Answer
The pressure drop across the aorta is $26.7~kPa$
Work Step by Step
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 pressure drop $\Delta P$:
$\Delta P = \frac{8~Q~\eta~L}{\pi~r^4}$
$\Delta P = \frac{(8)~(4.1\times 10^{-3}~m^3/s)~(4.0\times 10^{-3}~Pa~s)~(0.40~m)}{(\pi)~(5.0\times 10^{-3}~m)^4}$
$\Delta P = 26.7~kPa$
The pressure drop across the aorta is $26.7~kPa$.
