Answer:
The velocity of the pulling plate is 14.98 m/s.
Explanation:
The shear stress caused by the 20 Newton force on the plate is given by
[tex]\tau =\frac{20}{Area}=\frac{20}{0.3\times 1.0}=66.66N/m^2[/tex]
Now by newton's law of viscosity we have
[tex]\tau =\mu\cdot \frac{\Delta u}{\Delta y}[/tex]
where
[tex]\mu [/tex] is the dynamic viscosity of the water at 20 degree
[tex]\frac{\Delta u}{\Delta y}[/tex] is the velocity gradient
Applying values we get
[tex]66.66=8.90\times 10^{-4}\times \frac{U}{0.2\times 10^-3}\\\\\therefore u=14.98m/s[/tex]
