College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 360: 2

Answer

The frictional force exerted on the cork by the neck of the bottle is $143~N$

Work Step by Step

We can convert $4.5~atm$ to units of Pascals: $4.5~atm \times \frac{1.013\times 10^5~Pa}{1~atm} = 4.56\times 10^5~Pa$ We can find the force exerted on the cork by the pressure difference: $F = P~A$ $F = P~\pi~r^2$ $F = (4.56\times 10^5~N/m^2)(\pi)(0.010~m)^2$ $F = 143~N$ The frictional force exerted on the cork by the neck of the bottle must be equal in magnitude to this force. The frictional force exerted on the cork by the neck of the bottle is $143~N$
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