course Phy 232 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: ** 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 F_net = P * A acts thru distance L doing work `dW = F_net * L = P * A * L. If the initial velocity of the plug 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). Your Self-Critique: I have the same solution I just didnt break it down as much, so my solution isnt as simplified.
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Given Solution: 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 of the air in that room is close to the mass of an average-sized person. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person. ********************************************* Your Solution: I am in university physics Confidence Rating:
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Given Solution: 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. The density of blood is actually a bit higher than that of water; if you use a more accurate value for the density of blood you will therefore get a slightly great result. Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force ********************************************* Your Solution: I am in university physics Confidence Rating:
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Given Solution: ** 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 Net force = 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.. ** Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: 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. ********************************************* Your Solution: ρwater = 1000 kg/m3 c = 13600 kg/m3 Part a: Pgauge = P - Patm Pleft side = Patm + Pwater + PHg Pright side = Patm + PHg Gauge pressure at the interface is just going to be the pressure of the water pushing down on the mercury: Pgauge = 1000 kg/m3 * 9.8 m/s2 * 0.15 m = 1470 Pa Part b: For this problem I am going to use the Bernoulli equation and set the Pleft side = Pright side Since there is a certain height of mercury on both the left and right side that is equal the pressure exerted by each side cancels out and is not included in the equation below: ρwater * g * hwater = ρHg * g * hHg ρwater * g * hwater = ρHg * g * hHg 1000 kg/m3 * 9.8 m/s2 * 0.15 m = 13600 kg/m3 * 9.8 m/s2 * h h = 0.011 m = 1.1 cm This is the height of the mercury on the right side that goes into the water height on the left side. The difference in height = 15 cm 1.1 cm = 13.9 cm The water column is 13.9 cm higher than the mercury Confidence Rating: 3
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Given Solution: ** 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. ** Your Self-Critique: For part a, the book asks for the gauge pressure which means you would only find the pressure of the water column and not add in atmospheric pressure. This is the difference in our numbers for part a. For part b, I used the Bernoulli Equation and found the same answer. Your Self-Critique Rating: 2