PHYS 210 Homework, Spring 2008

1,2,3,7,11,12,13,16, 8,20,22,23,24, 25, 27, 30, 31, 32, 33, 38,40,43,46,49,51,54,61,62,71 1. Calculate the mass of a solid iron sphere that has a diameter of 3.00 cm. M ironV  7 860 kg m3   4 3  0.015 0 m   3     M  0.111 kg The density of the nucleus is of the same order of magnitude as that of one proton, according to the assumption of close packing:  m V ~ 1.67 10 27 kg 4 3  10 15 m   3 ~ 1018 kg m3 . With vastly smaller average density, a macroscopic chunk of matter or an atom must be mostly empty space. In a cube of 1m side: 1018 Kg 2. Find the order of magnitude of the density of the nucleus of an atom. What does this result suggest concerning the structure of matter? Model a nucleus as protons and neutrons closely packed together. Each has mass 1.67  10–27 kg and radius on the order of 10–15 m. 3. A 50.0-kg woman balances on one heel of a pair of high-heeled shoes. If the heel is circular and has a radius of 0.500 cm, what pressure does she exert on the floor? P F A  50.0 9.80    0.50010 2   2  6.24106 N m2 Fel Ffluid kx ghA h  kx gA h  1 000 N m2  5.0010 3 m   103 kg m3  9.80 m s2   1.00 10 2 m   2      1.62 m 7. The spring of the pressure gauge shown in Figure 14.2 has a force constant of 1 000 N/m, and the piston has a diameter of 2.00 cm. As the gauge is lowered into water, what change in depth causes the piston to move in by 0.500 cm? 11. the excess water pressure (over air pressure) halfway down is     3 2 4 gauge 1 000 kg m 9.80 m s 1.20 m 1.18 10 Pa P gh    The force on the wall due to the water is     4 5 gauge 1.18 10 Pa 2.40 m 9.60 m 2.71 10 N F P A      horizontally toward the back of the hole. For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.40 m. A concrete foundation wall is built all the way across the 9.60-m width of the excavation. This foundation wall is 0.183 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by 2.40 m  9.60 m  0.183 m  1 000 kg/m3  9.80 m/s2 = 41.3 kN. The pressure of the bottom due to the water is Pb = gz = 1.96 x 104 Pa. So Fb = Pb Ab = 5.88 N. On each end, Fe = Pav.Aend = (Pb/2).Aend = 9.8 x103 x 20 = 196 kN On each side, Fs = Pav Aside = 9.8 x103 x 60 = 588 kN Aend 1 2 3 Aside a g bubble gf 12. A swimming pool has dimensions 30.0 m  10.0 m and a flat bottom. When the pool is filled to a depth of 2.00 m with fresh water, what is the force caused by the water on the bottom? On each end? On each side? 13. .A sealed spherical shell of diameter d is rigidly attached to a cart, which is moving horizontally with an acceleration a as in Figure P14.13. The sphere is nearly filled with a fluid having density P P0  g eff h  P0  1 2 d g2  a2 . In the reference frame of the fluid, the cart’s acceleration causes a fictitious force to act backward, as if the acceleration of gravity where  tan  1 a g       from 2 2 a g  Directed downward at From the vertical. The center of the spherical shell is at depth d/2 Below the air buble and the pressure is: 16. We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the water surface in the basin and point 2 at the water surface in the straw: 1 1 2 2 P gy P gy         5 2 3 2 2 1.013 10 N m 0 0 1 000 kg m 9.80 m s y     2 10.3 m y  (b) No atmosphere can lift the water in the straw through zero height difference. Figure P14.16 shows Superman attempting to drink water through a very long straw. With his great strength he achieves maximum possible suction. The walls of the tubular straw do not collapse. (a) Find the maximum height through which he can lift the water. (b) What If? Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside the straw. Figure P14.16 At the level mercury-water interface in the right tube: 18. Mercury is poured into a U-tube as in Figure P14.18a. The left arm of the tube has cross-sectional area A of 10.0 cm2, and the right arm has a cross-sectional area A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm as in Figure P14.18b. (a) Determine the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm? hw  mwater A2water  100 g 5.00 cm2 1.00 g cm 3    20.0 cm A1h (a) Using the definition of density, we have (b) sketch (b) at the right represents the situation after the water is added. A volume A2h2 mercury has been displaced by water in the right tube. The additional of mercury now in the left tube is A1 h. . Since the total volume of mercury has not changed: h A A h A h A h ) ( 2 1 2 1 2 2    P P0  water gh w The pressure at this same level in the left tube is given by P P0  Hg g h  h2  P0  water ghw Hg h 1 A1 A2    water hw Then: And h  water h w Hg 1  A 1  A 2    0 Hg 2 0 1 2 and . A w B w P P gh gh P P g h h h              Hg 1 2 1 13.6 1 1.00 cm 12.6 cm w h h              B A h h 1 h 2 water Mercury Let h be the height of the water column added to the right side of the U–tube. Then when equilibrium is reached, the situation is as shown in the sketch at right. Now consider two points, A and B shown in the sketch, at the level of the e water–mercury interface. By Pascal’s Principle, the absolute pressure at B is the same as that at A. But, 1 2 Hg 2 w w w w h h h h h          , , or   B A p p 20. A U-tube of uniform cross-sectional area, open to the atmosphere, is partially filled with mercury. Water is then poured into both arms. If the equilibrium configuration of the tube is as shown in Figure P14.20, with h2 = 1.00 cm, determine the value of h1. 22. (a) A light balloon is filled with 400 m3 of helium. At 0C, the balloon can lift a payload of what mass? (b) What If? In Table 14.1, observe that the density of hydrogen is nearly one-half the density of helium. What load can the balloon lift if filled with hydrogen? a) The balloon is nearly in equilibrium:     helium payload 0 y y g g F ma B F F       or air helium payload 0 gV gV m g           3 3 3 payload air helium payload 1.29 kg m 0.179 kg m 400 m 444 kg m V m        (b) Similarly,      3 3 3 payload air hydrogen payload 1.29 kg m 0.089 9 kg m 400 m 480 kg m V m             3 3 3 payload air hydrogen payload 1.29 kg m 0.089 9 kg m 400 m 480 kg m V m        23.A Ping-Pong ball has a diameter of 3.80 cm and average density of 0.084 0 g/cm3. What force is required to hold it completely submerged under water? At equilibrium  F 0 Fapp mg B   where B is the buoyant force. Fapp B mg    water  Vol B g     ball Vol m         3 water ball water ball 4 Vol 3 Fapp g r g               3 2 2 3 3 3 4 1.90 10 m 9.80 m s 10 kg m 84.0 kg m 0.258 N 3 Fapp        24. A Styrofoam slab has thickness h and density. When a swimmer of mass m is resting on it, the slab floats in fresh water with its top at the same level as the water surface. Find the area of the slab.   g s F m  V g   b w F  Vg must be equal to V Ah  s w m Ah Ah     ,   w s m A h     25. A piece of aluminum with mass 1.00 kg and density 2 700 kg/m3 is suspended from a string and then completely immersed in a container of water (Figure P14.25). Calculate the tension in the string (a) before and (b) after the metal is immersed. Fig 14.25 (a) Before the metal is immersed 1 0 Fy T Mg        2 1 1.00 kg 9.80 m s 9.80 N T Mg   (b) After the metal is immersed: 2 0 Fy T B Mg        2 w T Mg B Mg  V g     3 1.00 kg 2 700 kg m M V        3 2 2 3 1.00 kg 9.80 N 1 000 kg m 9.80 m s 6.17 N 2 700 kg m T Mg B            27. 