A small sphere is allowed to fall through air,and when the sphere has a speed of 4.8 × 10^-4 m/s,the viscous force on the sphere is 3.0 × 10^-15 N.If the viscosity of air is 1.4 × 10^-5 N·s/m²,what is the radius of the drop? A)2.4E-8 m B)8.4E-6 m C)2.3E-7 m D)8.5E-7 m

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Quizplus · Accepted Answer

The viscous force on a sphere falling through a fluid is given by the Stokes' law:$$F_v=6\pi\eta rv$$where:* $F_v$ is the viscous force* $\eta$ is the viscosity of the fluid* $r$ is the radius of the sphere* $v$ is the speed of the sphereWe can rearrange this equation to solve for the radius of the sphere:$$r=\frac{F_v}{6\pi\eta v}$$Plugging in the given values, we get:$$r=\frac{3.0	imes10^{-15}	ext{ N}}{6\pi(1.4	imes10^{-5}	ext{ N}\cdot	ext{s/m}^2)(4.8	imes10^{-4}	ext{ m/s})}$$$$r=2.4	imes10^{-8}	ext{ m}$$Therefore, the radius of the sphere is 2.4E-8 m.

A Small Sphere Is Allowed to Fall Through Air,and When