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: 26


The balloon can lift a cargo of 1045 kg

Work Step by Step

We can find the volume $V$ of the balloon as: $V = \frac{4}{3}\pi~r^3$ $V = \frac{4}{3}\pi~(7.15~m)^3$ $V = 1531~m^3$ Let $M_c$ be the mass of the cargo. To find the maximum possible mass $M_c$, we can let the buoyant force be equal to the total weight of the balloon and the cargo. Let $\rho_a$ be the density of the air. Therefore; $(M_c+930~kg)~g = F_B$ $(M_c+930~kg)~g = \rho_a~V~g$ $(M_c+930~kg) = \rho_a~V$ $M_c = \rho_a~V - 930~kg$ $M_c = (1.29~kg/m^3)(1531~m^3) - 930~kg$ $M_c = 1045~kg$ The balloon can lift a cargo of 1045 kg.
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