Respuesta :
Answer:
[tex]\nu =0.000108\ \frac{m^2}{sec}[/tex]
[tex]\nu =0.001161\ \frac{ft^2}{sec}[/tex]
Explanation:
Given that
[tex]\mu =0.1\ Pa-s[/tex]
Specific gravity of fluid S=0.92
So that density of fluid
[tex]\rho=920\ \frac{kg}{m^3}[/tex]
As we know that ,kinematic viscosity of fluid is given as
[tex]\nu =\dfrac{\mu }{\rho}[/tex]
Now by putting the values
[tex]\nu =\dfrac{0.1 }{920}[/tex]
[tex]\nu =0.000108\ \frac{m^2}{sec}[/tex]
So the kinematic viscosity of fluid in SI
[tex]\nu =0.000108\ \frac{m^2}{sec}[/tex]
The kinematic viscosity of fluid in SI
[tex]\nu =0.001161\ \frac{ft^2}{sec}[/tex]
