#$&* course Phy 232 If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: In this experiment, the drag force from the water was measured on a marble to see how increasing the velocity affected it. From the results, we can see that increasing the velocity of the marble as it moves through the water creates a larger drag force. The more weight added to increase the velocity, over time, the less affect it had on continuing to increase that velocity. This is because the faster something is dragged through the water, the more water resistance force the object encounters. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** When weights were repetitively added the velocity of the sphere repetitively increased. As the velocities started to aproach 0.1254 m/sec the added weights had less and less effect on increasing the velocity. We conclude that as the velocity increased so did the drag force of the water. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: `q001. If you know the pressure drop of a moving liquid between two points in a narrowing round pipe, with both points at the same altitude, and you know the speed and pipe diameter in the section of pipe with the greater diameter, how could you determine the pipe diameter at the other point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Given the pressure difference in between the two different sized pipes, along with the other givens, we can find the final or exiting velocity of the liquid. This can be done by using the equation, pressure difference = (1/2)*(density)*(v2^2) - (1/2)*(density)*(v1^2). Rearranging the formula we get, [pressure difference + (1/2)*(density)*(v1^2)] / [(1/2)*(density)] = v2^2. Using this we can get the final velocity of the liquid assuming that we can find the density of the liquid. The next step is to use the exiting velocity and initial velocity and the diameter of the larger pipe section to find the diameter of the smaller pipe section. Using the expression, (v1 / v2) = (d2 / d1)^2, we can solve for the diameter of the smaller pipe section. Rearranging this formula we get, d2 = square root(v2 / v1) * (d1). This will give us the diameter of the smaller pipe section. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: query univ phy problem 12.93 / 14.91 11th edition14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: To start this problem we can say that the velocity of the fluid at the point where the fluid would exit from the top of the tube can be described by the equation, exit velocity = square root(2*g*h). We know that in the constricted area, or narrow section of the tube, the diameter is half of the diameter where the fluid exits from. Using that knowledge we can say that the velocity of the fluid in the smaller diameter section is twice as fast as the velocity of the fluid in the larger diameter section. Using Bernoulli’s equation, conservation shows that p1 + (1/2)*(density)*(v1^2) = p2 + (1/2)*(density)*(v2^2). Since the pressure of the system is atmospheric, the new pressure equation turns into pressure = 1 atm + [(1/2)*(density)*(v2^2 - v1^2)]. I am not sure how to connect this to the height h2 in the lower tube from here. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The fluid exits the narrowed part of the tube at atmospheric pressure. The widened part at the end of the tube is irrelevant--it won't be filled with fluid and the pressure in this part of the tube is 1 atmosphere. So Bernoulli's Equation will tell you that the fluid velocity in this part is vExit such that .5 rho vExit^2 = rho g h1. However the fact that the widened end of the tube isn't full is not consistent with the assumption made by the text. So let's assume that it is somehow full, though that would require either an expandable fluid (which would make the density rho variable) or a non-ideal situation with friction losses. We will consider a number of points: • point 0, at the highest level of the fluid in the top tank; • point 1, in the narrowed tube; • point 2 at the point where the fluid exits; • point 3 at the top of the fluid in the vertical tube; and • point 4 at the level of the fluid surface in the lower container. At point 2 the pressure is atmospheric so the previous analysis holds and velocity is vExit such that .5 rho vExit^2 = rho g h1. Thus v_2 = vExit = sqrt(2 g h1). At point 1, where the cross-sectional area of the tube is half the area at point 2, the fluid velocity is double that at point 1, so v_1 = 2 v_2 = 2 sqrt( 2 g h1 ). Comparing points 1 and 2, there is no difference in altitude so the rho g y term of Bernoulli's equation doesn't change. It follows that P_1 + 1/2 rho v_1^2 = P_2 + 1/2 rho v_2^2, so that P_1 = 1 atmosphere + 1/2 rho (v_2^2 - v_1^2) = 1 atmosphere + 1/2 rho ( 2 g h1 - 8 g h1) = 1 atmosphere - 3 rho g h1. There is no fluid between point 1 and point 3, so the pressure at point 3 is the same as that at point 1, and the fluid velocity is zero. There is continuous fluid between point 3 and point 4, so Bernoulli's Equation holds. Comparing point 3 with point 4 (where fluid velocity is also zero, but where the pressure is 1 atmosphere) we have P_3 + rho g y_3 = P_4 + rho g y_4 where y_3 - y_4 = h_2, so that h_2 = y_3 - y_4 = (P_4 - P_3) / (rho g) = (1 atmosphere - (1 atmosphere - 3 rho g h1) ) / (rho g) = 3 h1. ------------------------------------------------ Self-Critique Rating: I felt confident with my work leading up to that point and now I see how you used the new pressure in the equation to find the height h_2. I did not finish the last step to connect the two but now I understand the process. This is the book's answer. Again I don't have the problem in front of me and I might have missed something, but the idea of the fluid expanding to refill the larger pipe doesn't seem consistent with the behavior of liquids, or with the implicit assumption that rho remains constant. However note that I am often (though not always) wrong when I disagree with the textbook's solution. ** ********************************************* Question: `q002. If you know the pressure drop of a moving liquid between two points in a narrowing round pipe, with both points at the same altitude, and you know the speed and pipe diameter in the section of pipe with the greater diameter, how could you determine the pipe diameter at the other point? If a U tube containing mercury articulates with the pipe at the two points, how can you find the difference between the mercury levels in the two sides of the pipe? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Given the pressure difference in between the two different sized pipes, along with the other givens, we can find the final or exiting velocity of the liquid. This can be done by using the equation, pressure difference = (1/2)*(density)*(v2^2) - (1/2)*(density)*(v1^2). Rearranging the formula we get, [pressure difference + (1/2)*(density)*(v1^2)] / [(1/2)*(density)] = v2^2. Using this we can get the final velocity of the liquid assuming that we can find the density of the liquid. The next step is to use the exiting velocity and initial velocity and the diameter of the larger pipe section to find the diameter of the smaller pipe section. Using the expression, (v1 / v2) = (d2 / d1)^2, we can solve for the diameter of the smaller pipe section. Rearranging this formula we get, d2 = square root(v2 / v1) * (d1). This will give us the diameter of the smaller pipe section. To find the difference in height between mercury levels at the large and small pipe diameter sections, we can use a formula given all the known values expressed above. The variable y represents the change in height of a liquid between two different diameter sections of a pipe. Before using this equation however, we must find the height that the liquid in the pipe starts at. This can be done by using the formula, h = (v2^2) / (2*g). This is derived from the original equation to find exiting velocity. The formula to use is, change in y = (change in height)*(1- ((d2 / d1)^4). The y value which we obtain here is the difference in height levels of the mercury. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!