fully suberged when steel = 0.0 kg. The ass of the steel object can increase slightly above this value without causing it and the block to sink to the botto. As the ass of the steel object is gradually in- Page 9.
creased above 0.0 kg, the steel object begins to suberge, displacing additional water, and providing a slight increase in the buoyant fce. With a density of about eight ties that of water, the steel object will be able to displace approxiately 0.0 kg/8 = 0.09 kg of additional water befe it becoes fully suberged. At this point, the steel object will have a ass of about 0.49 kg and will be unable to displace any additional water. Any further increase in the ass of the object causes it and the block to sink to the botto. In conclusion, the block + steel syste will sink if stee 0.50 kg. 9.4 (a) Since the balloon is fully suberged in air, V suberged = V b = 5, and B gv b air.9 kg 9.80 s 5 4. 0 N (c) Fy B wb whe B bg HegVb B b HeV g 4. 0 N 6 kg 0.79 kg 5 9.80 s. 0 N Since F y = a y 0, a y will be positive (upward), and the balloon rises. F B w w w and (d) If the balloon and load are in equilibriu, y b He load 0 wload B wb whe. 0 N. Thus, the ass of the load is w g 9.80 s load. 0 N load 6 kg (e) If load 6 kg, then the net fce acting on the balloon + load syste is upward and the balloon and its load will accelerate upward. (f) The density of the surrounding air, teperature, and pressure all decrease as the balloon rises. Because of Page 9.
these effects, the buoyant fce will decrease until at soe height the balloon will coe to equilibriu and go no higher. 4r 4 air balloon air 9.5 (a) B gv g.9 kg 9.80 s.00.4 0 N.4 kn Fy B wtotal.4 0 N 5.0 kg 9.80 s.8 0 N.8 kn upward (c) The balloon expands as it rises because the external pressure (atospheric pressure) decreases with increasing altitude. 9.6 (a) Taking upward as positive, F B g a, y y a gv g. y w Since = V, we have V a y w gv V g, ay w g.00 0 kg 050 kg (c) a y 9.80 s 0.467 s 0.467 s downward (d) Fro y v t a t with 0y = 0, we find 0y y t y 8.00 a 0.467 s y 5.85 s Page 9.4
9.7 4 (a) Btotal 600 Bsingle 600 air gvballoon 600 air g r balloon 4 600.9 kg 9.80 s 0.50 4.0 0 N 4.0 kn Fy B g total total 4.0 0 N 600 0.0 kg 9.8 s. 0 N. kn (c) Atospheric pressure at this high altitude is uch lower than at Earth s surface, so the balloons expanded and eventually burst. 9.8 Note: We deliberately violate the rules of significant figures in this proble to illustrate a point. (a) The absolute pressure at the level of the top of the block is P P gh top 0 water top 5 kg.00 0 Pa 0 9.80 5.00 0 s.079 05 Pa and that at the level of the botto of the block is P P gh botto 0 water botto 5 kg.00 0 Pa 0 9.80 7.0 0 s.097 05 Pa Thus, the downward fce exerted on the top by the water is 5 Ftop Ptop A.079 0 Pa 0.00 07.9 N and the upward fce the water exerts on the botto of the block is 5 Fbot Pbot A.097 0 Pa 0.00 09.7 N The scale reading equals the tension, T, in the cd suppting the block. Since the block is in equilibriu, F T F F g, y bot top 0 Page 9.5
