## College Physics (4th Edition)

The absolute pressure at the bug's end of the feeding tube is $-107~kPa$
We can find the flow rate $Q$: $Q = \frac{V}{t} = \frac{3.0\times 10^{-7}~m^3}{(25)(60~s)} = 2.0\times 10^{-10}~m^3/s$ 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 $\Delta P$: $\Delta P = \frac{8~Q~\eta~L}{\pi~r^4}$ $\Delta P = \frac{(8)~(2.0\times 10^{-10}~m^3/s)~(0.0013~Pa~s)~(2.0\times 10^{-4}~m)}{(\pi)~(5.0\times 10^{-6}~m)^4}$ $\Delta P = 212~kPa$ Since the pressure difference is $212~kPa$ and the pressure in the arm is $105~kPa$, the absolute pressure at the bug's end of the feeding tube is $105~kPa-212~kPa$ which is $-107~kPa$.