0 P P gh   5 2 0 1.013 10 N m P   h 5.00 cm 5 2 top 1.017 9 10 N m P    For h=1.70 m, one gets 5 2 bot 1.029 7 10 N m P   Since the areas of the top and bottom are   2 2 2 0.100 m 10 m A    3 top top 1.017 9 10 N F P A    3 bot 1.029 7 10 N F   One finds 0 T B Mg    b)     3 3 3 3 2 10 kg m 1.20 10 m 9.80 m s 11.8 N w B  Vg       10.0 9.80 98.0 N Mg   98.0 11.8 86.2 N T Mg B         3 bot top 1.029 7 1.017 9 10 N 11.8 N F  F     c) which is equal to B found in part (b). A 10.0-kg block of metal measuring 12.0 cm  10.0 cm  10.0 cm is suspended from a scale and immersed in water as in Figure P14.25b. The 12.0-cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the forces acting on the top and on the bottom of the block? (Take P0 = 1.013 0  105 N/m2.) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block. 30. A spherical aluminum ball of mass 1.26 kg contains an empty spherical cavity that is concentric with the ball. The ball just barely floats in water. Calculate (a) the outer radius of the ball and (b) the radius of the cavity. (a) The weight of the ball must be equal to the buoyant force of the water: m r g r mg shell shell water .6 70 4 3      (b) The mass of the ball is determined by the density of aluminum: cm r r r V m c c shell Al Al .5 74 ) ( 3 4 3 3         Let A represent the horizontal cross-sectional area of the rod, which we presume to be constant. The rod is in equilibrium: 0 y  F  0 whole rod fluid immersed 0 mg B V g V g          0ALg A L h g    0L L h     the density of the liquid is . 31. Determination of the density of a fluid has many important applications. A car battery contains sulfuric acid, for which density is a measure of concentration. For the battery to function properly the density must be inside a range specified by the manufacturer. Similarly, the effectiveness of antifreeze in your car’s engine coolant depends on the density of the mixture (usually ethylene glycol and water). When you donate blood to a blood bank, its screening includes determination of the density of the blood, since higher density correlates with higher hemoglobin content. A hydrometer is an instrument used to determine liquid density. A simple one is sketched in Figure P14.31. The bulb of a syringe is squeezed and released to let the atmosphere lift a sample of the liquid of length L 0 floats partially immersed in the liquid of density. A length h of the rod Protrudes above the surface of the liquid. Show that the density of the liquid is given by h L L   0  Fig. 14.31 33. How many cubic meters of helium are required to lift a balloon with a 400-kg payload to a height of 8 000 m?. (Take =o.180 kg/m). Assume that the balloon maintains a constant volume and that the density of air decreases with the altitude z according to the expression 8000 / 0 z air  e  where z is the height measured in meters and 0 =1.25 Kg/m3 is the density at sea level. The balloon stops when : 3 1430 ) ( m M V He air       M Mg  airVg  HeVg  Mg gV gV He air    It is in equilibrium 38. A horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter. If the pressure of the water in the larger pipe is 8.00  104 Pa and the pressure in the smaller pipe is 6.00  104 Pa, at what rate does water flow through the pipes? By Bernoulli’s equation,           4 2 2 4 2 2 4 2 2 2 2 1 1 8.00 10 N m 1 000 6.00 10 N m 1 000 16 2 2 1 2.00 10 N m 1 000 15 2 1.63 m s 1 000 5.00 10 1.63 m s 12.8 kg s v v v v dm Av dt                40. A village maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 6.60 cm. The hose ends with a nozzle of diameter 2.20 cm. A rubber stopper is inserted into the nozzle. The water level in the tank is kept 7.50 m above the nozzle. (a) Calculate the friction force exerted on the stopper by the nozzle. (b) The stopper is removed. What mass of water flows from the nozzle in 2.00 h? (c) Calculate the gauge pressure of the flowing water in the hose just behind the nozzle. Take point at the free