Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.3 Applications to Physics and Engineering - 8.3 Exercises: 2

Answer

(a) $12,054Pa \approx 12 kPa$ (b) $\approx 3.86 \times 10^{5} N$ (c) $\approx 3.62 \times 10^{4} N$

Work Step by Step

(a) $P = \rho gd = (820 kg/m^{3})(9.8 m/s^{2})(1.5m) = 12,054Pa \approx 12 kPa$ (b) $F = PA = (12,054Pa) (8m)(4m) \approx 3.86 \times 10^{5} N$ (c) $F = \int ^{1.5}_{0} \rho g x 4dx = (820)(9.8) 4\int^{1.5}_{0} x dx = 32,144[\frac{1}{2} x^{2}]^{3/2}_{0} \approx 3.62 \times 10^{4} N$
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