#$&* course phy 242 Question: query introset plug of water from cylinder given gauge pressure, c-s hole area, length of plug.
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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). STUDENT SOLUTION AND QUESTION From looking at my notes from the Intro Problem Set, it was a lot of steps to determine the velocity in that situation. The Force was determined first by using F = (P * cross-sectional area). With that obtained Force, use the work formula ‘dW = F * ‘ds using the given length as the ‘ds. Then we had to obtain the Volume of the ‘plug’ by cross-sectional area * length. That Volume will then be using to determine the mass by using m = V * d, and the density is a given as 1000kg/m^3. With that mass, use the KE equation 1/2(m * v^2) to solve for v using the previously obtained ‘dW for KE since ‘dW = ‘dKE. Some of my symbols are different than yours. I’m in your PHY 201 class right now and am in the work and energy sections. So is it wrong to use 1/2(m * v^2) instead of 1/2(‘rho * A * L * v^2)?? Aren’t they basically the same thing?? INSTRUCTOR RESPONSE You explained the process very well, though you did miss a step. m = rho * V (much better to use rho than d, which isn't really a good letter to use for density or distance, given its use to represent the prefix 'change in'). You covered this in your explanation. V isn't a given quantity; it's equal to the length of the plug multiplied by its cross-sectional area and should be expressed as such. You didn't cover this in your explanation. However, other than this one missing step, your explanation did cover the entire process. With that one additional step, your solution would be a good one. In any case, V would be A * L, and m would be rho * V = rho * A * L. So 1/2 m v^2 is the same as 1/2 rho A L v^2. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: `q001. Water pressure exerts a force of .8 Newtons on a water 'plug' with cross-sectional area 3 cm^2 and length 5 cm. The 'plug' is forced out of the side of the container by this force as it moves through its 5 cm length, starting from rest. • How much work will the force do on the 'plug'? • What will be the KE of the 'plug' as it exits the container? • How fast will the 'plug' be moving as it leaves the container? Answer the analogous series of questions for a 'plug with cross-sectional area 1 cm^2 and length 2 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: W=F*D = F*L W=.8 N* .05m= 0.04 J KE= ˝*m*v^2=P*A*L=W KE= 0.04 J 0.04= ˝ *rho*A*l*v^2 Or V=sqrt(2P/rho) P= 0.04 J/(.03m^2*.05m)= 26.7 pa V= sqrt(2*26.7pa/(1000kg/m^3) V= 0.23 m/s W=F*D = F*L W=.8 N* .02m= 0.016 J KE= ˝*m*v^2=P*A*L=W KE= 0.016 J 0.04= ˝ *rho*A*l*v^2 Or V=sqrt(2P/rho) P= 0.016 J/(.01m^2*.02m)= 26.7 pa V= sqrt(2*80 pa/(1000kg/m^3) V= 0.4 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:ok ********************************************* Question: `q002. A closed glass jar with a half-liter capacity has a mass of 200 grams. If it is submerged in water what will be the buoyant force acting on it, and at the instant it is released from rest what will be the net force on it and its acceleration? The drag force of water on the jar is 1 N s^2 / m^2 then at what speed will it be rising when the net force acting on it becomes zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: FB=rho*V*g Rho = 1000kg/m^3 FB=1000kg/m^3*500ml *9.81 m/s^2 = 4.905 N Fnet= FB-Wobj Fnet=4.905- rho*V*g(object) Fnet= 4.905 N-((200g/500ml)*500ml*9.81m/s^2) Fnet= 4.905 N- 1.962 N Fnet= 2.943 N F= m*a 2.943N = .2 kg* a a= 14.715 m/s^2 Fdrag=k*v^2 k= 1 N s^2 / m^2 Fdrag= FB+mg Fdrag= 4.905+(9.81m/s^2)*(.2kg) Fdrag=6.867 N 6.867 N = (1 N s^2 / m^2)* v^2 6.867 m^2/s^2 = v^2 V= 2.62 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok ********************************************* Question: univ 12/58 / 14.57 was 14.51 (14.55 10th edition) U tube with Hg, 15 cm water added, pressure at interface, vert separation of top of water and top of Hg where exposed to atmosphere. Give your solution to this problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We are going to use Bernoulli’s Equation: Since v=0 P1 + rho g x1 = P2 + rho g x2 Pinterface= Patm+ 1000kg/m^3*9.81m/s^2 * 0.15 m Pinterface= Patm+ 1470 pa P2-P1=1470 pa rho g x1= rho g x2+1470 rho*g*(x1-x2) = 1470 pa 13534kg/m^3*9.81m/s^2 *( x1-x2)=1470 pa (x1-x2) = 0.011 m .15-0.011 = .139 m. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ok ** 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. ********************************************* Question: `q003. The cap is screwed onto jar half-full of water, and the jar is place on a level surface, on its side. The cap of the jar has diameter 8 cm. How much force will water pressure exert on the cap? Note that it is necessary to set up an integral to solve this problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: By the definition of pressure P=F/A, we have dF = P*dA. In order to attain the total force, we must integrate P*dA over the entire area where the water touches the cap, which happens to be a semicircle. So now to further simplify it we use P= rho*g*h, which we get F = integral of (p*g*h*dA). We now need expression for dA In terms of other parts of the integral. So I set dA= 2*sqrt(r^2-h^2)*dh and integrate is from h= 0 to h=r. Putting these equation together we get integral of 2*sqrt(.04^2-h^2)*pgh*dh from 0 to .04 m, which I got to equal to 0.41856 N. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!