course Phy 232 RiUzٚassignment #014
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00:15:01 query introset plug of water from cylinder given gauge pressure, c-s hole area, length of plug. Explain how we can determine the velocity of exiting water, given the pressure difference between inside and outside, by considering a plug of known cross-sectional area and length?
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RESPONSE --> We find the velocity of the exiting water by first calculating the change in potential energy that occurs due to the exiting water. The water column loses potential energy equal to the energy used to produce the net pressure. The gauge pressure plus the difference between the pressure inside and outside the container is the net pressure. The force is product of net pressure and cross-sectional area of the hole. This force is multiplied by the distance traveled by the plug (which is equal to the plug length) to find the change in potential energy. In this case, the potential energy will decrease. The change in kinetic energy is equal and opposite to the change in potential energy. The product of the cross-sectional area of the hole and length of plug gives us the volume of water. From this volume and the known density of water (1000kg/m^3) we calculate the mass of the water. We can now substitute our calculated values into the equation for velocity : v = `sqrt(2KE / mass)
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00:18:23 ** The net force on the plug is P * A, where A is its cross-sectional area and P the pressure difference. If L is the length of the plug then the net force P * A acts thru dist L doing work P * A * L. If init vel is 0 and there are no dissipative forces, this is the kinetic energy attained by the plug. The volume of the plug is A * L so its mass is rho * A * L. Thus we have mass rho * A * L with KE equal to P * A * L. Setting .5 m v^2 = KE we have .5 rho A L v^2 = P A L so that v = sqrt( 2 P / rho). **
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00:18:26 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
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00:18:28 The density of air at common atmospheric temperature and pressure it about 1.3 kg / m^3. The volume of the room is 4.8 m * 3.8 m * 2.8 m = 51 m^3, approximately. The mass of the air in this room is therefore mass = density * volume = 1.3 kg / m^3 * 51 m^3 = 66 kg, approximately. This is a medium-sized room, and the mass is close to the average mass of a medium-sized person.
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00:18:31 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
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00:18:33 The density of blood is close to that of water, 1000 kg / m^3. So the pressure difference corresponding to a height difference of 1.6 m is pressure difference = rho g h = 1000 kg/m^3 * 9.80 m/s^2 * 1.60 m = 15600 kg / ( m s^2) = 15600 (kg m / s^2) / m^2 = 15600 N / m^2, or 15600 Pascals. 1 mm of mercury is 133 Pascals, so 15,600 Pa = 15,600 Pa * ( 1 mm of Hg / 133 Pa) = 117 mm of mercury. Blood pressures are measured in mm of mercury; e.g. a blood pressure of 120 / 70 stands for a systolic pressure of 120 mm of mercury and a diastolic pressure of 70 mm of mercury.
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00:18:36 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
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00:18:38 ** The volume of the balloon is about 4/3 pi r^3 = 1660 cubic meters and mass of air displaced is about 1.3 kg / m^3 * 1660 m^3 = 2160 kg. The buoyant force is equal in magnitude to the force of gravity on the displaced air, or about 2160 kg * 9.8 m/s^2 = 20500 Newtons, approx.. If the total mass of the balloon, including helium, is 930 kg then the net force is about buoyant force - weight = 20,500 N - 9100 N = 11,400 N If the 930 kg doesn't include the helium we have to account also for the force of gravity on its mass. At about .18 kg/m^3 the 1660 m^3 of helium will have mass about 300 kg on which gravity exerts an approximate force of 2900 N, so the net force on the balloon would be around 11,400 N - 2900 N = 8500 N approx. The mass that can be supported by this force is m = F / g = 8500 N / (9.8 m/s^2) = 870 kg, approx.. **
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00:36:28 univ 14.51 (14.55 10th edition) U tube with Hg, 15 cm water added, pressure at interface, vert separation of top of water and top of Hg where exposed to atmosphere. Give your solution to this problem.