T 0.0 kg 9.80 s 09.7 07.9 N 86. N (c) Fro Archiedes s principle, the buoyant fce on the block equals the weight of the displaced water. Thus, B V g water block 0 kg 0.00 0.0 9.80 s.8 N Fro part (a), F F above. bot top 09.7 07.9 N.8 N, which is the sae as the buoyant fce found 9.9 Constant velocity eans that the subersible is in equilibriu under the gravitational fce, the upward buoyant fce, and the upward resistance fce: F a y y 0.0 04 kg g sea watergv 00 N 0 where is the ass of the added sea water and V is the sphere s volue. Thus, kg 4 00 N.0 0.50.0 04 kg 9.80 s.67 0 kg 9.40 At equilibriu, F B Fspring g 0, so the spring fce is y Fspring B g water Vblock g where V 5.00 kg 7.69 0 block wood 650 kg Page 9.6
Thus, F spring 0 kg 7.69 0 5.00 kg 9.80 s 6.4 N. The elongation of the spring is then Fspring 6.4 N x 0.65 6.5 c k 60 N 9.4 (a) The buoyant fce is the difference between the weight in air and the apparent weight when iersed in the alcohol, B = 00 N 00 N = 00 N. But, fro Archiedes s principle, this is also the weight of the displaced alcohol, so B = ( alcohol V)g. Since the saple is fully suberged, the volue of the displaced alcohol is the sae as the volue of the saple. This volue is V B 00 N.46 0 g alcohol 700 kg 9.80 s The ass of the saple is weight in air 00 N g 9.80 s 0.6 kg and its density is 0.6 kg.0 0 kg V.46 0 9.4 The difference between the weight in air and the apparent weight when iersed is the buoyant fce exerted on the object by the fluid. (a) The ass of the object is weight in air 00 N g 9.80 s 0.6 kg The buoyant fce when iersed in water is the weight of a volue of water equal to the volue of the object, B w = ( w V)g. Thus, the volue of the object is Page 9.7
V Bw 00 N 65 N.57 0 g w 0 kg 9.80 s and its density is 0.6 kg 8.57 0 kg V.57 0 object The buoyant fce when iersed in oil is equal to the weight of a volue V =.57 0. of oil. Hence B oil = ( oil V)g, the density of the oil is B 00 N 75 N oil oil 74 kg Vg.57 0 9.80 s 9.4 The volue of the iron block is V.00 kg iron.54 0 4 iron 7.86 0 kg and the buoyant fce exerted on the iron by the oil is V g 4 B oil 96 kg.54 0 9.80 s.8 N Applying F y = 0 to the iron block gives the suppt fce exerted by the upper scale (and hence the reading on that scale) as Fupper iron g B 9.6 N.8 N 7. N Fro Newton s third law, the iron exerts fce B downward on the oil (and hence the beaker). Applying F y = 0 to the syste consisting of the beaker and the oil gives F B g lower oil beaker 0 The suppt fce exerted by the lower scale (and the lower scale reading) is then Flower B oil beaker g.8 N.00.00 kg 9.80 s.7 N Page 9.8
9.44 (a) The cross-sectional area of the hose is (volue per unit tie) is A =5.0 L/.50 in. Thus, A r d 4.74 c 4, and the volue flow rate v 5.0 L.50 in 5.0 L 4 in 0 c A.50 in.74 c 60 s L 47. c s 0.47 s 0 c A d 4 d A 4 d d 9 A A 9 Then fro the equation of continuity, A v A v, we find A v v 9 0.47 s 4.4 s A 9.45 (a) The volue flow rate is A, and the ass flow rate is Av.0 g c.0 c 40 c s 80 g s Fro the equation of continuity, the speed in the capillaries is A ata.0 c capiliaries ata A capillaries.0 0 c v v 40 c s v capiliaries.7 0 c s 0.7 s. 9.46 (a) Fro the equation of continuity, the flow speed in the second pipe is A 0.0 c A.50 c v v.75 s.0 s Using Bernoulli s equation and choosing y = 0 along the centerline of the pipes gives v v.0 05 Pa.65 0 kg.75 s.0 s P P P 4.64 0 Pa. Page 9.9
