#$&* course phy 242 june 22 Question: query introset How do we find the change in pressure due to diameter change given the original velocity of the flow and pipe diameter and final diameter?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . ** Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: As the object is pulled in the water, the faster one drags it (the higher the velocity), the more force of water resistance will be apparent and therefore the more drag force needed. The one apparent evidence is that it takes more strength to pull the object in the water when one tries to increase the velocity. Another evidence is every time the mass is added the velocity seems to decrease. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Using p = F/A , we can set up a proportionality where v1/a1 = V2/a2 or v2/v1 = a1/a2 which is also equal to (d1/d2)^(2), therefore v2 = (d1/d2)^(2)*v1. Since h remains constant the formula only consists of the following variables: P2-P1 = rho*0.5*(v1^2-v2^2 ). Now given the value of d, its speed and pressure change , we just need to find v2 using P2-P1 = rho*0.5*(v1^2-v2^2 ) and plug that value in v2 = (d1/d2)^(2)*v1 and obtain d2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok ********************************************* Question: query univ phy problem 12.93 / 14.91 11th edition14.85 (14.89 10th edition) half-area constriction then open to outflow at dist h1 below reservoir level, tube from lower reservoir into constricted area, same fluid in both. Find ht h2 to which fluid in lower tube rises. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I am not understanding the scenario that this problem is setting. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: