(d) Rossby Number Estimates ( $f = 10^{-4}\,\mathrm{s}^{-1}$)

(i)

Midlatitude Ocean Current: $U = 0.1\,\mathrm{m/s}$, $L = 100\,\mathrm{km} = 10^5\,\mathrm{m}$

$$\mathrm{Ro}_{(i)} = \frac{0.1\,\mathrm{m/s}}{(10^{-4}\,\mathrm{s}^{-1})(10^5\,\mathrm{m})} = \frac{0.1}{10} = \mathbf{0.01}$$

(ii)

Tornado: $U = 50\,\mathrm{m/s}$, $L = 100\,\mathrm{m}$

$$\mathrm{Ro}_{(ii)} = \frac{50\,\mathrm{m/s}}{(10^{-4}\,\mathrm{s}^{-1})(100\,\mathrm{m})} = \frac{50}{10^{-2}} = \mathbf{5000}$$

The **midlatitude ocean current** ( $\mathrm{Ro} \approx 0.01$) is more strongly influenced by rotation. This is because its $\mathrm{Ro} \ll 1$, meaning the Coriolis force significantly exceeds the inertial forces, forcing the flow into an approximate **Geostrophic Balance**. The tornado's $\mathrm{Ro} \gg 1$, indicating that inertial forces dominate, and planetary rotation is negligible.