9.47 Fro Bernoulli s equation, choosing at y = 0 the level of the syringe and needle, so the flow speed in the needle, v P P v is v v P P In this situation, F.00 N P P P P P 8.00 0 Pa 4 ato gauge A 5.50 0 Thus, assuing 0, v 8.00 04 Pa 0.6 s.00 0 kg 9.48 We apply Bernoulli s equation, igning the very sall change in vertical position, to obtain P P v v v v v, P.9 kg 5 0 s 4.4 0 Pa 9.49 (a) Assuing the airplane is in level flight, the net lift (the difference in the upward and downward fces exerted on the wings by the air flowing over the) ust equal the weight of the plane, ( P P ) A g. This yields lower upper wings surface surface P 8.66 0 4 kg 9.80 s g P lower upper surface surface Awings 90.0 9.4 0 Pa Neglecting the sall difference in altitude between the upper and lower surfaces of the wings and applying Bernoulli s equation yields P v P v lower air lower upper air upper Page 9.0
so P lower P upper 9.4 0 Pa upper lower air.9 kg v v 5 s 55 s (c) The density of air decreases with increasing height, resulting in a saller pressure difference. Beyond the axiu operational altitude, the pressure difference can no longer suppt the aircraft. 9.50 F level flight, the net lift (difference between the upward and downward fces exerted on the wing surfaces by air flowing over the) ust equal the weight of the aircraft, ( Plower Pupper ) A Mg. This gives the air pressure at the upper surface as surface surface P upper surface P lower surface Mg A 9.5 (a) Since the pistol is fired hizontally, the eerging water strea has initial velocity coponents of ( 0x = nozzle, 0y = 0) Then, y = 0y t + a y t /, with a y = g, gives the tie of flight as t y.50 a 9.80 s y 0.55 s With a x = 0, and 0x = nozzle, the hizontal range of the eergent strea is x = nozzle t where t is the tie of flight fro above. Thus, the speed of the water eerging fro the nozzle is v nozzle x 8.00 t 0.55 s 4.5 s (c) Fro the equation of continuity,a = A, the speed of the water in the larger cylinder is = (A /A ) = (A /A ) nozzle, r r.00 nozzle nozzle r r 0.0 v v v 4.5 s 0.45 s (d) The pressure at the nozzle is atospheric pressure, P =.0 0 5 /s. (e) With the two cylinders hizontal, y = y and gravity ters fro Bernoulli s equation can be neglected, Page 9.
leaving P v P v so the needed pressure in the larger cylinder is w w w.00 0 kg P P v v.0 05 Pa 4.5 s 0.45 s P 5.06 0 Pa (f) To create an overpressure of P =.06 0 5 Pa =.05 0 5 Pa in the larger cylinder, the fce that ust be exerted on the piston is 5 F P A P r.05 0 Pa.00 0.0 N 9.5 (a) Fro Bernoulli s equation, v v P gy P gy w w w w P P g y y w v v Figure 9.9 and using the given data values, we obtain.75 0 Pa.0 0 Pa 9.80 s.50.00 0 kg 5 5 v v and [] v v 6.0 s Fro the equation of continuity, A r r A r r v v v v Page 9.
v.00 c.50 c v = 4 [] 6 v 6.0 s Substituting Equation [] into [] gives, v 6.0 s 5.0 s Equation [] above now yields = 4(.0 /s) = 8.08 /s. (c) The volue flow rate through the pipe is: flow rate = A = A. Looking at the lower point: v flow rate r.00 0.0 s 5.7 0 s 9.5 First, consider the path fro the viewpoint of projectile otion to find the speed at which the water eerges fro the tank. Fro 0y y y v t a t with 0y = 0, we find the tie of flight as t y.00 a 9.80 s y 0.45 s Fro the hizontal otion, the speed of the water coing out of the hole is x 0.600 v v 0x. s t 0.45 s We now use Bernoulli s equation, with point at the top of the tank and point at the level of the hole. With P = P = P ato and, 0. This gives gy v gy v. s h y y 9.00 0 9.00 c g 9.80 s Page 9.
