Answer:
[tex]\nu =0.01\frac{m^2}{s}[/tex]
Explanation:
Hello,
Kinematic viscosity is defined in terms of the dynamic viscosity as follows:
[tex]\nu =\frac{\mu}{\rho }[/tex]
Whereas [tex]\nu[/tex] is the kinematic viscosity, [tex]\mu[/tex] the dynamic viscosity and [tex]\rho[/tex] the density. In this case, the given temperature and pressure are defining both [tex]\mu[/tex] and [tex]\rho[/tex] so they are not included into the aforesaid equation, in such a way, the kinematic viscosity is straightforwardly computed as shown below:
[tex]\nu =\frac{0.1Pa*s}{10.0kg/m^3} =0.01\frac{kg*\frac{m}{s^2*m^2}*s}{kg/m^3} \\\nu =0.01\frac{m^2}{s}[/tex]
Best regards.