surface of the water in the tank and inside the nozzle (a) With the cork in place 2 2 1 1 1 2 2 2 1 1 2 2 P gy v P gy v          3 2 0 2 1 000 kg m 9.8 m s 7.5 m 0 0 0 P P      becomes 4 2 0 7.35 10 Pa P  P   For the stopper xF 0    water air 2 0 2 4 0 7.35 10 Pa 0.011 m 27.9 N F F f P A P A f f          (b) Now Bernoulli’s equation gives   4 3 2 0 0 2 2 1 7.35 10 Pa 0 0 1 000 kg m 2 12.1 m s P P v v        Continuation next page The quantity leaving the nozzle in 2 h is Continuation of problem 40       2 3 4 2 1 000 kg m 0.011 m 12.1 m s 7 200 s= 3.32 10 kg V Av t       (c) Take point 1 in the wide hose and 2 just outside the nozzle. Continuity :equation       1 1 2 2 2 2 1 1 2 2 1 1 1 2 2 2 2 2 3 3 1 0 4 2 4 1 0 6.6 cm 2.2 cm 12.1 m s 2 2 12.1 m s 1.35 m s 9 1 1 2 2 1 1 0 1 000 kg m 1.35 m s 0 1 000 kg m 12.1 m s 2 2 7.35 10 Pa 9.07 10 Pa 7.26 10 Pa A v A v v v P gy v P gy v P P P P                                        43. The Figure shows a stream of water in steady flow from a kitchen faucet. At the faucet the diameter of the stream is 0.960 cm. The stream fills a 125-cm3 container in 16.3 s. Find the diameter of the stream 13.0 cm below the opening of the faucet. Stream fills a tank of volume 125 cm3 in 13.6 s, therefore: cm s s cm r A / 10 6. v 16 3. 125 )v ( v 1 3 1 2 1      At 13 cm below, we obtain the velocity by using the Bernoulli's equation: s m gy p gy p / 6.1 v v 2 1 v 2 1 2 2 2 2 0 2 1 1 0            Using the continuity equation, we obtain the diameter of the flow at 13 cm: cm d m s d s cm or A A 247 .0 / ) 6.1( / 2) ( 16 3. 125 v v 2 3 2 2 1 1      46. Old Faithful Geyser in Yellowstone Park (Fig. P14.46) erupts at approximately 1-h intervals, and the height of the water column reaches 40.0 m. (a) Model the rising stream as a series of separate drops. Analyze the free-fall motion of one of the drops to determine the speed at which the water leaves the ground. (b) What If? Model the rising stream as an ideal fluid in streamline flow. Use Bernoulli’s equation to determine the speed of the water as it leaves ground level. (c) What is the pressure (above atmospheric) in the heated underground chamber if its depth is 175 m? You may assume that the chamber is large compared with the geyser’s vent. Solution on next slide Solution for 46 a) For upward flight of a water-drop projectile from geyser vent to fountain-top: 2 2 2 2 . 0 2( 9.8)( 40) 28 / yf yi y yi yi v v a y v v m s           b) Between geyser vent and fountain-top: 2 2 1 1 1 2 2 2 1 1 2 2 p v gy p v gy          Air so low in density, that very nearly p1 = p2 = 1atm. Then 2 1 1 1 0 0 9.8 40 28 / 2 v v m s       c) Between the chamber and fountain-top: 2 2 1 1 1 2 2 2 1 1 2 2 p v gy p v gy          1 0 1 0 0 1000 9.8( 175) 0 1000 9.8(40) 2.11 p p p p MPa           49. A Pitot tube can be used to determine the velocity of air flow by measuring the difference between the total pressure and the static pressure (Fig. P14.49). If the fluid in the tube is mercury, density Hg = 13 600 kg/m3, and h = 5.00 cm, find the speed of air flow. (Assume that the air is stagnant at point A and take air = 1.25 kg/m3.) Figure P14.49 2 air Hg 2 v P g h      Hg air 2 103 m s g h v      51. A siphon is used to drain water from a tank, as illustrated in Figure P14.51. The siphon has a uniform diameter. Assume steady flow without friction. (a) If the distance h = 1.00 m, find the speed of outflow at the end of the siphon. (b) What If? What is the limitation on the height of the top of the siphon above the water surface? (For the flow of the liquid to be continuous, the pressure must not drop below the vapor pressure of the liquid.) Figure P14.51 2 0 0 3 1 0 0 2 P gh P v        3 2 v gh  h 1.00 m 3 4.43 m s v  2 2 2 0 3 1 1 0 