9.54 (a) Apply Bernoulli s equation with point at the open top of the tank and point at the opening of the hole. Then, P = P = P ato and we assue 0. This gives v gy gy, v y g y 9.80 s 6.0 7.7 s The area of the hole is found fro A flow rate.50 0 in in.5 0 v 7.7 s 60 s 6 The diaeter is then d 6 4 A 4.5 0.7 0.7 9.55 First, deterine the flow speed inside the larger ptions fro v flow rate.80 0 4 s A.50 0 4 0.67 s The absolute pressure inside the large section on the left is P = P 0 + gh, where h is the height of the water in the leftost standpipe. The absolute pressure in the constriction is P = P + gh, so P P g h h g 5.00 c The flow speed inside the constriction is found fro Bernoulli s equation with y = y. This gives v v v P P g h h v 0.67 s 9.80 s 5.00 0.06 s The cross-sectional area of the constriction is then A flow rate.80 0 4 s.7 0 4, v.06 s Page 9.4
and the diaeter is d 4 4 A 4.7 0.47 0.47 c 9.56 (a) F iniu pressure, we assue the flow is very slow. Then, Bernoulli s equation gives P v gy = P v gy river ri P 0 at 0 g y y river in ri river kg s P.0 05 Pa 0 9.80 096 564 river in P river in.0 05.50 07 Pa =.5 07 Pa 5. MPa The volue flow rate is flow rate =A=( d /4). Thus, the velocity in the pipe is v flow rate 4 4500 d d 0.50 4 d 86 400 s.95 s (c) We iagine the pressure being applied to stationary water at river level, so Bernoulli s equation becoes v P 0 at g y y river ri river ri kg Priver Priver vri Priver 0.95 in in s P river in 4.5 kpa Page 9.5
The additional pressure required to achieve the desired flow rate is P = 4.5 kpa 9.57 (a) F upward flight of a water-drop projectile fro geyser vent to fountain-top, 0 a y = 0, when y = y ax, gives vy v y y, with y v 0y y ax y 0 a 9.8 s 40.0 8.0 s Because of the low density of air and the sall change in altitude, atospheric pressure at the fountain top will be considered equal to that at the geyser vent. Bernoulli s equation, with top = 0, then gives v 0 g y y vent top vent v y g y 9.80 s 40.0 8.0 s vent top vent (c) Between the chaber and the geyser vent, Bernoulli s equation with chaber 0 yields P gy P gy chaber 0 v at vent vent P P v g y y at vent vent chaber s kg 8.0 s 0 9.80 75. MPa 6 5 Pgauge P Pato. 0 Pa at.0 0 Pa 0.8 atospheres 9.58 (a) Since the tube is hizontal, y = y and the gravity Page 9.6
ters in Bernoulli s equation cancel, leaving P v P v Figure 9.0(a) P P.0 0 Pa v v 7.00 0 kg and v v [].4 s Fro the continuity equation, A = A, we find A r.40 c A r.0 c v v v v v 4v [] Substituting Equation [] into [] yields 5v.4 s and = 0.478 /s. Then, Equation [] gives = 4(0.478 /s) =.9 /s. The volue flow rate is A A r 4 v v v.0 0.9 s 8.64 0 s 9.59 Fro F y = T g F y = 0, the balance reading is found to be T = g + F y where F y is the vertical coponent of the surface tension fce. Since this is a two-sided surface, the surface tension fce is F = (L) and its vertical coponent is F = (L)cos where is the contact angle. Thus, T = g + Lcos. T 0.40 N when 0 g L 0.40 N [] T 0.9 N when 80 g L 0.9 N [] Subtracting Equation [] fro [] gives Page 9.7
0.40 N 0.9 N 0.40 N 0.9 N 4 L 4.0 0 8. 0 N 9.60 Because there are two edges (the inside and outside of the ring), we have F F = = L ( circuference ) total F.6 0 N = = 7. 0 N 4r 4.75 0 9.6 Fro h = cos/gr, the surface tension is hgr cos. 0 080 kg 9.80 s 5.0 0 4 cos0 5.6 0 N 9.6 The height the blood can rise is given by h 0.058 N cos 0 050 kg 9.80 s.0 0 6 cos gr 5.6 9.6 Fro the definition of the coefficient of viscosity, = FL/A, the required fce is F.79 0 N s 0.800.0 0.50 s A v L 0.0 0 8.6 N 9.64 Fro the definition of the coefficient of viscosity, = FL/A, the required fce is F 500 0 N s 0.00 0.040 0.0 s A v L.5 0 0. N Page 9.8
