09 homework JACOBS, AMBER Due: Apr 8 2008, 1:00 am Question 1, chap 14, sect 7. part 1 of 2 10 points A rock weighs 110 N in air and has a volume of 0.0029 m3 . The acceleration of gravity is 9.8 m/s2 . What is its apparent weight when submerged in water? Correct answer: 81.58 N (tolerance 1 %). Explanation: The buoyant force for any liquid is defined to be F = V g but it is equal to the weight lost by the object, so W - Wa = V g Wa = W - V g Dimensional analysis for F : kg m m kg m3 2 = =N 3 m s s2 apparent weight is Wa = W - w V g Question 2, chap 14, sect 7. part 2 of 2 10 points If it is submerged in a liquid with a density exactly 1.9 times that of water, what will be its new apparent weight? Correct answer: 56.002 N (tolerance 1 %). Explanation: liquid, the apparent weight is Wa2 = W - n V g Question 3, chap 14, sect 10. part 1 of 4 10 points Assume: The fluid is incompressible non-viscous. Shown below is a cross-section of a tical view of a pipe discharging a fluid the atmosphere at its highest elevation. and verinto The
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pipe diameter increases and then remains constant. Pi is the pressure and vi is the speed of the fluid, at locations i = w, u, z, and y. w u
z y
The relationship between the magnitude of the velocity v v at position w and u is 1. vu = vw . 2. vu > vw . 3. vu < vw . correct 4. indeterminable, not enough information available. Explanation: The fluid is incompressible, so vw > vu . Question 4, chap 14, sect 10. part 2 of 4 10 points The relationship between the pressure P at position u and z is 1. Pz = Pu . 2. Pz > Pu . correct 3. Pz < Pu . 4. indeterminable, not enough information available. Explanation: Bernoulli's equation 1 P + v 2 + g y = constant. 2 At a lower elevation the pressure P = g y is greater. Thus, Pz > Pu .
homework 09 JACOBS, AMBER Due: Apr 8 2008, 1:00 am Question 5, chap 14, sect 10. part 3 of 4 10 points The relationship between the pressure P at position w and y is 1. Py < Pw . 2. Py > Pw . 3. indeterminable, not enough information available. correct 4. Py = Pw . Explanation: The pressure at w and y may be comparable. Not enough information is available to determine which is larger, if they are not equal. Both Pw < Pu and Py < Pu , see Parts 4 & 2. Question 6, chap 14, sect 10. part 4 of 4 10 points The relationship between the pressure P at position w and u is 1. Pu < Pw . 2. Pu > Pw . correct 3. Pu = Pw . 4. indeterminable, not enough information available. Explanation: Using Bernoulli's equation, we have P+ 1 2 v + g y = constant. 2 4.5 m/s 3.2 cm v2 2.1 cm
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Given: The viscosity is negligible. Atmospheric pressure is 101300 Pa. Water flows at speed of 4.5 m/s through a horizontal pipe of diameter 3.2 cm. The gauge pressure P1 of the water in the pipe is 2 atm. A short segment of the pipe is constricted to a smaller diameter of 2.1 cm. 2 atm P1 P2
What is the speed v2 of the water flowing through the constricted segment? Correct answer: 10.449 m/s (tolerance 1 %). Explanation: Let : Patm = 101300 Pa , = 1000 kg/m3 , = 0, v1 = 4.5 m/s , d1 = 3.2 cm , d2 = 2.1 cm , and gauge P1 = 2 atm . To a good approximation, water is incompressible (at ordinary pressures) and thus must obey the continuity equation F = Av = In other words, d2 v2 = d2 v1 , 2 1 and hence the speed of the water flow through the constricted segment of the pipe is v2 = v1 d2 1 d2 2 (3.2 cm)2 (2.1 cm)2 d 2
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v = constant.
