Solution
= Solution
The normal-incidence <acoustic impedance> is $Z=\rho V_P$. Therefore
$$
\boxed{Z_1=1500(1500)=2.25\times10^6\ \mathrm{kg\,m^{-2}\,s^{-1}},}
$$
$$
\boxed{Z_2=2000(3500)=7.00\times10^6\ \mathrm{kg\,m^{-2}\,s^{-1}}.}
$$
These are $2.25$ and $7.00$ MRayl. They combine the material’s density and wave speed; using $V_S$ would give shear rather than acoustic P impedance.