course phy202 ???????????assignment #014014. `query 4 Physics II 02-27-2008
......!!!!!!!!...................................
15:32:27 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?
......!!!!!!!!...................................
RESPONSE --> the equation of continuity states: p1A1V1=p2A2V2 solving this for V2 gives us: p1A1V1 /p2A2 = V2
.................................................
......!!!!!!!!...................................
15:35:09 ** 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). **
......!!!!!!!!...................................
RESPONSE --> I misunderstood the problem, partley because I did not complete this specific intro problem. However, I've read chapter 10, taken notes, and worked several fluids problems and understand the equation of continuity and Bernoulli's equation.
.................................................
......!!!!!!!!...................................
15:37:59 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
......!!!!!!!!...................................
RESPONSE --> n/a
.................................................
......!!!!!!!!...................................
15:38:14 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
......!!!!!!!!...................................
RESPONSE --> n/a Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
.................................................
......!!!!!!!!...................................
15:40:08 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.
......!!!!!!!!...................................
RESPONSE --> Was this the answer to a question? No question appeared in my screen.
.................................................
......!!!!!!!!...................................
15:56:05 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
......!!!!!!!!...................................
RESPONSE --> the density of helium is .18 Kg/m^3 while the density of air is 1.29Kg/m^3 the balloon has a radius of 7.35m and a volume of 4/3 pi r^3 = 1663 m^3 The balloons bouyant force will be equal to the weight of the air it displaces minus the weight of the helium and the structure of the balloon. weight of displaced air is 1663 m^3 * 1.29Kg/m^3 = 2145 Kg weight of the helium is 1663 m^3*.18 = 299 Kg weight of apparatus is: 930 Kg 2145 -(299+930)=916Kg of net lifting force
.................................................
......!!!!!!!!...................................
16:01:21 ** 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.. **
......!!!!!!!!...................................
RESPONSE --> Since the problem in the book asked for the max payload it could lift am I correct that we can simply view this using Archimedes principle and not include the accel of g in the problem since the densities already take this into account because they are expresed as weight per unit volume? ((mg)
.................................................
......!!!!!!!!...................................
16:01:31 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.
......!!!!!!!!...................................
RESPONSE --> n/a
.................................................
......!!!!!!!!...................................
16:01:37 ** 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. **
......!!!!!!!!...................................
RESPONSE --> n/a
.................................................