## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 9 - Problems - Page 363: 61

#### Answer

(a) The pressure of the fluid in the syringe must be $51.4~mm~Hg$ (b) A force of $0.685~N$ must be applied to the plunger.

#### Work Step by Step

(a) We can use Poiseuille's law to find an expression for the flow rate: $Q = \frac{\pi~\Delta P~r^4}{8~\eta~L}$ $Q$ is the flow rate ($m^3~s^{-1}$) $\Delta P$ is the pressure difference ($Pa$) $r$ is the radius ($m$) $\eta$ is the fluid viscosity $L$ is the length of the tube ($m$) We can find the required pressure difference $\Delta P$: $\Delta P = \frac{8~Q~\eta~L}{\pi~r^4}$ $\Delta P = \frac{(8)~(2.50\times 10^{-7}~m^3/s)~(2.00\times 10^{-3}~Pa~s)~(0.030~m)}{(\pi)~(3.00\times 10^{-4}~m)^4}$ $\Delta P = 4715.7~Pa$ We can convert the pressure difference to units of mm Hg: $\Delta P = (4715.7~Pa) \times \frac{1~mm~Hg}{133.3~Pa} = 35.4~mm~Hg$ Since the pressure difference is $35.4~mm~Hg$ and the pressure in the vein is $16.0~mm~Hg$, the pressure of the fluid in the syringe must be $51.4~mm~Hg$ (b) We can convert the pressure in the syringe to units of $Pa$: $51.4~mm~Hg \times \frac{133.3~Pa}{1~mm~Hg} = 6851.6~Pa$ We can find the required force: $F = P~A = (6851.6~N/m^2)(1.00\times 10^{-4}~m^2) = 0.685~N$ A force of $0.685~N$ must be applied to the plunger.

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