ρ B ρ A Physics: Principles and Applications, 6e Giancoli Chapter 10 Fluids = m B V B m A = m A V A = ρ A = ρ B m B Answer B = 4 3


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1 Physics: Principles and Applications, 6e Giancoli Chapter 10 Fluids Conceptual Questions 1) The three common phases of matter are A) solid, liquid, and as. B) solid, liquid, and vapor. C) solid, plasma, and as. D) condensate, plasma, and as. It is common knowlede that the three common phases of matter are solid, liquid and as. You should be familiar with the concepts related to the models of these three phases of matter as outlined in the document Models of the Phases of Matter. Especially be familiar with how the properties of each phase are related to the strenth of interparticle forces and kinetic enery of particles. Answer A ) Density is A) proportional to both mass and volume. B) proportional to mass and inversely proportional to volume. C) inversely proportional to mass and proportional to volume. D) inversely proportional to both mass and volume. The definition of density is ρ = m V. From this formula you can see that if mass is increased, density increases, so density is proportional to mass. You can also see that if V increases, density decreases, so, density is inversely proportional to volume. Answer B 3) Substance A has a density of 3.0 /cm3 and substance B has a density of 4.0 /cm3. In order to obtain equal masses of these two substances, the ratio of the volume of A to the volume of B will be equal to A) 1:3. B) 4:3. C) 3:4. D) 1:4. Manipulatin ratios is formulas is not intuitive enouh to most people that you can see the answer to this question riht away. Because you will be able to see the manipulation visually, it is more likely that you will be successful answerin this question quickly if you immediately write down the relationships involved, manipulate the formulas to et the correct ratio, and plu the values in: ρ A = m A V A ρ B = m B V B m A = ρ A V A m B = ρ B V B ρ A V A = ρ B V B Answer B V A V B = ρ B ρ A = 4 3 4) Pressure is A) proportional to both force and area. B) proportional to force and inversely proportional to area. C) inversely proportional to force and proportional to area. D) inversely proportional to both force and area. The definition of pressure is P = F A = N m From this formula, you can see that if you increase force, pressure increases, so pressure is proportional to force. You can also see that if you increase area, pressure decreases, so pressure is inversely proportional to A. Answer B 1
2 5) Which of the followin is not a unit of pressure? A) atmosphere B) N/m C) Pascal D) mm of mercury Units of pressure are either in units of force per units of area, relate to the heiht of a column of mercury supported by the weiht of the atmosphere, or are iven an arbitrary value of 1 atmophere. (A) is a statement of the unit of an atmosphere; (C) is the name iven for the unit representin one N per cubic meter; (D) is a statement of the heiht of a column of mercury supported by the weiht of the atmosphere in mm. (C) is a unit that would represent a unit of force alon a one dimensional lenth, not per unit of area. Answer B 6) Consider three drinkin lasses. All three have the same area base, and all three are filled to the same depth with water. Glass s cylindrical. Glass B is wider at the top than at the bottom, and so holds more water than A. Glass C is narrower at the top than at the bottom, and so holds less water than A. Which lass has the reatest liquid pressure at the bottom? A) Glass A B) Glass B C) Glass C D) All three have equal pressure. Always remember that the pressure in fluid only depends on the depth (the heiht in the formula), accordin to the formula, P =ρh. This means that it does not matter what the shape of the volume of water is, because all three lasses have the same depth, all three will have the same pressure at the bottom. This is because pressure is exerted uniformly in all directions in a fluid. Remember the sequence of loic we used to derive this formula P = F A = m A = ρv A = ρ Ah A = ρh Answer D 7) What is the difference between the pressures inside and outside a tire called? A) absolute pressure B) atmospheric pressure C) aue pressure D) N/m Whenever there is a situation in which a valve separates a pressure inside a container and the atmospheric pressure, if the pressure is measured when the valve is open, the atmospheric pressure will be pushin aainst the pressure comin throuh the valve, and so, the measured pressure in this circumstance will actually be lessened