91Ó°ÊÓ

First, we need to find the gravitational force acting on a single water droplet. The gravitational force is given by the formula: \(F_g = m \times g\) where \(F_g\) is the gravitational force, \(m\) is the mass of the water droplet, and \(g\) is the acceleration due to gravity (approximately \(9.81 \frac{\text{m}}{\text{s}^2}\)). We must find the mass of a water droplet, which can be determined using the volume and density of water. The volume \(V\) of a sphere is given by: \(V = \frac{4}{3} \pi r^3\) where \(r\) is the radius of the droplet, and the density \(\rho\) of water is approximately \(1000 \frac{\text{kg}}{\text{m}^3}\). The mass of the droplet is then: \(m = \rho V = \rho \cdot \frac{4}{3} \pi r^3\)

The viscous drag force acting on the droplet can be estimated using Stokes' law, which states: \(F_d = 6 \pi \eta r v_t\) where \(F_d\) is the viscous drag force, \(\eta\) is the dynamic viscosity of air, \(r\) is the radius of the droplet, and \(v_t\) is the terminal velocity. The dynamic viscosity of air at \(20^{\circ} \mathrm{C}\) is approximately \(1.82 \times 10^{-5} \frac{\text{kg}}{\text{m s}}\).

At terminal velocity, the droplet is no longer accelerating, and the gravitational force acting on it is equal to the viscous drag force, so: \(F_g = F_d \Rightarrow m \cdot g = 6 \pi \eta r v_t\)

Now, we need to solve the equation for terminal velocity. Substituting the mass from step 1, we get: \(\rho \cdot \frac{4}{3} \pi r^3 g = 6 \pi \eta r v_t\) Now, we can simplify the equation and solve for \(v_t\): \(v_t = \frac{\rho \cdot \frac{4}{3} \pi r^3 g}{6 \pi \eta r} = \frac{2}{9} \frac{\rho r^2 g}{\eta}\) Plug in the given values for the radius of water droplets, temperature-related dynamic viscosity, and water density: \(v_t = \frac{2}{9} \frac{(1000 \frac{\text{kg}}{\text{m}^3}) (5.0 \times 10^{-6} \text{m})^2 (9.81 \frac{\text{m}}{\text{s}^2})}{1.82 \times 10^{-5} \frac{\text{kg}}{\text{m s}}} \approx 2.69 \times 10^{-4} \frac{\text{m}}{\text{s}}\) So, the terminal velocity of water droplets in a cumulus cloud at \(20^{\circ} \mathrm{C}\) is approximately \(2.69 \times 10^{-4} \frac{\text{m}}{\text{s}}\).