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For problem 1 of challenge problems for units and dimensional analysis, im confused on section 1.10 of the solution. How does Y = -1 and Z = -1?

MIT OCW Physics
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Hi! I won't be on too much longer, but maybe I can help. I don't know what the question or solution are, though, since I'm not in the course. Feel free to provide that information, so that other people on OpenStudy can help!
Problem 1: An ideal (non-viscous) liquid with a density of ! is poured into a cylindrical vessel with a cross-sectional area of A1to a level at a height h from the bottom. The bottom has an opening with a cross-sectional area A2. Find the time it takes the k=liquid to flow out. I'll post the link to the solution because copying it would take up too much space. http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/introduction-to-mechanics/units-and-dimensional-analysis/MIT8_01SC_problems01_soln.pdf
Hi, I just finished the set of questions. Well, to answer your question as to how is Y = -1 and Z = -1, Considering equation 1.10 and analyzing using dimensions, we obtain, [T] = [L]^y/[L]^(-z) Which leads to y+z = 0 since there is no [L] dimension on the LHS Further implies y =-z which implies that these quantities cancel out giving y=1 and -z=1. This is how I have understood it, let me know what you think?

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Ah ha! That makes sense, I forgot to use dimensional analysis for that last equation. Thank you for clarifying :)
Yes is it easy to miss :)

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