by an amount equal to the atmospheric pressure. If this lessened pressure is measured by aue, the measured pressure will be called the aue pressure, and it is required to remember that the real pressure in the container will be the measured pressure plus the atmospheric pressure. The real pressure in the container is called the absolute pressure. Therefore, the difference between the pressure inside and outside the tire will be the aue pressure. Answer C 8) When atmospheric pressure chanes, what happens to the absolute pressure at the bottom of a pool? A) It does not chane. B) It increases by a lesser amount. C) It increases by the same amount. D) It increases by a reater amount. As the pool is open to the atmospheric pressure, the additional pressure of the weiht of the atmosphere will be transferred throuhout the water, so the pressure at the bottom of the pool will be the pressure exerted by the water plus the pressure of the atmosphere. As the atmospheric pressure is uniformly transmitted throuhout the water, however, the atmospheric pressure chanes, that also will be the chane in pressure at the bottom of the pool. Answer C
3 9) You are oriinally 1.0 m beneath the surface of a pool. If you dive to.0 m beneath the surface, what happens to the absolute pressure on you? A) It quadruples. B) It more than doubles. C) It doubles. D) It less than doubles. If the depth of the water doubles, the pressure due to the weiht of the water will double however, the pressure due to the weiht of the atmosphere overlyin the pool will not have underone a correspondin doublin. Therefore, the pressure in the pool will increase, but won t undero the full doublin. Answer D 10) State Pascal's principle. If an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. 11) State Archimedes' principle. The buoyant force on an object immersed in a fluid is equal to the weiht of the fluid displaced by the object. 1) 50 cm3 of wood is floatin on water, and 50 cm3 of iron is totally submered. Which has the reater buoyant force on it? A) the wood B) the iron C) Both have the same buoyant force. D) cannot be determined without knowin their densities. Many will immediately jump to the conclusion that because the block of wood is floatin, the wood must be under the influence of a reater buoyant force. But the question is NOT askin, what is the likelihood that the buoyant force will be able to hold up either block. Remember that buoyant force is equal to the weiht of the water displaced. Both objects have the same volume, so the maximum buoyant force would equal the weiht of 50 cm 3 of water which, BTW, is 50cm 3 ρ w = (1000k / m 3 ) 1,000,000 cm3 m (9.8m / s ) =.49N 3 However, only the cube of iron will feel this buoyant force because not all of the cube of wood is submered. The amount of water displaced, and so, the buoyant force will be less for the wood. Answer B 13) As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? A) It increases. B) It remains constant. C) It decreases. D) It may increase or decrease, dependin on the shape of the rock. Many will think that as the pressure becomes reater and reater as the rock descends, this increase in pressure will result in an increase in buoyant force. However, remember that buoyant force depends only on the weiht of the water displaced. Therefore, as the volume of the water displaced by the rock does not chane durin the rocks descent, the buoyant force will not chane. Answer B 14) Salt water has reater density than fresh water. A boat floats in both fresh water and in salt water. Where is the buoyant force reater on the boat? A) salt water B) fresh water C) Buoyant force is the same in both. D) impossible to determine from the information iven 3
4 The maximum buoyant force will occur when an object is completely submered because that is the point where the object will displace the most fluid. However, when the object is not completely submered, the buoyant force will equal the weiht of the object, no matter what the density of the fluid is. What the density of the fluid will affect is how much of the object is submered so the boat will sink deeper in fresh water. But, the buoyant force will be the same in both. Answer C 15) Salt water is more dense than fresh water. A ship floats in both fresh water and salt water. Compared to the fresh water, the volume of water displaced in the salt water is A) more. B) less. C) the same. D) cannot be determined from the information iven The maximum buoyant force will occur when an object is completely submered because that is the point where the object