course Phy 122 Could you please tell me how to solve #9 and 22 in Chapter 10 of the text? For both of these problems I just didn't know how to start because I didn't think I had enough information.#9: Calculate total force of atmostphere on table that measure 1.6m x 2.9m.
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10:59:05 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 --> To find the velocity of the exiting water, we must first find the force on the plug, then we calculate the work done on the plug which is equal to its kinetic energy. We must also calculate the mass of the plug from the volume and density of water. Then from the mass and KE we can solve for the velocity. We are given: Pressure = 15000 N/m^2 Area = .00073 m^2 Displacement = .06m Density (p) = 1000 kg/m^3 Force = P * A = 15000 N/m^2 * .00073m^2 = 10.95 N Work = F / `ds = 10.95 N / .06m = 0.657J Volume = A * `ds = .00073m^2 * .06m = 4.84*10^-5 m^3 Mass = V * p = (4.84*10^-5 m^3)(1000 kg/m^3) = .0438 kg We know that the water was initially at rest so the initial velocity and KE are zero. Work = KE = 0.657 J KE = 1/2 m*v^2 0.657J = 1/2(.0438kg)(v^2) 0.657J = .0219kg * v^2 v = `sqrt (0.657J / .0219 kg) = 5.477 m/s
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10:59:34 ** 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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RESPONSE --> I understand the process for solving this problem.
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11:01:57 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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RESPONSE --> This problem is actually chapter 10.2, not #3. First I calculated the volume: V = 4.8m * 3.8m * 2.8m = 51.072 m^3 The mass of air is the product of its volume and density. The density of air is 1.29kg/m^3: mass = V * p = (51.072m^3)(1.29kg/m^3) = 65.9 kg
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11:02:13 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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RESPONSE --> I understand this problem.
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11:07:45 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
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RESPONSE --> The following equations and relationships are used to solve this problem: P = F/A = (p*A*h*g)/A = p*h*g `dP = p*`dh*g density of whole blood = 1.05*10^3 kg/m^3 h is the distance from head to feet = 1.6m g is the gravitational acceleration = 9.8m/s^2 `dP = (1.05*10^3 kg/m^3)(1.6m)(9.8m/s^2) = 16464 N/m^2 To convert the pressure from N/m^2 to mmHg: 1.013*10^5 N/m^2 = 1 atm = 760 mmHg (16464 N/m^2)(760mmHg / 1.013*10^5N/m^2) = 123.5 mmHg
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11:08:19 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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RESPONSE --> I answered this problem correctly, however my values are slightly different due to rounding.
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11:17:33 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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RESPONSE --> First I found the mass of the air that was displaced: mass = volume * density volume of the balloon = 4/3*pi*r^3 = 4/3*pi*(7.35m)^3 = 1663 m^3 density of air = 1.3 kg/m^3 Mass = (1663m^3)(1.3kg/m^3) = 2162 kg of air The buoyant force can now be calculated. It is the product of the mass of air displaced and its gravitational acceleration. F = m*g = 2162 kg * 9.8 m/s^2 = 21186 N The net force on the balloon is the difference in the buoyant force and the weight of the balloon: Fnet = buoyant force - (m balloon * g) Fnet = 21186N - (930kg * 9.8m/s^2) = 21186N - 9114N = 12072 N Now to find the buoyant force on the helium, we first find the mass of the helium. The density of helium is .179 kg/m^3: m = V*p = 1663m^3 * .179kg/m^3 = 297 kg F = m*g = 297kg * 9.8m/s^2 = 2917 N The net force on the balloon will be the difference in the buoyant force of the helium and the net force of the balloon itself. 12072N - 2917N = 9155N To calculate the mass that can be lifted by this force, we divide the force by the gravitational acceleration: m = F/g = 9155N / 9.8m/s^2 = 934 kg
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11:19:51 ** 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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RESPONSE --> I used the same methods to solve this problem, but my values are slightly different. I don't think I made any error in procedure.
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11:20:04 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 --> I am not a university physics student
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11:20:23 ** 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 --> Ok.
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