Velocity profile of a moving fluid over a flat (fixed) plate is given as: = from the fixed plate (yy = 0) in mm, u is the velocity in xx direction in mm/. Calculate the shear stress on the fluid at yy = 0.15 m if the viscosity of the fluid 8.6 poise (1 Poise= 1 yy.) 2

The question has some details missing which is the velocity profile .The velocity profile is given as u = (8y - 0.3y^2) mm/s

the shear stress on the fluid at y = 0.15 m = 6.80 x 10^-3N/m^2

Explanation:

The detailed steps is the use of the relationship between the shear stress, viscosity and the velocity profile. appropriate differentiation and substitution were made to calculate the shear stress as is shown in the attachment.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s