will displace the most fluid. However, when the object is not completely submered, the buoyant force will equal the weiht of the object, no matter what the density of the fluid is. What the density of the fluid will affect is how much of the object is submered the reater the density of the fluid, the less of the object submered (remember, in salt water is less than in fresh water. Answer B V o = ρ o ). So, the volume displaced 16) A steel ball sinks in water but floats in a pool of mercury. Where is the buoyant force on the ball reater? A) floatin on the mercury B) submered in the water C) It is the same in both cases. D) cannot be determined from the information iven Knowin that when an object is submered the buoyant force equals the mass of the weiht of the fluid displaced, we can reason that this will ultimately equal the weiht of the object times the ratio of the density of the fluid to the density of the object. If the object is sinkin this ratio must be less than one, so the buoyant force would be less than the objects weiht for water. In contrast, if the object is floatin on the mercury, the buoyant force will be equal to the weiht of the object, which in this circumstance would be reater than the buoyant force when the object is submered in water. When the object is submered: When the object is floatin: F B = m fl disp = V o = m o ρ o F B = m o 17) A 10k piece of aluminum sits at the bottom of a lake, riht next to a 10k piece of lead. Which has the reater buoyant force on it? A) the aluminum B) the lead C) Both have the same buoyant force. D) cannot be determined without knowin their volumes Remember that for submered objects, the buoyant force will equal weiht of the water displaced. As you are not asked for a quantitative answer you can simply reason this out. The object that displaces more water (the reater volume) will have the reater buoyant force. As both masses are equal, the mass with the reater volume will be the one with the lesser density, which will be the aluminum. This means that the aluminum will have the reater buoyant force. Answer A 18) A piece of iron rests on top of a piece of wood floatin in a bathtub. If the iron is removed from the wood, what happens to the water level in the tub? A) It oes up. 4
5 B) It oes down. C) It does not chane. D) impossible to determine from the information iven The overall density of the block of wood plus the iron must not have been so reat that the combined mass sank. But it should make sense that when you remove the iron from the block of wood that the block of wood will displace less water. Therefore, the level of the water must decrease. Answer B 19) A piece of wood is floatin in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. If the top piece is taken off and placed in the water, what happens to the water level in the tub? A) It oes up. B) It oes down. C) It does not chane. D) cannot be determined from the information iven As the amount of water displaced when one block is on top of the other will be equal to the weiht of the blocks combined, when the top block is removed and placed in the water, the weiht of water displaced will still be equal to the weiht of both block, so the level of the water will not chane. Answer C 0) Water flows throuh a pipe. The diameter of the pipe at point B is larer than at point A. Where is the speed of the water reater? A) point A B) point B C) same at both A and B D) cannot be determined from the information iven Remember that accordin to the equation of continuity/volume flow equation ( or A 1 v 1 = A v ΔV 1 = ΔV Δt Δt ), the reater the crosssectional area of a tube, the slower the rate of flow throuh the tube. Therefore, the speed of the water will be the reatest at point A. Answer A 1) An ideal fluid flows at 1 m/s in a horizontal pipe. If the pipe widens to twice its oriinal radius, what is the flow speed in the wider section? A) 1 m/s B) 6.0 m/s C) 4.0 m/s D) 3.0 m/s (An ideal fluid would be a fluid that has no internal resistanceits viscosity would be stated to be 0 these do not actually exist in nature.) This would be a question where instead of tryin to reason out the answer because you cannot remember how the actual relationship between velocity and diameter works without seein it visually, you simply write out the relationship, showin what happens to the result when you chane a variable as described in the problem. Make sure you remember in this case that if you double the radius you multiply the area by 4. A 1 = v so v A v = A A 1 v 1 1 A 1 v A 4 v So, when you double the radius you divide the velocity by 4, which means 3.0 m/s. Answer D ρ 1 A 1 v 1 = ρ A v ) An ideal fluid flows at 1 m/s in a horizontal pipe. If the pipe narrows to half its oriinal radius, what is the flow speed in the narrower section? A) 1 m/s B) 4 m/s 5
