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# RFinal - Problem 2 Solution Part a We will use Bernoullis...

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Problem 2 Solution Part a We will use Bernoulli’s Equation (BE) to relate what is happening at the surface of the water in the tank (A), to what is happening at the end of the tube (B). P A + ρgh + 1 2 ρv 2 A = P B + ρgL sin θ + 1 2 ρv 2 B The two pressures will be the same since both are exposed to the atmosphere. Also, since the radius of the tank is much larger than that of the tube, we can take v A = 0. Making these simplifications gives: gh = gL sin θ + 1 2 v 2 B This yields us v B easily. v B = 2 g ( h - L sin θ ). To get the maximum height to which the stream rises (H), we must consider that the stream will still have some velocity horizontally at the top of its motion. This is going to be the horizontal component of the speed at B. So using BE again gives: gL sin θ + 1 2 v 2 B = gH + 1 2 v 2 B cos θ 2 We have already set the presures equal and divided by a common factor of the density. If we plug in our expression for v B and solve for H we find: H = h - ( h - L sin θ ) cos θ 2 We are happy with this result for two reasons. The first is that the units are length on both sides. Also, if we let the angle go to special values (zero and ninety) we get the expected results ( H = 0 and H = h ). Part b When the radius of the tank and the tube are no longer drastically different we must think about the continuity equation as well. We start with BE again, but
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