Question

Kinematic viscosity of water at 27 degrees celsius

Answer 1

Answer:

8.90 ×10^-3. dyne.s/cm^2

Answer 2

Dear Student,

◆ Answer -

ν = 8.539×10^-7 m^2/s

◆ Explaination -

# Required information -

μ = 8.509×10^-4 Ns/m^2

ρ = 996.5 kg/m^3

# Solution -

Kinematic viscosity of water is calculated by -

Kinematic viscosity = dynamic viscosity / density of water

ν = μ/ρ

ν = 8.509×10^-4 / 996.5

ν = 8.539×10^-7 m^2/s

Hence, kinematic velocity of water at 27 ℃ is 8.539×10^-7 m^2/s.

Thanks dear..