6 C) 36 m/s D) 48 m/s (An ideal fluid would be a fluid that has no internal resistanceits viscosity would be stated to be 0 these do not actually exist in nature.) This would be a question where instead of tryin to reason out the answer because you cannot remember how the actual relationship between velocity and diameter works without seein it visually, you simply write out the relationship, showin what happens to the result when you chane a variable as described in the problem. Make sure you remember in this case that if you halv the radius you divide the area by 4. A 1 = v so v A v = A A 1 v 1 1 A 1 1 A v 1 4v So, when you halve the radius you 4 multiply the velocity by 4, which means 48 m/s. Answer D 3) State Bernoulli's principle. Where the velocity of fluid is hih, the pressure is low; and where the velocity is low, the pressure is hih. 4) Which one of the followin is associated with the law of conservation of enery in fluids? A) Archimedes' principle B) Bernoulli's principle C) Pascal's principle D) equation of continuity Remember, Pascals s principle states that if an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. Archimede s principle states that the buoyant force on an object immersed in a fluid is equal to the weiht of the fluid displaced by the object. Bernoulli s principle states that where the velocity of fluid is hih, the pressure is low; and where the velocity is low, the pressure is hih. In analyzin Bernoulli s principle to create Bernoulli s equation, we find that P + 1 ρv + ρy = P ρv + ρy 1 1. This essentially shows that the forces plus the mechanical enery present at one point in a fluid system equal the forces plus mechanical enery plus the mechanical enery at every other point conservation of enery. Answer B 5) As the speed of a movin fluid increases, the pressure in the fluid A) increases. B) remains constant. C) decreases. D) may increase or decrease, dependin on the viscosity. Remember, Bernoulli s equation states P + 1 ρv + ρy is constant therefore, if velocity of the fluid increases, pressure must undero a correspondin decrease. Answer C 6) Water flows throuh a pipe. The diameter of the pipe at point B is larer than at point A. Where is the water pressure reatest? 6
7 A) point A B) point B C) same at both A and B D) cannot be determined from the information iven Remember, Bernoulli s equation states P + 1 ρv + ρy is constant therefore, the pressure will be reatest wherever velocity is the lowest. Remember also that accordin to the equation of continuity, velocity will be the lowest wherever the diameter is reatest. Therefore, the pressure will be reatest wherever the diameter is the reatest. Answer B 7) When you blow some air above a paper strip, the paper rises. This is because A) the air above the paper moves faster and the pressure is hiher. B) the air above the paper moves faster and the pressure is lower. C) the air above the paper moves slower and the pressure is hiher. D) the air above the paper moves slower and the pressure is lower. Remember that Bernoulli s principle states, where velocity of fluid is hih, pressure is low; where velocity of fluid is low, pressure is hih. If you blow air above a paper strip, as the air above the strip will be movin faster, the pressure above the strip will be lower than underneath the strip therefore, the paper rises. Answer B 8) A sky diver falls throuh the air at terminal velocity. The force of air resistance on him is A) half his weiht. B) equal to his weiht. C) twice his weiht. D) cannot be determined from the information iven When a sky diver falls throuh the air at terminal velocity, this means that he is no loner acceleratin the force due to ravity must be balanced by the upward force due to the buoyancy of the fluid atmosphere. (Keep in mind, however, that the atmosphere is compressible and its density increases the closer you et to the earth in addition, as a skydiver oes faster they will compress the air particles beneath them at a reater and reater rate. So not only does the air density increase as one moves toward the surface of the earth, the faster the velocity, the reater this effect. Also, the more surface area exposed to the atmosphere the reater this effect. The actual relationship that determines terminal velocity includes these and other factors and is quite complex.) Under