course Phy 232 9/14 11 005. `query 5vvvv
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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 forgot the last part, but I think I got it pretty much right other than that part. Your Self-Critique Rating: 2 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For this problem I used Bernoullis Equation which is rho*g*y1=rho*g*y2. With values plugged in you get 1 *10^3*9.8*.15=13.6*10^3*9.8*y2. Then solving for y2 we get y2=0.011meters. We can then find h by h = y1-y2 which would be 0.15-0.011 = 0.139 meters. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** GENERAL STUDENT QUESTION I have completely confused myself on the equations for liquids and gas and cant figure out if they are interchangeable or I have just been leaning the wrong equations for different variables. I mentioned a few throughout the excercise. For example P= F/A = mg/A = rhoAgh/A = rhogh but this is the equation for PE as well? However in some notes PE = rho A g L then other times it = rho g h Is the first equation only used for fluids and the second for gas? "" INSTRUCTOR RESPONSE P = F / A is the definition of pressure (force per unit of area) In a fluid, the fluid pressure at depth h is rho g h. This quantity can also be interpreted as the PE per unit volume of fluid at height h, and is what I call the 'PE term' of Bernoulli's Equation • Bernoulli's Equation that equation applies to a continuous moving fluid, and one version reads 1/2 rho v^2 + rho g h + P = constant.. The 1/2 rho v^2 term is called the 'KE term' of the equation, and represents the KE per unit of volume. This equation represents conservation of energy, in a way that is at least partially illustrated by the exercises below. The equation you quote, PE = rho A g L, could apply to a specific situation with a fluid in a tube with cross-sectional area A and length L, but that would be a specific application of the definition of PE. This is not a generally applicable equation. "