Given "u=\\dfrac{3}{2}y-y^{3\/2}." Then
"\\dfrac{du}{dy}=\\dfrac{3}{2}-\\dfrac{3}{2}y^{1\/2}" At "y=9 cm=0.09m"
"\\dfrac{du}{dy}=\\dfrac{3}{2}-\\dfrac{3}{2}(0.09)^{1\/2}=1.05(s^{-1})" Shear stress is given
"\\tau=\\mu\\cdot\\dfrac{du}{dy}" Assume dynamic viscosity as "\\mu=8\\ \\text{poise}=0.8\\ N\\cdot s\/m^2"
Then
"\\tau=0.8\\ N\\cdot s\/m^2\\cdot1.05s^{-1}=0.84 N\/m^2"
The shear stress at y = 9 cm is "0.84N\/m^2."
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If u =2/3y - y2 then please help
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