the conditions described, even thouh we may not know all the factors that determine the force of air resistance, we know that to balance ravity, it must equal the weiht of the skydiver. Answer B Questions 9 throuh 37 require information that is not part of the AP Curriculum Quantitative Problems 1) A plastic block of dimensions.00 cm 3.00 cm 4.00 cm has a mass of What is its density? ρ = m V =.0300k (.000m)(.0300m)(.0400m) = 1.5 x 103 k / m 3 You are asked to find density so write out the formula for this. You have been iven the mass so you can plu this in directly, rememberin to convert to.0300 k. You have not been iven the volume but have the dimensions so you can plu these in the formula to find V, rememberin to convert to meters. Check your answer by knowin that in eneral, densities rane in the 1000 s of k per cubic meter. 3SF ) A liquid has a specific ravity of What is its density? Remember that specific ravity is defined as the ratio of the density of the iven substance to the density 7
8 of water, and that as the density of water is 1 (1.00 x 10 3 k/m 3 ), the value of the specific ravity is 3.57 x 10 k / m 3 3) A brick weihs 50.0 N, and measures 30.0 cm 10.0 cm 4.00 cm. What is the maximum pressure it can exert on a horizontal surface? As P = F A, pressure is inversely proportional to area, and the face of the brick throuh which the reatest pressure will be exerted is the smallest face, 10.0 cm x 4.00 cm. Therefore, the reatest pressure that can be exerted is: P = F A = 50.0N (.10m)(.04m) = 1.5x104 Pa = 1.5kPa A drawin may be useful. How does the pressure exerted by the brick compare to that of atmospheric pressure. 4) A person weihin 900 N is standin on snowshoes. Each snowshoe has area 500 cm. What is the pressure on the snow? You are asked for pressure so write down the formula for pressure. You have been iven the force, and you have been iven the area. Reconize that the force of the body weiht is spread over the area of both shoes combined. Also, you can either convert 500 cm cm = 5000 cm in your mind to.50 m, or you can convert in the calculation usin a conversion factor I have shown both ways. P = F A = 900N.50m = 1800Pa = 1.8x103 Pa = 1.8kPa P = F A = 900N (100cm x 100cm) x = 1800Pa = 1.8x10 3 Pa = 1.8kPa 5000cm 1m 5) How much pressure (absolute) must a submarine withstand at a depth of 10.0 m in the ocean? The term absolute is a clue to remind you that in this calculation you need to include the atmospheric pressure in the calculation, so you will need to add in 101,35 Pa. Additionally, the density of seawater is reater than that of fresh water so the density of seawater will be required. Remember that the formula for determinin the pressure at a iven depth h is: P = ρh. So, the calculation would be: P = ρh + 101,35 N m = 1.05 x 103 k m 3 (10.0m)(9.8m / s ) + 101,35 N m = 1.307x106 Pa = 1.307x10 3 kpa 6) A circular window of 30 cm diameter in a submarine can withstand a maximum force of N. What is the maximum depth in a lake to which the submarine can o without damain the window? The formula to find the pressure of water at a certain depth is P = ρh. This problem involves a simple rearranement of the pressure of water formula to isolate h. F/A will need to be substituted in for P, and A of a circle is πr. Atmospheric pressure will also need to be added in. P = ρh + P atm h = P water P atm ρ = F P A atm = ρ (5.0x10 5 N ) ( (π )(.15m) ) 101,35 N m (1.000x10 3 k / m 3 )(9.8m / s ) = 740. m 7) What is the aue pressure if the absolute pressure is 300 kpa? Whenever there is a situation in which a valve separates a pressure inside a container and the atmospheric pressure, if the pressure is measured when the valve is open, the atmospheric pressure will be pushin aainst the pressure comin throuh the valve, and so, the measured pressure in this 8
9 circumstance will actually be lessened by an amount equal to the atmospheric pressure. If this lessened pressure is measured by aue, the measured pressure will be called the aue pressure, and it is required to remember that the real pressure in the container will be the measured pressure plus the atmospheric pressure. The real pressure in the container is called the absolute pressure. Therefore, the difference between the pressure inside and outside the tire will be the aue pressure. This means that to find the aue pressure we need to subtract P atm from P abs : P aue = P abs P atm = 300kPa 101kPa = 199kPa 8) What is the aue pressure at a location 15.0 m below the surface of sea? (The density of seawater is k/m3.) Because the ocean is under the influence