#$&* course Phy 202 007. `query 6
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Given Solution: ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . ** Your Self-Critique: 3 Your Self-Critique Rating: ok ********************************************* Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Adding the weights increased the velocity of the sphere. However, as time progressed, the added weight had less effect on increasing the velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: With the information given, we can find the final velocity through Bernoulli's equation. Since the height is equal, we can take out the (density)(gravity)(height) part of the equation. (1/2)(density)(v1)^2 + P1 = (1/2)(density)(v2)^2 + P2 P2 - P1 = (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 Since we are given the pressure drop, and the first velocity, we can solve for the final velocity. v2 = sqrt([(1/2)(density)(v1)^2 - change in pressure]/[(1/2)(density)]) With the seconds velocity, we can use the Continuity Equation: A1 v1 = A2 v2 to solve for the second area. We can get the first area using the diameter from the first hole, so A1 = pi*(d/2)^2 Then, we can find the second area, A2 = (A1*V1)/(V2) Once we have the value for A2, we can set up: A2 = pi*(d/2)^2 and then solve for d to get the second diameter. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok ********************************************* Question: `q002. If you know the pressure drop of a moving liquid between two points in a narrowing round pipe, with both points at the same altitude, and you know the speed and pipe diameter in the section of pipe with the greater diameter, how could you determine the pipe diameter at the other point? If a U tube containing mercury articulates with the pipe at the two points, how can you find the difference between the mercury levels in the two sides of the pipe? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: You can use Bernoulli's equation, without the (density)(gravity)(height) part since the altitude does not change. (1/2)(density)(v1)^2 + P1 = (1/2)(density)(v2)^2 + P2 P2 - P1 = (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 Then solve for v2 since that is the only thing not given. v2 = sqrt([(1/2)(density)(v1)^2 - change in pressure]/[(1/2)(density)]) With the second velocity, we can use the Continuity Equation: A1 v1 = A2 v2 to solve for the second area. (the first area can be found using the first diameter, in the equation A = pi*(d/2)^2) A2 = (A1*V1)/(V2) Then set A2 = pi*(d/2)^2 and solve for d to get the second diameter. To find the difference between the mercury levels in the two sides of the pipe, you can use Bernoulli's equation: (1/2)(density)(v1)^2 + (density)(gravity)(height1) + P1 = (1/2)(density)(v2)^2 + (density)(gravity)(height2) + P2 (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 = (density)(gravity)(height2) - (density)(gravity)(height1) + P2 - P1 (1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 = (density)(gravity)(height2-height1) + P2 - P1 change in altitude = [(1/2)(density)(v1)^2 - (1/2)(density)(v2)^2 - change in pressure ] / [(density)(gravity)] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: