Physics: Principles with Applications (7th Edition)

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

Chapter 7 - Linear Momentum - Problems - Page 192: 13

Answer

The average force of the air on the building is $4.9\times 10^6~N$

Work Step by Step

$v = (120~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 33.3~m/s$ Each second, the volume of air that hits the building is $V = 33.3~m\times 45~m \times 75~m$ We can use this volume to calculate the mass of the air that hits the building each second. $mass = (1.3~kg/m^3)(33.3~m\times 45~m \times 75~m)$ $mass = 1.46\times 10^5~kg$ We can use the change in momentum of the air to find the magnitude of the force that the building exerts on the air: $F\cdot t = \Delta p = m \times \Delta v$ $F = \frac{(1.46\times 10^5~kg)(33.3~m/s)}{1~s}$ $F = 4.9\times 10^6~N$ The force of the air on the building must have the same magnitude as the force of the building on the air. Therefore, the average force of the air on the building is $4.9\times 10^6~N$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.