of the atmosphere, the pressure at a certain depth in the ocean will be the pressure of the weiht of the water at that depth, plus the atmospheric pressure. The aue pressure is considered to be the pressure at that depth without includin the pressure due to the weiht of the atmosphere. Therefore, the aue pressure at 15.0 m can be found by simply applyin the formula to determine water pressure: P aue = ρh = 1.03x103 k m 3 (15.0m)(9.8m / s ) = 151,410 Pa = 151kPa 9) What is the absolute pressure at a location 15.0 m below the surface of sea? (The density of seawater is k/m3.) In conjunction with the explanation for question 8, the absolute pressure would be considered the sum of the aue pressure and the atmospheric pressure: P abs = ρh + P atm = 1.03x103 k m 3 (15.0m)(9.8m / s ) + 101,35Pa = 5735 Pa = 53kPa 10) A 500N weiht sits on the small piston of a hydraulic machine. The small piston has area.0 cm. If the lare piston has area 40 cm, how much weiht can the lare piston support? Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throuhout the contained fluid, includin at the output piston. Therefore, P i = P o = F o A o F o F i = A o You will not be iven this formula on the test, you will have to derive it if you need it. For this problem we are bein asked for the weiht that can be supported by the output piston, which is the output force, so isolate F o, plu in the values and solve: F o A o = (500N ) 40 = 10,000N If the units of area are the same for both pistons, it does not matter if the units are not m, it is the ratio of the output piston to the input piston that matters. For multiple choice questions it is probably easy enouh to remember that the hydraulic advantae is simply this ratio times the input force. However, for free response questions you would still have to show all the work. 11) In a hydraulic arae lift, the small piston has a radius of 5.0 cm and the lare piston has a radius of 15 cm. What force must be applied on the small piston in order to lift a car weihin 0,000 N on the lare piston? Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throuhout the contained fluid, includin at the output piston. Therefore, 9
10 P i = P o = F o A o F o F i = A o You will not be iven this formula on the test, you will have to derive it if you need it. For this problem we are bein asked for the input force, so isolate this, plu in values and solve. F i = F o A o = F o π r i π r o 5 = (0000N ) 15 = N = 00N Make sure you are payin attention to reconize that you are iven the radius, not the area. 1) A 13,000 N vehicle is to be lifted by a 5 cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter piston to accomplish this? Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throuhout the contained fluid, includin at the output piston. Therefore, P i = P o = F o A o F o F i = A o You will not be iven this formula on the test, you will have to derive it if you need it. For this problem we are bein asked for the input force, so isolate this, plu in values and solve. As you are only concerned about the ratio of areas, as lon as the units are the same you do not need to convert to m. You have been iven diameters, however, so make sure these are converted to radii. = V o A F i π r o A o o π r = (13000N ).5 o 1.5 = 50N F B = w w = ρ o V o w w = ρ o w = ρ V o o = ρ o 13) The small piston of a hydraulic lift has a diameter of 8.0 cm, and its lare piston has a diameter of 40 cm. The lift raises a load of 15,000 N. (a) Determine the force that must be applied to the small piston. Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throuhout the contained fluid, includin at the output piston. Therefore, P i = P o = F o A o F o F i = A o You will not be iven this formula on the test, you will have to derive it if you need it. For this problem we are bein asked for the input force, so isolate this, plu in values and solve. As you are only concerned about the ratio of areas, as lon as the units are the same you do not need to convert to m. You have been iven diameters, however, so make sure these are converted to radii. 10
