The terminal velocity of a copper ball of rad | vedclass

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Question

(B) The formula for terminal velocity $v_t$ is given by Stokes' Law: $v_t = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$,where $r$ is the radius,$g

Vedclass · Accepted Answer

$9.9 	imes 10^{-1} \; kg \; m^{-1} \; s^{-1}$

Vedclass · Answer

(B) The formula for terminal velocity $v_t$ is given by Stokes' Law: $v_t = \frac{2}{9} \frac{r^2 g (ho - \sigma)}{\eta}$,where $r$ is the radius,$g$ is acceleration due to gravity,$ho$ is the density of the ball,$\sigma$ is the density of the oil,and $\eta$ is the coefficient of viscosity.Rearranging for $\eta$: $\eta = \frac{2 r^2 g (ho - \sigma)}{9 v_t}$.Given values: $r = 2.0 	imes 10^{-3} \; m$,$v_t = 6.5 	imes 10^{-2} \; m \; s^{-1}$,$g = 9.8 \; m \; s^{-2}$,$ho = 8.9 	imes 10^{3} \; kg \; m^{-3}$,$\sigma = 1.5 	imes 10^{3} \; kg \; m^{-3}$.$ho - \sigma = (8.9 - 1.5) 	imes 10^{3} = 7.4 	imes 10^{3} \; kg \; m^{-3}$.Substituting the values: $\eta = \frac{2 	imes (2.0 	imes 10^{-3})^2 	imes 9.8 	imes 7.4 	imes 10^{3}}{9 	imes 6.5 	imes 10^{-2}}$.$\eta = \frac{2 	imes 4.0 	imes 10^{-6} 	imes 9.8 	imes 7.4 	imes 10^{3}}{58.5 	imes 10^{-2}} = \frac{579.04 	imes 10^{-3}}{58.5 	imes 10^{-2}} \approx 9.9 	imes 10^{-1} \; kg \; m^{-1} \; s^{-1}$.

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