A. F 2 > F 1 > F 3 B. F 2 > F 1 = F 3 C. F 3 > F 2 > F 1 D. F 3 > F 1 > F 2 E. F 1 = F 2 = F 3 Rank in order, from largest to smallest, the magnitudes.

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A. F 2 > F 1 > F 3 B. F 2 > F 1 = F 3 C. F 3 > F 2 > F 1 D. F 3 > F 1 > F 2 E. F 1 = F 2 = F 3 Rank in order, from largest to smallest, the magnitudes of the forces required to balance the masses. The masses are in kilograms.

A. F 2 > F 1 > F 3 B. F 2 > F 1 = F 3 C. F 3 > F 2 > F 1 D. F 3 > F 1 > F 2 E. F 1 = F 2 = F 3 Rank in order, from largest to smallest, the magnitudes of the forces required to balance the masses. The masses are in kilograms. In each of the three pictures, the left and right pistons are at the same height in a continuous fluid so the pressure is the same. P = F left /A left = F right /A right, so F left = F right *A left /A right = m right *g*A left /A right. Since the areas are apparently the same, the individual mass will determine the necessary force.

The figure shows volume flow rates (in cm 3 /s) for all but one tube. What is the volume flow rate through the unmarked tube? Is the flow direction in or out? A. 1 cm 3 /s, out B. 1 cm 3 /s, in C. 10 cm 3 /s, out D. 10 cm 3 /s, in E. It depends on the relative size of the tubes.

The figure shows volume flow rates (in cm 3 /s) for all but one tube. What is the volume flow rate through the unmarked tube? Is the flow direction in or out? A. 1 cm 3 /s, out B. 1 cm 3 /s, in C. 10 cm 3 /s, out D. 10 cm 3 /s, in E. It depends. There cannot be a continuous increase in the volume of fluid in the pipes, nor can there be a continuous decrease in the volume of fluid. The inputs and outputs must match.

Rank in order, from highest to lowest, the liquid heights h 1 to h 4 in tubes 1 to 4. The air flow is from left to right. A. h 1 > h 3 > h 4 > h 2 B. h 1 > h 2 = h 3 = h 4 C. h 2 = h 3 = h 4 > h 1 D. h 2 > h 4 > h 3 > h 1 E. h 3 > h 4 > h 2 > h 1

Rank in order, from highest to lowest, the liquid heights h 1 to h 4 in tubes 1 to 4. The air flow is from left to right. A. h 1 > h 3 > h 4 > h 2 B. h 1 > h 2 = h 3 = h 4 C. h 2 = h 3 = h 4 > h 1 D. h 2 > h 4 > h 3 > h 1 E. h 3 > h 4 > h 2 > h 1 Larger cross sectional area results in lower speed, according to the equation of continuity. Lower speed results in higher pressure, according to Bernoulli’s equation.