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RESPONSE --> Gauge pressure is the pressure above atomspheric pressure. So pressure at the interface is equal to the pressure of water at a depth of .15 m. This is the product of the density, the depth, and the acceleration due to gravity: `rho * g * y = 1000 kg / m^3 * 9.81 m/sec^2 * .15 m = 1.47 kPa The height difference between the columns can be calculated by recognizing that the density of each fluid is constant, and that the pressure at the level of the interface is the same as the pressure at the same level (.15 - h) in the other column: (`rho * g * y)_water = (`rho * g * y)_mercury value of g cancels from each side `rho_water * .15= `rho_mercury * (15-h) this is solved for h: h= .15 m - ((150 kg/ m^2) / 13600 kg/m^3) = 13.9cm
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00:38:52 ** At the interface the pressure is that of the atmosphere plus 15 cm of water, or 1 atm + 1000 kg/m^3 * 9.8 m/s^2 * .15 m = 1 atm + 1470 Pa. The 15 cm of water above the water-mercury interface must be balanced by the mercury above this level on the other side. Since mercury is 13.6 times denser than water there must therefore be 15 cm / (13.6) = 1.1 cm of mercury above this level on the other side. This leaves the top of the water column at 15 cm - 1.1 cm = 13.9 cm above the mercury-air interface. } Here's an alternative solution using Bernoulli's equation, somewhat more rigorous and giving a broader context to the solution: Comparing the interface between mercury and atmosphere with the interface between mercury and water we see that the pressure difference is 1470 Pa. Since velocity is zero we have P1 + rho g y1 = P2 + rho g y2, or rho g (y1 - y2) = P2 - P1 = 1470 Pa. Thus altitude difference between these two points is y1 - y2 = 1470 Pa / (rho g) = 1470 Pa / (13600 kg/m^2 * 9.8 m/s^2) = .011 m approx. or about 1.1 cm. The top of the mercury column exposed to air is thus 1.1 cm higher than the water-mercury interface. Since there are 15 cm of water above this interface the top of the water column is 15 cm - 1.1 cm = 13.9 cm higher than the top of the mercury column. NOTE BRIEF SOLN BY STUDENT: Using Bernoullis Equation we come to: 'rho*g*y1='rho*g*y2 1*10^3*9.8*.15 =13.6*10^3*9.8*y2 y2=.011 m h=y1-y2 h=.15-.011=.139m h=13.9cm. **
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RESPONSE --> I'm getting better at seeing and using the proportional relationships, but missed this one.
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Ĩe{bω{̡ʆ[Z~cv assignment #016 016. `query 6 Physics II 04-15-2008
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00:08:58 query introset change in pressure from diameter change given original vel and diameter
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RESPONSE --> The volume flow rate does not change in a flow tube. This does not mean that velocity of the fluid does not change, but that the same total volume per unit time remains constant. Therefore, a change in cross-sectional area (determined by the change in diameter) will produce an inversely proportional change in velocity. The relationship between pressure, velocity, density and change in height is given by Bernoulli's equation: The following sum remains constant in an ideal fluid (incompressible, no viscocity): pressure + (1/2 density * velocity^2) + (density * acceleration due to gravity * height) = constant For a constant height, the change in pressure can be found: p_1 - p_2 = 1/2 `rho (v_2^2 - v_1^2)
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00:09:54 ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . **
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RESPONSE --> I had to review this idea a few times. I think I've finally gotten it.
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00:15:25 query video experiment 4 terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity?
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RESPONSE --> When additional weight is used to accelerate an object through the water, there appears to be a limiting value to the velocity the object attains. The force opposing the motion of a object moving through the water is a friction force. In order for it to limit velocity against an increasing force, it must also be increasing.
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00:15:36 ** When weights were repetitively added the velocity of the sphere repetively increased. As the velocities started to aproach 0.1254 m/sec the added weights had less and less effect on increasing the velocity. We conclude that as the velocity increased so did the drag force of the water. **
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00:53:39 query univ phy problem 14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises.
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RESPONSE --> I could not quite finish this one. I started at the end open to the atmosphere, because I had a known pressure ( assuming 1 atmosphere ) and known velocity (using speed of efflux = `sqrt(2*g*h1)). Streamline flow and zero viscosity are assumed, so Bernoulli's equation is valid. At point C, where tube has narrowed to one-half the cross-sectional area of D, the velocity must have doubled to 2* `sqrt(2*g*h1). This will result in a lower pressure at C than at D. Point E is the fluid level in the lower tube. It is separated from point C by air only, so the pressure can be considered the same at both points. Substituting values into Bernoulli's equation: v1^2 = (`sqrt(2*g*h1))^2 = 2 g h1 v2^2 = (2 *`sqrt(2*g*h1))^2 = 8 g h1 p1 + `rho g h1 + .5 `rho (2 g h1) = p2 + `rho g h2 + .5 `rho (8 g h1) I'm including more details than usual since I could not identify my error: p1 + `rho g h1 + `rho ( g h1) = p2 + `rho g h2 + `rho (4 g h1) p1 - p2 + 2 * `rho g h1 = `rho g h2 + `rho (4 g h1) p1 - p2 - 2 * `rho g h1 = `rho g h2 ((p1 - p2)/ `rho g) - 2 * h1 = h2 I know that if I consider p2 to be gauge pressure, I can eliminate p1 from the equation, but I don't see how h2 can be 3h1.
