#$&* course Phy 232 007. `query
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Given Solution: ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The evidence from the experiment shows that the drag force increases with velocity. The major pinpointed of this is that the added weight continued to have a lessening effect. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: v2=sqrt((pres dif+(1/2)*(density)*(v1^2))/((1/2)*(dens))) After solving for v2, we use (v1/v2)=(d2/d1)^2 d2=sqrt(v1/v2)d1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:OK ********************************************* Question: query univ phy problem 12.93 / 14.91 11th edition14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I understood to use Bernoulli's Equation, but I did not pick up on, at first, the use of several points. The solution given greatly helps my understanding and I am glad that it was there for help. After seeing the response I better understand how to relate the factors and understand how to achieve the correct answer. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 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: Once again, we use the same equation we have been using before. v2=sqrt((pres dif+(1/2)*(density)*(v1^2))/((1/2)*(dens))) After solving for v2, we use (v1/v2)=(d2/d1)^2 d2=sqrt(v1/v2)d1 We can use v2 to find h by using h=(v2^2)/(2g) y=(dh)*(1-((d2/d1)^4) The y value solved for is the change in mercury levels confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!