11 A F i π r o A o o π r = (15000N ) 4 o 0 = 600N (b) Determine the pressure applied to the fluid in the lift. Pressure equals force divided by area. In this case we are talkin about the input force and the input area. We know the input force is 600 N as we just calculated it. You have been told that the input area is a piston with a radius of.04m, so calculate the input area usin πr. Plu the values in and solve: P = F A = F πr = 600N π(.040m) = 1.x105 Pa 14) A block of metal weihs 40 N in air and 30 N in water. What is the buoyant force of the water? You are aware that the weiht in water is also known as the apparent weiht. You know a simple formula to determine the apparent weiht apparent weiht is the difference between the actual weiht and the buoyant force. Rearranin you can isolate F B. You have been iven both the actual and apparent weiht. Plu in the values and solve. w = w F B F B = w w = 40N 30N = 10N 15) An object has a volume of 4.0 m3 and weihs 40,000 N. What will its weiht be in water? You are aware that the weiht in water is also known as the apparent weiht. You know a simple formula to determine the apparent weiht apparent weiht is the difference between the actual weiht and the buoyant force. You have been iven the actual weiht and you know the buoyant force is equal to the weiht of the displaced water, which is also equal to ρ w. You have been iven the volume of the object and this will be the volume of the displaced water. You know the density of water. Plu the values in and solve: w = w F B = w ρ w = (40000N ) (1000k / m 3 )(4.0m 3 )(9.8m / s ) = 800N 16) A 4.00k cylinder of solid iron is supported by a strin while submered in water. What is the tension in the strin? (The specific ravity of iron is 7.86.) From readin the problem you know that as there is no acceleration, you will have the force due to ravity balanced by the sum of the buoyant force and tension in the strin. Drawin a free body diaram leads you to a net force equation, from which you can isolate the force of tension, which will equal the force due to ravity minus the buoyant force. You know how to find the force due to ravity. The buoyant force will be ρ w. However, you do not have the volume of water displaced. But, you have been iven the mass of the iron cylinder and its density so you can find the volume of the cylinder and so that of the water displaced. F n = F T + F B F G = 0N F T = F G F B = m Fe ρ W = m Fe ρ W m Fe ρ Fe 4.0k (4.0k)(9.8m / s ) (1000k / m 3 ) 7860k / m 3 (9.8m / s ) = 34.N 17) If the density of old is k/m3, what buoyant force does a 0.60k old crown experience when it is immersed in water? 11
12 Remember that the buoyant force equals the weiht of the displaced fluid which also equals. You know the density of water. As the volume displaced will equal the volume of the crown, you can find this by knowin that the density of the old will equal m/v, and so, V = m/d. We have been iven the both the mass and the density of the crown so we can determine V. Plu the values in and solve: m = ρ crown.60k fl = (1000k / m 3 ) ρ crown 19300k / m 3 (9.8m / s ) =.30N 18) A crane lifts a steel submarine (density = k/m3) of mass 0,000 k. What is the tension in the liftin cable (1) when the submarine is submered, and () when it is entirely out of the water? From readin the problem you know that as there is no acceleration, you will have the force due to ravity balanced by the sum of the buoyant force and tension in the strin. Drawin a free body diaram leads you to a net force equation, from which you can isolate the force of tension, which will equal the force due to ravity minus the buoyant force. You know how to find the force due to ravity. The buoyant force will be. However, you do not have the volume of water displaced. But, you have been iven the mass of the iron cylinder and its density so you can find the volume of the cylinder and so that of the water displaced. F net = F T + F B F G = 0N F T = F G F B = m sub = m sub m sub ρ sub 0000k = (0,000k)(9.8m / s ) (1000k / m 3 ) 7800k / m 3 (9.8m / s ) 1.7x10 5 N ρ w When the sub is out of water the weiht will only be balanced by the tension so the tension will simply equal the force due to ravity: F T = m = (0000k)(9.8) =.0x10 5 N 19) An object weihs 7.84 N when it is in air and 6.86 N when it is immersed in water. What is the specific ravity of the object? Remember that if we know the weiht and apparent weiht we can easily determine the specific ravity usin the followin relationship: w w w = ρ V o o and the apparent weih plu the values in and solve: = ρ o ρ o = We have been iven both the weiht w w w = 7.84N = ) A container of water is placed on a scale, and the scale reads 10. Now a 0 piece of copper (specific ravity = 8.9) is suspended from a thread and lowered into the water, not touchin the bottom of the container. What will the scale now read? We cannot assume that any water is lost from the pail as we are not iven this information we must assume all the water is still in the pail. Reconize that the water will enerate a buoyant force on the cube of copper. Therefore, the cube will exert an equal and opposite force on the water, so this must be added to the weiht the pail already has. This extra force will equal the weiht of the water displaced by the cube, and we will want to divide by to et the actual mass. We can find the volume 1
