Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 10 - Fluids - Problems - Page 286: 13



Work Step by Step

The pressure difference is due to the depth of water. Use equation 10–3b, and round the answer to 2 significant figures. $$\Delta P=\rho g \Delta h $$ $$\Delta h= \frac{\Delta P}{\rho g}$$ $$=\frac{(85 mm-Hg)((133 N/m^2)/(1mm-Hg))}{(1.00 \times 10^3 kg/m^3)(9.80 m/s^2)} \approx 1.2m$$ His maximum depth would be about 1.2 m.
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