Two parallel plates, one moving at 4 m/s and the other stationary, are separated by a 5mm thick layer of oil with specific gravity of 0.80 and kinematic viscosity 1.25E-4 m2/s. What is the average shear stress in the oil? Show your work.

Average shear stress;τ = 80 Pa

Explanation:

We are given;

Thickness;t = 5mm = 0.005 m

Velocity;u = 4 m/s

Specific Gravity;Sg = 0.8

Kinematic viscosity;ν = 1.25 × 10^(-4) m²/s

We are not given density of the oil, but given specific gravity of 0.8. we know that formula for specific gravity of an object = Density of object/Density of water

Density of water = 1000 kg/m³

Thus;

0.8 = density of oil/1000

Density of oil = 800 kg/m³

Now, let's calculate dynamic viscosity of the oil. It's given by the formula;

μ = ρν

Where;

μ is dynamic viscosity

ρ is density of oil

ν is kinematic viscosity

So,

μ = 800 × 1.25 × 10^(-4)

μ = 0.1 kg/m.s

Now, formula for the average shear stress is given as;

τ = μ × u/t

Plugging in the relevant values, we have;

τ = 0.1 × 4/0.005

τ = 80 Pa

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.