13 displaced because we know the mass of copper and its density. Therefore, the mass added to the existin mass of the pail will equal density of the water displaced times the mass of the copper divided by its density: m final mass = m init + m = m disp init + ρ W = m init + ρ Cu 0 W = 10 + (1.00 / cm 3 ) ρ Cu 8.9 / cm 3 = 1 1) A piece of aluminum with a mass of 1.0 k and density of 700 k/m3 is suspended from a strin and then completely immersed in a container of water. The density of water is 1000 k/m3. (a) Determine the volume of the piece of aluminum. (b) Determine the tension in the strin after the metal is immersed in the container of water. ) A cylindrical rod of lenth 1 cm and diameter.0 cm will just barely float in water. What is its mass? 3) A rectanular box of neliible mass measures 5.0 m lon, 1.0 m wide, and 0.50 m hih. How many kilorams of mass can be loaded onto the box before it sinks in a lake? 4) A 1.0m3 object floats in water with 0% of it above the waterline. What does the object weih out of the water? 5) An object floats with half its volume beneath the surface of the water. The weiht of the displaced water is 000 N. What is the weiht of the object? 13
14 6) A solid object floats in water with threefourths of its volume beneath the surface. What is the object's density? 7) A 00N object floats with threefourths of its volume beneath the surface of the water. What is the buoyant force on the object? 8) A polar bear of mass 00 k stands on an ice floe 100 cm thick. What is the minimum area of the floe that will just support the bear in saltwater of specific ravity 1.03? The specific ravity of ice is ) Liquid flows throuh a pipe of diameter 3.0 cm at.0 m/s. Find the flow rate. 30) Liquid flows throuh a 4.0 cm diameter pipe at 1.0 m/s. There is a.0 cm diameter restriction in the line. What is the velocity in this restriction? 14
15 31) Water flows at 1 m/s in a horizontal pipe with a pressure of N/m. If the pipe widens to twice its oriinal radius, what is the pressure in the wider section? 3) How much pressure does it take for a pump to supply a drinkin fountain with 300 kpa, if the fountain is 30.0 m above the pump? 33) A hole of radius 1.0 mm occurs in the bottom of a water storae tank that holds water at a depth of 15 m. At what rate will water flow out of the hole? 34) Water flows throuh a horizontal pipe of crosssectional area 10.0 cm at a pressure of 0.50 atm. The flow rate is m3/s. At a valve, the effective crosssectional area of the pipe is reduced to 5.00 cm. What is the pressure at the valve? 35) SAE No. 10 oil has a viscosity of 0.0 Pas. How lon would it take to pour 4.0 L of oil throuh a funnel with a neck 15 cm lon and.0 cm in diameter? Assume the surface of the oil is kept 6 cm above the top of the neck, and nelect any dra effects due to the upper part of the funnel. 36) Suppose that the buildup of fatty tissue on the wall of an artery decreased the radius by 10%. By how much would the pressure provided by the heart have to be increased to maintain a constant blood flow? 15
16 37) Two narrow tubes are placed in a pan of water. Tube A has twice the diameter of tube B. If water rises 4 cm in tube A, how hih will it rise in tube B? 38) The surface tension of water is N/m. How hih will water rise in a capillary tube of diameter 1. mm? 39) To what heiht would water at 0 deree Celsius rise in a lass capillary tube with a diameter of 0.50 mm? 40) A narrow tube is placed vertically in a pan of water, and the water rises in the tube to 4.0 cm above the level of the pan. The surface tension in the liquid is lowered to onehalf its oriinal value by the addition of some soap. What happens to the heiht of the liquid column in the tube? 41) Consider a rectanular frame of lenth 0.10 m and width m with a soap film formed within its confined area. If the surface tension of soapy water is N/m, how much force does the soap film exert on the m side? 16
17 4) An adjustable rectanular frame has lenth 1 cm and width 5 cm, and is built in such a way that one of the loner sides can be moved without burstin the soap film contained in the frame. Soapy water has surface tension 0.06 N/m. How much work is required to move the adjustable lon side so that the frame is 1 cm square? 17
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