9.65 Poiseuille s law gives flow rate P P 4 R 8L and P = P at in this case. Thus, the desired gauge pressure is 8 L flow rate P P 80. N s 5 50 8.6 0 s R4 0.50 0 at 4 P 6 Pat. 0 Pa. MPa 9.66 Fro Poiseuille s law, the flow rate in the artery is flow rate 4 P R4 400 Pa.6 0 8L 8.7 0 N s 8.4 0. 0 5 s Thus, the flow speed is v flow rate. 0 5 s A.6 0.5 s 9.67 If a particle is still in suspension after hour, its terinal velocity ust be less than v c h t 5.0.4 0 s ax h 600 s 00 c. 5 Thus, fro = r g( f )/9, we find the axiu radius of the particle: r ax 9 v t ax g f 5 9.00 0 N s.4 0 s 9.80 s 800 000 kg.8 0 6.8 Page 9.9
9.68 Fro Poiseuille s law, the excess pressure required to produce a given volue flow rate of fluid with viscosity through a tube of radius R and length L is P 8L V t R 4 If the ass flow rate is (/t) =.0 0 kg/s, the volue flow rate of the water is V t.0 0 kg s t.0 0 kg.0 0 6 s and the required excess pressure is 6 4 0.5 0 8.0 0 Pa s.0 0.0 0 s P.5 05 Pa 9.69 With the IV bag elevated.0 above the needle, the pressure difference across the needle is P gh.0 0 kg 9.8 s.0 9.8 0 Pa and the desired flow rate is V t 500 c 06 c 0 in 60 s in.8 0 7 s Poiseuille s law then gives the required diaeter of the needle as D 4 7 9.8 0 Pa 8L V t 8.0 0 Pa s.5 0.8 0 s R P 4 D 4. 0 4 0.4 9.70 We write Bernoulli s equation as Page 9.0
P v gy P v gy out out out in in in P P P g y y v v gauge in out out in out in Approxiating the speed of the fluid inside the tank as in 0, we find P gauge.00 0 kg 0.0 s 9.80 s 0.500 P 5 gauge 4.55 0 Pa 455 kpa 9.7 The Reynolds nuber is RN 050 kg 0.55 s.0 0 vd 4. 0.7 0 N s In this region (RN > 000), the flow is turbulent. 9.7 Fro the definition of the Reynolds nuber, the axiu flow speed f strealined ( lainar) flow in this pipe is v RN.0 0 N s 000 ax d 000 kg.5 0 0.080 s 8.0 c s ax 9.7 The observed diffusion rate is 8.0 0 4 kg/5 s = 5. 0 5 kg/s. Then, fro Fick s law, the difference in concentration levels is found to be C C diffusion rate L DA 5. 0 5 kg s 0.0 0 4 5.0 0 s 6.0 0.8 0 kg Page 9.
9.74 Fick s law gives the diffusion coefficient as D = diffusion rate/a.(c/l), where C/L is the concentration gradient. Thus, D 5.7 0 5 kg s.0 0 4 4.0 0 kg 9.5 0 0 s 9.75 Stokes s law gives the viscosity of the air as F.0 0 N.4 0 N s 6rv 6.5 0 4.5 0 s 6 4 5 9.76 Using =r g( f )/9, the density of the droplet is found to be = f + 9 t /r g. Thus, if r = d/ = 0.500 0 t =.0 0 /s and when falling through 0 C water ( =.00 0 N.s/ ) the density of the oil is 4 kg 9.00 0 N s.0 0 s 000.0 0 kg 5.00 0 9.80 s 9.77 (a) Both iron and aluinu are denser than water, so both blocks will be fully suberged. Since the two blocks have the sae volue, they displace equal aounts of water and the buoyant fces acting on the two blocks are equal. Since the block is held in equilibriu, the fce diagra at the right shows that F 0 T g B y The buoyant fce B is the sae f the two blocks, so the spring scale reading T is largest f the iron block, which has a higher density, and hence weight, than the aluinu block. (c) The buoyant fce in each case is B V g water.0 0 kg 0.0 9.8 s.0 0 N F the iron block: Tiron ironv g B 7.86 0 kg 0.0 9.8 s B Page 9.
