Posted by Anonymous on Sunday, September 2, 2012 at 11:39pm.
ρ =2 g/cm³= 2•10³kg/m³
The terminal speed of the particle is
v<h/t =0.037/3600=1.0228•10^-5 m/s.
The terminal speed of the particle of density ρ and radius R which is falling down in the water od density ρ1 =1000 kg/m³ and viscosity η is
v=2•R²•g• (ρ - ρ1)/9•η.
R=sqrt[9• η•v/2•g•(ρ - ρ1)] =
= sqrt[9•0.001•1.0228•10^-5/2•9.8•1•10³]=
=2.17•10^-6 m =2.17 μm
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