#$&* course Phy 122 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . ** Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: query video experiment terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The velocity of the sphere increased when the weights kept being added. At around .1255 m/ sec velocity, the weights started to not have as high of an effect on making the velocity increase. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok ********************************************* Question: `q002. If you know the pressure drop of a moving liquid between two points in a narrowing round pipe, with both points at the same altitude, and you know the speed and pipe diameter in the section of pipe with the greater diameter, how could you determine the pipe diameter at the other point? If a U tube containing mercury articulates with the pipe at the two points, how can you find the difference between the mercury levels in the two sides of the pipe? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: use Bernoulli's equation, without (density) (gravity) (height) (altitude does not change) (1/2) (den) (v1) ^2 + P1 = (1/2) (den) (v2) ^2 + P2 P2 - P1 = (1/2) (den) (v1) ^2 - (1/2) (den) (v2) ^2 v2 = sqrt((1/2) (den)(v1) ^2 - change in pressure]/ [(1/2)(den)]) Continuity Equation: A1 v1 = A2 v2 A2 = (A1*V1) /(V2) Then set A2 = pi* (d/2) ^2 Bernoulli's equation: (1/2) (den) (v1 )^2 + (den) (gravity) (height1) + P1 = (1/2) (den)(v2) ^2 + (den) (gravity) (height2) + P2 (1/2) (density) (v1) ^2 - (1/2) (den) (v2) ^2 = (den) (gravity) (height2) - (den) (gravity)(height1) + P2 - P1 (1/2)(den)(v1)^2 - (1/2)(den) (v2) ^2 = (den) (gravity) (height2-height1) + P2 - P1 change in altitude = [(1/2)(den)(v1) ^2 - (1/2)(den)(v2)^2 - change in pressure ] / [(den)(gravity)] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: use Bernoulli's equation, without (density) (gravity) (height) (altitude does not change) (1/2) (den) (v1) ^2 + P1 = (1/2) (den) (v2) ^2 + P2 P2 - P1 = (1/2) (den) (v1) ^2 - (1/2) (den) (v2) ^2 v2 = sqrt((1/2) (den)(v1) ^2 - change in pressure]/ [(1/2)(den)]) Continuity Equation: A1 v1 = A2 v2 A2 = (A1*V1) /(V2) Then set A2 = pi* (d/2) ^2 Bernoulli's equation: (1/2) (den) (v1 )^2 + (den) (gravity) (height1) + P1 = (1/2) (den)(v2) ^2 + (den) (gravity) (height2) + P2 (1/2) (density) (v1) ^2 - (1/2) (den) (v2) ^2 = (den) (gravity) (height2) - (den) (gravity)(height1) + P2 - P1 (1/2)(den)(v1)^2 - (1/2)(den) (v2) ^2 = (den) (gravity) (height2-height1) + P2 - P1 change in altitude = [(1/2)(den)(v1) ^2 - (1/2)(den)(v2)^2 - change in pressure ] / [(den)(gravity)] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!