T 4 iron.5 0 N.0 0 N 0 N F the aluinu block: Taluinu aluinuv g B.70 0 kg 0.0 9.8 s B T aluinu 5. 0 N.0 0 N. 0 N 9.78 In going fro the ocean surface to a depth of.40 k, the increase in pressure is P P P gh 7 0.05 0 kg 9.80 s.40 0.4 0 Pa The fractional change in volue of the steel ball is given by the defining equation f bulk odulus, P =B(V/V), as V P.4 07 Pa.5 0 V B 6.0 0 0 Pa steel 4 9.79 (a) Fro Archiedes s principle, the granite continent will sink down into the peridotite layer until the weight f the displaced peridotite equals the weight of the continent. Thus, at equilibriu, At g Ad g g p t d. g p If the continent sinks 5.0 k below the surface of the peridotite, then d = 5.0, and the result of part (a) gives the first approxiation of the thickness of the continent as t p. 0 kg d g.8 0 kg 5.0 k 5.9 k 9.80 (a) Starting with P = P 0 + gh, we choose the reference level at the level of the heart, so P 0 = P H. The pressure at the feet, a depth h H below the reference level in the pool of blood in the body is P F = P H + gh H. The pressure Page 9.
difference between feet and heart is then P F = P H + gh H. Using the result of part (a), P F H P.06 0 kg 9.80 s.0.5 04 Pa 4 9.8 The cross-sectional area of the ata is A d and that of a single capillary is A d 4 If the circulaty syste has N such capillaries, the total cross-sectional area carrying blood fro the ata is c A NA c N d 4 Fro the equation of continuity, A v A v, Nd v d 4 v 4, which gives N d.0 s 0.50 0.5 07 6 d.0 0 s 0 0 v v 9.8 (a) We iagine that a superhero is capable of producing a perfect vacuu above the water in the straw. Then P = P 0 + gh, with the reference level at the water surface inside the straw and P being atospheric pressure on the water in the cup outside the straw, gives the axiu height of the water in the straw as h P 0 P.0 0 N 5 at at ax waterg waterg.00 0 kg 9.80 s 0. The oon has no atosphere so P at = 0, which yields.h ax = 0. 9.8 (a) P 60 of H O g60 H O kg s 0 9.80 0.60.57 kpa Page 9.4
P at.0 0 Pa.57 0 Pa 5.55 0 at The pressure is P gh gh H O H O Hg Hg HO 0 kg Hg H O Hg.6 0 kg h h 60.8 of Hg (c) The fluid level in the tap should rise. Blockage of flow of the cerebrospinal fluid. 9.84 When the rod floats, the weight of the displaced fluid equals the weight of the rod, f gv displaced 0 gv rod. But, assuing a cylindrical rod, V r L. The volue of fluid displaced is the sae as the volue of the rod rod that is suberged, V r L h. Thus, displaced f g r L h 0 g r L, which reduces to L f 0 L h 9.85 Consider the diagra and apply Bernoulli s equation to points A and B, taking y = 0 at the level of point B, and recognizing that A 0. This gives PA 0 wg h L sin P B w vb 0 Recognize that P A = P B = P at since both points are open to the atosphere. Thus, we obtain v B g h L sin 9.80 s 0.0.00 sin 0.0º. s Now the proble reduces to one of projectile otion with v0y v B sin 0.0º 6.64 s Page 9.5
At the top of the arc, y = 0, and y = y ax. Then, vy v 0 y ay y gives 0 6.64 s 9.80 s y 0 y.5 above the level of point B. ax, ax 9.86 When the balloon coes into equilibriu, the weight of the displaced air equals the weight of the filled balloon plus the weight of string that is above ground level. If s and L are the total ass and length of the string, the ass of string that is above ground level is (h/l) s. Thus, h air gvballoon balloon g heliu gvballoon s g L which reduces to h V air heliu balloon balloon s L This yields h.9 kg 0.79 kg 4 0.40 0.5 kg.0.9 0.050 kg 9.87 When the balloon floats, the weight of the displaced air equals the cobined weights of the filled balloon and its load. Thus, gv g gv g, air balloon balloon heliu balloon load V 600 kg 4 000 kg balloon load balloon 4.4 0 air heliu.9 0.79 kg 9.88 Page 9.6
(a) Consider the pressure at points A and B in part of the figure by applying P = P 0 + f gh. Looking at the left tube gives P P gl h A at water, and looking at the tube on the right, B at oil P P gl. Pascal s principle says that P B = P A. Therefe, P gl P gl h, giving at oil at water h oil 750 kg L water 000 kg 5.00 c.5 c Consider part (c) of the diagra showing the situation when the air flow over the left tube equalizes the fluid levels in the two tubes. First, apply Bernoulli s equation to points A and B. This gives P v gy P v gy A air A air A B air B air B Since y y, v v, and v 0, this reduces to [] A B A B PB PA air v Now use P = P 0 + f gh to find the pressure at points C and D, both at the level of the oil water interface in the right tube. Fro the left tube, PC PA water gl, and fro the right tube, P P gl. D B oil Pascal s principle says that P D = Pc, and equating these two gives P gl P gl B oil A water B A water oil P P gl [] Cobining Equations [] and [] yields v water air oil gl 000 750 9.80 s 5.00 0.9.8 s 9.89 While the ball is suberged, the buoyant fce acting on it is B = ( w V)g. The upward acceleration of the ball while under water is Page 9.7
F B g 4 y w a y r g 000 kg 4 0.0 9.80 s s.0 kg Thus, when the ball reaches the surface, the square of its speed is y v v 0 a 0 s.0 5 s y y y When the ball leaves the water, it becoes a projectile with initial upward speed of v 0y 5 s and acceleration of ay g 9.80 s. Then, v v 0 a y gives the axiu height above the surface as y y y y 0 5 s 9.80 s ax 6.4 9.90 Since the block is floating, the total buoyant fce ust equal the weight of the block. Thus, 4.00 c water A 4.00 c oil A x g A x g g wood where A is the surface area of the top botto of the rectangular block. Solving f the distance x gives x wood oil 960 90 4.00 c 4.00 c.7 c 000 90 water oil 9.9 A water droplet eerging fro one of the holes becoes a projectile with 0y = 0 and 0x = 0. The tie f this droplet to fall distance h to the flo is found fro 0y y y v t a t to be Page 9.8
t h g The hizontal range is R h vt v. g If the two streas hit the flo at the sae spot, it is necessary that R = R, h h v v g g With h = 5.00 c and h =.0 c, this reduces to h.0 c v v v v v.40 [] 5.00 c h Apply Bernoulli s equation to points (the lower hole) and (the surface of the water). The pressure is atospheric pressure at both points and, if the tank is large in coparison to the size of the holes, 0. Thus, we obtain P gh P 0 gh v gh h at at v. [] Siilarly, applying Bernoulli s equation to point (the upper hole) and point gives P gh P 0 gh v gh h at at v. [] Square Equation [] and substitute fro Equations [] and [] to obtain g h h.40 g h h Solving f h yields h.40 h h.40.0 c 5.00 c 7.0 c,.40.40 so the surface of the water in the tank is 7.0 c above flo level. Page 9.9
9.9 When the section of walkway oves downward distance L, the cable is stretched distance L and the colun is copressed distance L. The tension fce required to stretch the cable and the copression fce required to copress the colun this distance is F cable Ysteel Acable L and L cable F colun Y A L Al L colun colun Cobined, these fces suppt the weight of the walkway section: Fcable Fcolun F g 8 500 N Ysteel Acable L YAl Acolun L L L cable colun 8 500 N giving L 8 500 N Y A Y A L L steel cable Al colun cable colun The cross-sectional area of the cable is A cable.7 0 D 4 4 and the area of aluinu in the cross section of the colun is A cable D 0.6 4 0.6 4 outer D inner D outer D inner 4 4 4 4 Thus, the downward displaceent of the walkway will be L 0 0 0 Pa.7 0 8 500 N 7.0 0 Pa 0.6 4 0.6 4 0 4 5.75 4.5 L 8.6 0 4 0.86 Page 9.40