Answer, Key Homework 2 David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. Chapter 15 problems. 001 (part 1 of 2 5 points A simple U-tube that is open at both ends is partially filled with a of density 553 kg/m 3. Water is then poured into one arm of the tube, forming a column 11 cm in height, as shown in the following diagram. The density of water is 1000 kg/m 3. h 1 water What is the difference, h 1, in the heights of the two surfaces? Correct answer: 8.8915 cm. Basic Concepts: gauge pressure, variation of pressure with depth Because the in the U-tube is static, the pressure exerted by the water column of height l in the right branch of the tube must balance the pressure exerted by the of height h 1 + l in the left branch. Therefore, P 0 + (l + h 1 g = P 0 + l ρ w g Solving for h 1, ( ρw h 1 = l 1 ( 1000 kg/m 3 = 11 cm 553 kg/m 3 1 = 8.8915 cm. l partially filled with water. A of density 553 kg/m 3 is then poured into one arm of the tube, forming a column 11 cm in height, as shown in the following diagram. water h 2 What is the difference, h 2, in the heights of the two surfaces? Correct answer: 4.917 cm. Because the in the U-tube is static, the pressure exerted by the water column of height l h 2 in the left branch of the tube must balance the pressure exerted by the of height l poured into the right branch. Therefore, Solving for h 2, P 0 + (l h 2 ρ w g = P 0 + l g ( h 2 = l 1 ρ l ρ w 553 kg/m3 = 11 cm (1 1000 kg/m 3 = 4.917 cm 003 (part 1 of 1 0 points Consider an object that floats in water but sinks in oil. When the object floats in a glass of water, half of the object is submerged. We now slowly pour oil into the glass so it completely covers the object. oil l 002 (part 2 of 2 5 points A simple U-tube that is open at both ends is
Answer, Key Homework 2 David McIntyre 2 Note: The second figure does not necessarily show the correct changes, only the experimental setup. Compared to the water level, the object h Wood s air 1. moves up correct 2. moves down 3. stays in the same place Consider the point where we have poured in an amount of oil exactly enough to cover the object: According to Archimedes principle, have ρ w s 3 g = h s 2 g, and solving for h, we have h = s ρ w = (12 mm (812 kg/m3 (1296 kg/m 3 = 7.51852 mm. we We can infer that it does not matter how much more oil we pour into the glass after this point, since it will exert an evenly distributed pressure on both water and object. Before this point, however, the oil added will be on top of the water only. Therefore, the oil will push the water down, while not pushing the object down. The extra pressure from the oil will be transferred to the bottom of the object, so it must push the object up. 004 (part 1 of 3 4 points A cube of wood whose edge is 12 mm is floating in a in a glass with one of its faces parallel to the surface. The density of wood is 812 kg/m 3, that of is 1296 kg/m 3. How far below the surface is the bottom face of the cube? Correct answer: 7.51852 mm. Given: s = 12 mm ρ w = 812 kg/m 3 = 1296 kg/m 3 005 (part 2 of 3 3 points A light oil is gently poured onto the immiscible surface to form a 4 mm thick layer above the. The density of the oil is 606 kg/m 3. When the wood cube attains hydrostatic equilibrium again, what will be the distance from the top of the to the bottom face of the cube? Correct answer: 5.64815 mm. Given: h a = 4 mm ρ o = 606 kg/m 3. air h w Wood alcohol We assume that the top of the cube is still above the oil s surface. If h a is the width of the oil layer, we have that the buoyant force is B = (ρ o s 2 h a + s 2 h w g. Once again, the buoyant force must be equal to the weight of the cube, so h a ρ w s 3 g = (ρ o s 2 h a + s 2 h w g.
Answer, Key Homework 2 David McIntyre 3 Solving for h w, we get ρ w s = ρ o h a + h w h w = ρ w s ρ o h a = (812 kg/m3 (12 mm (1296 kg/m 3 (606 kg/m3 (4 mm (1296 kg/m 3 = 5.64815 mm. 006 (part 3 of 3 3 points Additional light oil is poured onto the surface until the top surface of the oil coincides with the top surface of the wood cube (in hydrostatic equilibrium. How thick is the whole layer of the light oil? Correct answer: 8.41739 mm. h w air Wood alcohol If the width of the oil layer is h a, the buoyant force is now [ ] B = ρ o s 2 h a + s 2 (s h a g. Equating the buoyant force with the weight of the wood cube, we get [ ] ρ w s 3 g = ρ o s 2 h a + s 2 (s h a g h a ρ w s = ρ o h a + (s h a h a ( ρ o = s ( ρ w h a = s ρ w ρ o = (12 mm [ (1296 kg/m 3 (812 kg/m 3 ] (1296 kg/m 3 (606 kg/m 3 = 8.41739 mm. 007 (part 1 of 1 0 points A styrofoam slab has a thickness of 8.4 cm and a density of 247 kg/m 3. What is the area of the slab if it floats just awash (top of slab is even with the water surface in fresh water when a 98.3 kg swimmer is aboard? Correct answer: 1.5541 m 2. Since the slab floats awash, the buoyant force is B = ρ w V g = ρ w dag. This force is equal in magnitude to the weight of the slab and the swimmer, Solving for A, A = ρ w dag = ρ s dag + Mg M d(ρ w ρ s = 1.5541 m2 m 008 (part 1 of 2 0 points Calculate the flow rate of blood (ρ = 0.837 g/cm 3 in an aorta with a crosssectional area of 0.902 cm 2 if the flow speed is 46.9 cm/s. Correct answer: 35.4083 g/s. Flow Rate = ρ A v = (0.837 g/cm 3 (0.902 cm 2 (46.9 cm/s = (0.837 g/cm 3 (42.3038 cm 3 /s = 35.4083 g/s. 009 (part 2 of 2 0 points Assume: The aorta branches to form a large number of capillaries with a combined crosssectional area of 4550 cm 2. What is the flow speed in the capillaries? Correct answer: 0.00929754 cm/s. From the equation of continuity, we have v 2 = A 1 v 1 A ( 2 0.902 cm 2 = 4550 cm 2 (46.9 cm/s = 0.00929754 cm/s.
Answer, Key Homework 2 David McIntyre 4 010 (part 1 of 1 0 points You use a fan to blow air across the surface of the exposed mercury in a barometer. The level of mercury in the tube will. 1. drop correct 2. rise 3. be unaffected Solution: Bernoulli s principle says that forcing the air to move across the barometer will cause a decrease in the air s pressure. Since this pressure is holding the mercury up in the barometer tube, the level of mercury in the tube will drop. 011 (part 1 of 1 10 points A tall water cooler tank is standing on the floor. Some fool punched two small holes in the tank s wall, one hole at a height of 45 cm above the floor and the other hole 47 cm directly above the first hole and 92 cm above the floor. Each hole produces a jet of water that emerges in a horizontal direction but eventually hits the floor at some distance from the tank. If the two water jets (emerging from each hole hit the floor at exactly the same spot, how high H is the water level in the tank (relative to the room s floor? Correct answer: 137 cm. Apply Bernoulli s equation to the stream of water jetting from the first hole. This stream ultimately begins within the tank, at the surface level H where the pressure is equal to the atmospheric pressure and the speed of water flow is essentially zero. When this stream emerges from the hole, the pressure is again equal to the atmospheric pressure but the height h 1 is lower than H and the stream s velocity is v 1 0. Applying Bernoulli s equation P + ρ g h + 1 2 ρ v2 = const to this situation, we obtain v 1 = 2 g (H h 1. This formula is known as Torricelli s theorem. Now consider what happens to the water after it emerges from the hole with velocity v 1 in the horizontal direction. The Bernoulli equation tells us the speed of the water at every point of its trajectory, but it has nothing to say about the direction of the flow. Instead, we use the fact that the water droplets comprising the jet exert no net force on each other (the pressure is constant everywhere outside the tank, so each droplet is essentially in a free fall. A drop falling to the floor from height h 1 with no initial vertical component of the velocity (the jet emerges horizontally from the hole 2 h 1 takes time t 1 = to fall, during which g time it flies through a horizontal distance L 1 = v 1 t 1 = 4 (H h 1 h 1. In exactly the same way, the jet emerging from the second hole has horizontal initial velocity v 2 = 2 g (H h 2 and hits the floor at the distance L 2 = 4 (H h 2 h 2. If the two jets hit the floor at the same spot, it means that L 2 = L 1, L 2 2 = L 2 1 or 0 = L 2 2 L 2 1 = 4 (H h 2 h 2 4 (H h 1 h 1 = 4 H h 2 4 h 2 2 4 H h 1 + 4 h 2 1 = 4 (h 2 h 1 (H h 2 h 1, and since h 2 h 1 0, we should have H h 2 h 1 = 0 or H = h 1 + h 2 = 137 cm. And given the above result, we immediately find that the two jets hit the floor at the distance L 1 = L 2 = 4 h 1 h 2 = 128.686 cm from the tank.
Answer, Key Homework 2 David McIntyre 5 012 (part 1 of 3 4 points A siphon of circular cross section is used to drain water from a tank as shown below. The siphon has a uniform diameter of 3.09 cm and the surface of the reservoir is at a height of h 1 = 2.19 m (see figure. Assume: A steady flow, no friction for the water, and that the water cannot sustain a negative pressure. Note: Density of water is 10 3 kg/m 3. h 2 013 (part 2 of 3 3 points Assume: The reservoir is very large so that h 1 barely changes. Note: 1 liter = 10 3 m 3. How long would it take to siphon 486 L of water from the reservoir? Correct answer: 1.64865 min. Since the reservoir is assumed to be large, we assume that the height h 1 will not change appreciably; i.e., v is constant. In a time t, the volume of water pumped V will be V = A v t, What is the speed of the flow at the end of the siphon? Correct answer: 6.55164 m/s. Basic concept: Bernoulli s equation h 1 P + 1 2 ρ v2 + ρ g y = constant along a streamline. Solution: Define the potential energy to be zero at the outside end of the siphon. Bernoulli s equation says that along a streamline, P + ρ v2 2 + ρ g y = constant. So at the surface of the reservoir and at the end of the siphon, these sums have the same value P atm + 0 + ρ g h 1 = P atm + 1 2 ρ v2 + 0. So the velocity at the end of the siphon is v = 2 g h 1 = 2(9.8 m/s 2 (2.19 m = 6.55164 m/s. where A = cross sectional area of the siphon. t = V π d2 4 v = 4 V π d 2 v = 4 (486 L (0.001 m3 /L π [(3.09 cm (0.01 m/cm] 2 (0.0166667 min/s (6.55164 m/s = 1.64865 min. 014 (part 3 of 3 3 points Note: Instead of water, suppose the tank were filed with a with density 1400 kg/m 3. What is the maximum value h 2 can have in order for the siphon to still work? Hint: We note that the velocity v in the tube is a constant since the cross sectional area A is a constant. Correct answer: 5.19338 m. We note that in the siphon tube flows at constant speed and Q = v A is conserved. Now equating the expression for Bernoulli s equation at the end of the siphon to that at the height h 2, we have P atm + 1 2 ρ v2 + 0 = P ht + 1 2 ρ v2 + ρ g (h 1 + h 2,
where P ht is the pressure at height h 2 + h 1 and P ht 0 ( cannot sustain a negative pressure. Note: If a negative pressure occurs on the discharge side of the siphon, the velocity in the discharge side of the siphon will increase. Neglecting viscous considerations, the siphon will not work. Thus, (h 1 + h 2 (instead of simply h 2 must be included on the righthand side of Bernoulli s equation. Answer, Key Homework 2 David McIntyre 6 P atm ρ g (h 2 + h 1 = P ht 0 ρ g (h 2 + h 1 P atm h 2 P atm ρ g h 1. The limiting height is h 2 = P atm ρ g h 1 (101300 Pa = (1400 kg/m 3 (9.8 m/s 2 2.19 m = 5.19338 m.