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Water flows through a small horizontal pipe with a speed of 12 m/s into a larger part of the pipe for which the diameter of the pipe is doubled. What is the speed of the water in the larger part of the pipe?
  1. 0.75 m/s
  2. 3.0 m/s
  3. 6.0 m/s
  4. 12 m/s
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OpenStax College Physics for AP® Courses Solution, Chapter 12, Problem 1 (Test Prep for AP® Courses) (2:23)

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This is College Physics Answers with Shaun Dychko. The volume rate of flow in both portion of the pipe have to be the same. And, so this is the continuity equation, you could say it. So since the fluid is incompressible, the rate at which the fluid flows through one part of the pipe has the equal the rate at which it flows to the other pipe. So this is units of cubic meters per second here, on each side. Now the cross-sectional area of the pipe in the first case is Pi times this diameter squared over four multiplied by the first speed equals the cross-sectional area of the second pipe which is Pi times diameter two squared over four times the speed two. And we'll figure out <i>V2</i> by first getting rid of this common factors Pi and four of both sides. Said to, what we're going to do is multiply both sides by four over Pi <i>d2</i> squared. And then do the same to the other side. And the four is cancel and the Pi is cancel. And we're left with <i>d1</i> squared over <i>d2</i> squared on this side. And we're switching the sides around to isolate the unknown on the left. So we have <i>V2</i> is <i>d1</i> over <i>d2</i>, both of them are squared. And so we'll write this ratio in brackets and then square the result there dividing the two. I mean, you could write this as <i>d1</i> squared over <i>d2</i> squared, that would also be fine. But I chose to do it like this. Now we're told that diameter two is twice diameter one. So the larger part of the pipe has a diameter that is doubled. And so we'll substitute two times <i>d1</i> in place of <i>d2</i>, and we do that substitution here. And the <i>d1s</i> are common factor there that cancel on top and bottom. And we're left with one half squared times <i>V1</i> is going to be <i>V2</i>. So yeah, we have <i>V2</i> is going to be <i>V1</i> divided by four 'coz one half squared is one quarter. So that's 12 meter per second initially we're told is the speed, divided by four gives us three meters per second and the answer is b.