A dinoflagellate takes 5.0 s to travel 1.0 mm. Approximate a dinoflagellate as a sphere of radius \(35.0 \mu \mathrm{m}\) (ignoring the flagellum). (a) What is the drag force on the dinoflagellate in seawater of viscosity $0.0010 \mathrm{Pa} \cdot \mathrm{s} ?$ (b) What is the power output of the flagellate?

Question: Calculate the drag force and power output of a dinoflagellate in seawater, given that it travels a distance of 1.0 mm in 5.0 seconds and has a radius of 35.0 μm. The viscosity of seawater is 0.0010 Pa·s. Step 1: Calculate the velocity of the dinoflagellate \(v = \frac{d}{t} = \frac{1.0\,\text{mm}}{5.0\,\text{s}} = 0.2\,\text{mm/s}\) Convert the distance from mm to m: \(0.2\,\text{mm/s} \times \frac{1\,\text{m}}{1000\,\text{mm}} = 2.0 \times 10^{-4}\, \text{m/s}\) Step 2: Apply Stokes' Law to find the drag force Convert radius from μm to m: \(35.0\,\mu \text{m} \times \frac{1\,\text{m}}{10^6\, \mu \text{m}} = 3.5 \times 10^{-5}\, \text{m}\) Use Stokes' Law to calculate the drag force: \(F_d = 6\pi r\eta v = 6\pi (3.5 \times 10^{-5}\, \text{m})(0.0010\, \mathrm{Pa}\cdot\mathrm{s})(2.0 \times 10^{-4}\, \text{m/s}) \approx 1.31 \times 10^{-10}\,\text{N}\) Step 3: Calculate the power output of the dinoflagellate \(P = F_d \cdot v = (1.31 \times 10^{-10}\,\text{N})(2.0 \times 10^{-4}\, \text{m/s}) \approx 2.62 \times 10^{-14}\,\text{W}\) Answer: The drag force acting on the dinoflagellate is approximately \(1.31 \times 10^{-10}\, \text{N}\), and its power output is approximately \(2.62 \times 10^{-14}\, \text{W}\).