At a greater velocity the pressure 1 P = v 2 is smaller. 2 Since vw > vu , then Pu > Pw . Question 7, chap 14, sect 10. part 1 of 2 10 points
= (4.5 m/s)
= 10.449 m/s .
homework 09 JACOBS, AMBER Due: Apr 8 2008, 1:00 am Question 8, chap 14, sect 10. part 2 of 2 10 points What is the gauge pressure of the water flowing through the constricted segment? Correct answer: 1.56105 atm (tolerance 1 %). Explanation: In the absence of viscosity, the water pressure and speed obey the Bernoulli equation
1 P + g h + 2 v 2 = constant,
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Correct answer: 3640.29 kg/m3 (tolerance 1 %). Explanation: Basic Concepts: Fapparent = Fg - FB Fg,object metal = FB water Given: Fg = 50.6 N Fapparent,water = 36.7 N water = 1.00 103 kg/m3 Solution: FB = Fg - Fapparent,water so metal = Fg water FB Fg water = Fg - Fapparent,water (50.6 N) 1000 kg/m3 = 50.6 N - 36.7 N = 3640.29 kg/m3
which for a horizontal pipe (h1 = h2 ) gives us
1 1 2 2 P2 + 2 v2 = P1 + 2 v1 ,
or P2 = P1 -
1 2 2 v2
-
2 v1
.
Strictly speaking, the pressures P1 and P2 in the Bernoulli equation are the absolute pressures rather than gauge pressures. However, since the air pressure outside the pipe is congauge gauge stant, the difference P2 - P1 is the same as the difference between the absolute pressures P2 - P1 . Therefore,
gauge P2
1 gauge 2 2 = P1 - v2 - v1 2 1000 kg/m3 = (202600 Pa) - 2 2 (10.449 m/s) - (4.5 m/s)2 = [(202600 Pa) - (44465.6 Pa)] (9.87167 10-6 atm/Pa) = 1.56105 atm .
Question 10, chap 14, sect 7. part 2 of 2 10 points b) Find the density of the unknown liquid. Correct answer: 683.453 kg/m3 (tolerance 1 %). Explanation: Basic Concept: Fg,object metal = FB liquid Given: Fapparent,liquid = 41.1 N Solution: FB metal liquid = Fg Fg - Fapparent,liquid metal = Fg (50.6 N - 41.1 N) (3640.29 kg/m3 ) = 50.6 N 3 = 683.453 kg/m
Dimensional analysis of v kg m 3 m s
2
2
Pa
1 atm atm 1.013 105 Pa
Question 9, chap 14, sect 7. part 1 of 2 10 points A piece of metal weighs 50.6 N in air, 36.7 N in water, and 41.1 N in an unknown liquid. a) Find the density of the metal.
homework 09 JACOBS, AMBER Due: Apr 8 2008, 1:00 am Question 11, chap 14, sect 7. part 1 of 4 10 points A 10.5 kg block of metal is suspended from a scale and immersed in water as in the figure. The dimensions of the block are 11.7 cm 9.1 cm 10.3 cm. The 11.7 cm dimension is vertical, and the top of the block is 4.84 cm below the surface of the water. The acceleration of gravity is 9.8 m/s2 .
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Thus, the downward force exerted on the top by the water is Ftop = Ptop A = Ptop l w = (101774 Pa) (0.091 m) (0.103 m) = 953.931 N .
Question 12, chap 14, sect 7. part 2 of 4 10 points What is the force exerted by the water on the bottom of the block? Correct answer: 964.678 N (tolerance 1 %). Explanation:
B
T2 Mg
Given : h = 11.7 cm = 0.117 m and hbottom = htop + h = 0.1654 m . The absolute pressure at the level of the bottom of the block is Pbottom = P0 + (water ) (g) (hbottom) = 101300 Pa + (1000 kg/m3 ) (9.8 m/s2 ) (0.1654 m) = 102921 Pa . Thus, the downward force exerted on the top by the water is Fbottom = Pbottom A = (102921 Pa) (0.091 m) (0.103 m) = 964.678 N .
