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n Water atao°C. Findl the terminal speed 0 per second, 1gnonng the and ass uming Stokes" law meed to notethat the dYa terminal velocity. Take l,4x108 kglm³. The visc 3. T.002x10° Kg|m•sand •S and

n Water atao°C. Findl the terminal speed 0 per second, 1gnonng the and ass uming Stokes" law meed to notethat the dYa terminal velocity. Take l,4x108 kglm³. The visc 3. T.002x10° Kg|m•sand •S and

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please help I have provided my answer which is wrong. The radius is given in micrometers, which is 10^-6.  Please also keep several sig figs when doing your calculations, besides this factor of 1000.

talling Consider a spherical bacter jum, with radius 0-l05mm in Water atao°c. Findl the terminal speed o7 the sphencal bacterium In mcters persecond, ignoring the buoyant 7orce on the baderium and assuming Stokes' law o7he viscous Aorce. You willAiot moneed to note that the drag 70rce iseaual TO the weight at Terminal velocity. Take the denst OA the bacterium To be 1.4x10° 3. Kg! Kalm. do°Cis The Viscosity 07 water at 20°CIS 1.002x10 03 .t Kg/m.Sand Yhe denoity 15 998 kg m V= चथाी ।
Transcribed Image Text:talling Consider a spherical bacter jum, with radius 0-l05mm in Water atao°c. Findl the terminal speed o7 the sphencal bacterium In mcters persecond, ignoring the buoyant 7orce on the baderium and assuming Stokes' law o7he viscous Aorce. You willAiot moneed to note that the drag 70rce iseaual TO the weight at Terminal velocity. Take the denst OA the bacterium To be 1.4x10° 3. Kg! Kalm. do°Cis The Viscosity 07 water at 20°CIS 1.002x10 03 .t Kg/m.Sand Yhe denoity 15 998 kg m V= चथाी ।
V4=a(0 uS X103° (1.9888 Kqlm³ 9.8 mli) V4=2(0.65X103 (1.988 Kglm³ 9.8mls y o1(1-00 a x10 3 kg|ms) Kg =1.835x103m|s nc w.t
Transcribed Image Text:V4=a(0 uS X103° (1.9888 Kqlm³ 9.8 mli) V4=2(0.65X103 (1.988 Kglm³ 9.8mls y o1(1-00 a x10 3 kg|ms) Kg =1.835x103m|s nc w.t
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