2 2 P gy v P v         b) 2 3 v v since 0 P P gy   0 P  since    5 0 3 3 2 1.013 10 Pa 10.3 m 10 kg m 9.8 m s P y g      54.Figure P14.54 shows a water tank with a valve at the bottom. If this valve is opened, what is the maximum height attained by the water stream coming out of the right side of the tank? Assume that h = 10.0 m, L = 2.00 m, and = 30.0, and that the cross-sectional area at A is very large compared with that at B. Figure P14.54 Consider the diagram and apply Bernoulli’s equation to points A and B, taking at the level of point B, and recognizing that is approximately zero. This gives:       2 2 1 0 sin 2 1 0 2 A w w B w B w P g h L P v g            Now, recognize that atmosphere A B P P P   since both points are open to the atmosphere (neglecting variation of atmospheric pressure with altitude). Thus, we obtain       2 2 sin 2 9.80 m s 10.0 m 2.00 m sin30.0 13.3 m s B B v g h L v            Now the problem reduces to one of projectile motion with sin 30.0 6.64 m s yi B v v     2 2 2 yf yi v v a y    then gives at the top of the arc max y y where 0 vyf       2 2 max 0 6.64 m s 2 9.80 m s 0 y     ymax= =2.25m, above the level where the water emerges 1 6 gwH 3 61. Review problem. With reference to Figure 14.5, show that the total torque exerted by the water behind the dam about a horizontal axis through O is Show that the effective line of action of the total force exerted by the water is at a distance H/3 above O. H h y dy w dF 0 0 0 3 ( ) ( ) 1 ( ) ( ) 6 H H H d ydF yp y dA yp y wdy p y g H y gwH                  Torque: The total force is 2 0 0 0 1 ( ) ( ) 2 H H H F dF p y dA p y wdy gwH        If this were applied at a height yeff such that the torque remains unchanged, we have 2 3 1 1 2 6 3 eff eff eff H Fy gwH y gwH y             62. In about 1657 Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemispheres. Two teams of eight horses each could pull the hemispheres apart only on some trials, and then "with greatest difficulty," with the resulting sound likened to a cannon firing (Fig. P14.62). (a) Show that the force F required to pull the evacuated hemispheres apart is R2(P0 – P), where R is the radius of the hemispheres and P is the pressure inside the hemispheres, which is much less than P0. (b) Determine the force if P = 0.100P0 and R = 0.300 m. Figure P14.62 (a) The pressure on the surface of the two hemispheres is constant at all points, and the force on each element of surface area is directed along the radius of the hemispheres. The applied force along the axis must balance the force on the “effective” area, which is the projection of the actual surface onto a plane perpendicular to the x axis, 2 A R   2 0 F P P R      2 4 0 0 0 0.100 0.300 m 0.254 2.58 10 N F P P P          (b) For the values given 71. A U-tube open at both ends is partially filled with water (Fig. P14.71a). Oil having a density 750 kg/m3 is then poured into the right arm and forms a column L = 5.00 cm high (Fig. P 14.71b). (a) Determine the difference h in the heights of the two liquid surfaces. (b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Fig. P14.71c). Determine the speed of the air being blown across the left arm. Take the density of air as 1.29 kg/m3. Fig. 14.71 Solution on next page 3 / 750 m kg oil  L=5.0 cm a) What is h? Consider interface water-oil (points A and B) gL p P h g L p P oil B w A        0 0 ) ( w oil w B A h P P       b) Right arm shielded (no air motion) Using Bernoulli's equation: 2 2 2 v 2 1 0) ( 2 1 v 2 1 air A B B A B air air B A air air A p p y y g y p g y p               3 / 750 m kg air   To determine A B p p  we use Pascal’s law At the Interface C and D: v ) (           A B oil air B D w air A C p p gL gH p p gL gH p p     Solution of 14.71