What is the force exerted by the water on the top of the block? (Take P0 = 101300 Pa .) Correct answer: 953.931 N (tolerance 1 %). Explanation: Given : m = 10.5 kg , w = 9.1 cm = 0.091 m , l = 10.3 cm = 0.103 m , htop = 4.84 cm = 0.0484 m ,
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water = 1000 kg/m , P0 = Pa 101300 , and g = 9.8 m/s . The absolute pressure at the level of the top of the block is Ptop = P0 + (water ) (g) (htop) = 101300 Pa + (1000 kg/m3 ) (9.8 m/s2 ) (0.0484 m) = 101774 Pa .
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Question 13, chap 14, sect 7. part 3 of 4 10 points What is the reading of the spring scale? Correct answer: 92.1529 N (tolerance 1 %). Explanation: The scale reading equals the tension, T , in the cord supporting the block. Since the block
homework 09 JACOBS, AMBER Due: Apr 8 2008, 1:00 am is in equilibrium, Fy = T + Fbot - Ftop - m g = 0 T = -Fbot + Ftop + m g = -964.678 N + 953.931 N + (10.5 kg) (9.8 m/s2 ) = 92.1529 N . Question 14, chap 14, sect 7. part 4 of 4 10 points What is the buoyant force? Correct answer: 10.7471 N (tolerance 1 %). Explanation: From Archimedes's principle, the buoyant force on the block equals the weight of the displaced water. Thus B = (water Vblock ) g = (water ) (h w d) g = (1000 kg/m3 ) (0.117 m) (0.091 m) (0.103 m) (9.8 m/s2 ) = 10.7471 N . From part(a), the difference between the forces at the top and bottom of the block is Fbot - Ftop = 964.678 N - 953.931 N = 10.7471 N , which is the same as the buoyant force. Question 15, chap 14, sect 7. part 1 of 1 10 points Assume you are floating in water with your lungs full of air. When you exhale, you sink lower in the water. Which of the following is a possible explanation of the sinking effect? I) Your volume decreases and so does the buoyant force. II) Your mass decreases as you let the air out of your lungs making it easier for you to sink. III) Your overall density decreases. IV) Your overall density increases. 1. II only 2. I only 3. I and IV only correct 4. I and II only 5. III only 6. IV only 7. I and III only 8. I, II, and III only 9. II and III only 10. II and IV only
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Explanation: Your mass decreases, but not significantly. Your volume decreases as your lungs deflate, causing your density to increase significantly (nearly the same mass divided by a smaller volume). A smaller volume means a smaller buoyant force. Density most directly affects how high in the water you float. Question 16, chap 14, sect 7. part 1 of 1 10 points The relative densities of water, ice, and alcohol are 1.0, 0.9, and 0.8 respectively. Which of the following is true about ice cubes floating in a mixed alcoholic drink? 1. In a drink that is predominantly alcohol ice cubes will float the highest. 2. Ice cubes will float in a mixed drink, but not as high as they would in water. correct 3. Ice cubes will sink to the bottom of a mixed drink. Explanation: The mixture of alcohol and water is less dense than water. A greater amount of this
homework 09 JACOBS, AMBER Due: Apr 8 2008, 1:00 am less dense liquid must be displaced to equal the weight of the floating ice. Thus ice cubes will float lower in a mixed drink than in water. Submerged ice cubes in a cocktail indicate that it is predominantly alcohol. Question 17, chap 14, sect 10. part 1 of 2 10 points A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 13 m below the water level. The rate of flow of water from the leak is 2.8 10-3 m3 /min. The acceleration of gravity is 9.81 m/s2 . a) Determine the speed at which the water leaves the hole. Correct answer: 15.9706 m/s (tolerance 1 %). Explanation: Basic Concept: 1 2 1 2 P1 + v1 + gh1 = P2 + v2 + gh2 2 2 Given: h2 - h1 = 13 m g = 9.81 m/s2 Solution: 1 2 P0 + v1 + gh1 = P0 + gh2 2 since, for the open top of the tank and the hole, P1 = P2 = P0 , and v2 0 m/s because the hole is small. 1 2 v = g(h2 - h1 ) 2 1 2 v1 = 2g(h2 - h1 ) v1 = = 2g(h2 - h1 ) 2(9.81 m/s2 )(13 m)