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01:04:20 ** The fluid exits the narrowed part of the tube at atmospheric pressure. The widened part at the end of the tube is irrelevant--it won't be filled with fluid. So Bernoulli's Equation will tell you that the fluid velocity in this part is vExit such that .5 rho vExit^2 = rho g h1. Now compare three points: point 1, in the narrowed tube; point 2 at the top of the fluid in the lower tube; and point 3 at the level of the fluid surface in the lower container. At point 1 the pressure is atmospheric and velocity is vExit. Pressure is atmospheric because there is no pressure loss within the tube, since friction and viscosity are both assumed negligible. At point 2 fluid velocity is zero and since there is no fluid between the narrowed tube and this point there is no net rho g h contribution to Bernoulli's equation. So we have .5 rho v1^2 + P1 = .5 rho v2^2 + P2. P1 is atmospheric and v1 is vExit from above, while v2 = 0 so P2 = atmospheric pressure + .5 rho vExit^2 = atmospheric pressure + rho g h1. Now comparing point 2 with point 3 we see that there is a difference h in the fluid altitude, with velocity 0 at both points and atmospheric pressure at point 3. Thus P2 + rho g h2 = P3 + rho g h3, or (atmospheric pressure + rho g h1) + rho g h2 = atmospheric pressure + rho g h3. Thus rho g (h3 - h2) = rho g h1 and h3 - h2, which is the height of the fluid in the lower tube, is just equal to h1. If we assume that somehow the fluid manages to expand on escaping the narrowed tube so that it fills the once-again-widened tube, and exits with vExit as above, then the velocity in the narrowed tube will be 2 * vExit. This leads to the conclusion that pressure change from small to large tube is .5 rho (2 vExit)^2 - .5 rho vExit^2 = .5 rho (3 vExit^2). Since pressure is atmospheric in the large tube, pressure in the small tube is atmospheric pressure + 3 ( .5 rho vExit^2). If we use this pressure for point 1 and follow the steps given above we conclude that h3 - h2, the height of the fluid column in the lower tube, is 3 h1. This is the book's answer. Again I don't have the problem in front of me and I might have missed something, but the idea of the fluid expanding to refill the larger pipe doesn't seem consistent with the behavior of even ideal liquids, which are pretty much incompressible at ordinary pressures. However note that I am sometimes wrong when I disagree with the textbook's solution. **
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RESPONSE --> It looks as though my error was in considering the exit tube to be full. The velocity won't change between the narrow part and the wider part because they're both at atmospheric pressure. And I basically ignored the lower tank altogether. I'm going to work this problem again and make sure I can do it without having just read your solution.
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01:04:25 query univ phy problem 14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises.
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01:04:31 ** The fluid exits the narrowed part of the tube at atmospheric pressure. The widened part at the end of the tube is irrelevant--it won't be filled with fluid. So Bernoulli's Equation will tell you that the fluid velocity in this part is vExit such that .5 rho vExit^2 = rho g h1. Now compare three points: point 1, in the narrowed tube; point 2 at the top of the fluid in the lower tube; and point 3 at the level of the fluid surface in the lower container. At point 1 the pressure is atmospheric and velocity is vExit. Pressure is atmospheric because there is no pressure loss within the tube, since friction and viscosity are both assumed negligible. At point 2 fluid velocity is zero and since there is no fluid between the narrowed tube and this point there is no net rho g h contribution to Bernoulli's equation. So we have .5 rho v1^2 + P1 = .5 rho v2^2 + P2. P1 is atmospheric and v1 is vExit from above, while v2 = 0 so P2 = atmospheric pressure + .5 rho vExit^2 = atmospheric pressure + rho g h1. Now comparing point 2 with point 3 we see that there is a difference h in the fluid altitude, with velocity 0 at both points and atmospheric pressure at point 3. Thus P2 + rho g h2 = P3 + rho g h3, or (atmospheric pressure + rho g h1) + rho g h2 = atmospheric pressure + rho g h3. Thus rho g (h3 - h2) = rho g h1 and h3 - h2, which is the height of the fluid in the lower tube, is just equal to h1. If we assume that somehow the fluid manages to expand on escaping the narrowed tube so that it fills the once-again-widened tube, and exits with vExit as above, then the velocity in the narrowed tube will be 2 * vExit. This leads to the conclusion that pressure change from small to large tube is .5 rho (2 vExit)^2 - .5 rho vExit^2 = .5 rho (3 vExit^2). Since pressure is atmospheric in the large tube, pressure in the small tube is atmospheric pressure + 3 ( .5 rho vExit^2). If we use this pressure for point 1 and follow the steps given above we conclude that h3 - h2, the height of the fluid column in the lower tube, is 3 h1. This is the book's answer. Again I don't have the problem in front of me and I might have missed something, but the idea of the fluid expanding to refill the larger pipe doesn't seem consistent with the behavior of liquids, which are pretty much incompressible at ordinary pressures. However note that I am sometimes wrong when I disagree with the textbook's solution. **
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