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Calculate the terminal velocity of two steel balls falling through water. The diameters of the two balls are a) cm and b)10cm. Also calculate the Reynolds number for the 10cm sphere. The forces acting on each sphere are gravity, buoyancy and drag Setting up the force equation mg At terminal velocity, the acceleration is zero, as is the net force. Vpsg-Vpwg- 0 1.003 x 10-3 Pa s for water at 20°C ps 7.9 x103 kg/m3 pw 1.0 x103 kg/m3 v-1.5042x107 r2 rl 5x10-3m r2-5x10-2m vl- 376 m/s v2-3.76x104 m/s Reynolds Number: Re 2rpv/n Re 2(5x10-2m)(1x103kg/m3)(3.76x104 m/s/ 1.003 x 10-3 Pa s -0.523 So the Stokes law drag force is appropriate. (Re<l) Ketchup has a density of about 1150kg/m and a viscosity of 50 Pa s. Calculate the drag force and Reynolds number for the 1 cm steel ball falling through Ketchup. Is Stokes Law still valid? What if we dropped the 10 cm steel sphere through peanut butter with a viscosity of 250 Pa s and a density of 2825 kg/m?
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