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b) Determine the diameter of the hole. Correct answer: 0.00192885 m (tolerance 1 %). Explanation: Basic Concept: flow rate = A1 v1 = Given: flow rate = 2.8 10-3 m3 /min Solution: flow rate D2 = 4 v1 D 2
2
v1
D= =
4(flow rate) v1 1 min 60 s
4(0.0028 m3 /min) (15.9706 m/s) = 0.00192885 m
Question 19, chap 14, sect 10. part 1 of 1 10 points Bernoulli's equation can be derived from the conservation of: 1. volume 2. mass 3. angular momentum 4. energy correct 5. pressure Explanation: Solution Bernoulli's Equation states that P+ 1 2 v + g y = constant 2
= 15.9706 m/s Question 18, chap 14, sect 10. part 2 of 2 10 points
along a streamline in incompressible, steady flow. This is a statement of the conservation
homework 09 JACOBS, AMBER Due: Apr 8 2008, 1:00 am of energy, where the energy takes three forms: energy gained from the work done by the pressure, kinetic energy, and potential energy. The fact that the sum is a constant expresses the conservation of energy. Question 20, chap 15, sect 3. part 1 of 3 10 points A 211 g mass is connected to a light spring of force constant 6 N/m and it is free to oscillate on a horizontal, frictionless track. The mass is displaced 4 cm from the equilibrium point and released form rest. 4 cm 6 N/m x 211 g
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Explanation: The velocity as a function of time is given by v(t) = - A sin( t) , so the maximum speed of the mass is equal to vmax = A = (5.33254 s-1 ) (0.04 m) = 0.213302 m/s . Question 22, chap 15, sect 3. part 3 of 3 10 points What is the maximum acceleration of the mass? Correct answer: 1.13744 m/s2 (tolerance 1 %). Explanation: The acceleration as a function of time is given by a(t) = - 2 A cos( t) , so the maximum acceleration of the mass is equal to amax = 2 A = (5.33254 s-1 )2 (0.04 m) = 1.13744 m/s2 . Question 23, chap 15, sect 4. part 1 of 1 10 points A block of unknown mass is attached to a spring of spring constant 4.63 N/m and undergoes simple harmonic motion with an amplitude of 9.63 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 36.4 cm/s. Calculate the mass of the block. Correct answer: 0.243048 kg (tolerance 1 %). Explanation: Basic Concepts Energy conservation: If K is kinetic energy and U is potential energy, Ki + Ui = Kf + Uf
x=0 Find the period of the motion. Correct answer: 1.17827 s (tolerance 1 %). Explanation: This situation corresponds to the special case x(t) = A cos t , with A = 4 cm = 0.04 m . Therefore, the frequency is = = k m 6 N/m 0.211 kg
= 5.33254 s-1 . For the period we find T = 2 = 1.17827 s .
Question 21, chap 15, sect 3. part 2 of 3 10 points What is the maximum speed of the mass? Correct answer: 0.213302 m/s (tolerance 1 %).
homework 09 JACOBS, AMBER Due: Apr 8 2008, 1:00 am Kinetic energy of particle with mass m and speed v: 1 K = m v2 2 Mass m on spring with constant k: = k m
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and potential energy of a spring at displacement x: 1 U = k x2 2 Period 2 T = Solution: Call the maximum displacement (amplitude) A. The halfway displacement is A/2. Energy conservation requires 0+ or k A2 = m v 2 + so m= 3 k A2 4 v2 (3) (4.63 N/m) (0.0963 m)2 = (4) (0.364 m/s)2 = 0.243048 kg . 1 1 1 k A2 = m v 2 + k 2 2 2 1 k A2 4 A 2
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