Two fluid layers with different viscosities ( u1= 0.1 pa ∙ s and u2 = 0.2 pa ∙ s) are sandwiched between two plates of area a = 1 m2. each layer is h = 1 mm thick. find the force f necessary to make the upper plate move at a speed u = 1 m/s. also, find the fluid velocity at the interface between the two fluids.

Explanation:

The given data is as follows.

            = 0.1 Pa.s,             = 0.2 Pa.s

            = = 1 mm =   (as 1 m = 1000 mm)

As the velocity gradients are linear so, the shear stress will be the same throughout.

                  = velocity at the interface

or,                

Now, putting the given values into the above formula as follows.

                          =        

                          = 0.667 m/s

Hence, force required will be F =          

or,                  F =

                         =

                         = 66.67 N

Thus, we can conclude that the fluid velocity is 0.667 m/s and force necessary to make the upper plate move at a speed U = 1 m/s is 66.67 N.

The answer to your question is   C₂H₆O

Explanation:

Data

Compound   CxHyOz

mass of the sample = 0.9827

mass of CO₂ = 1.90 g

mass of H₂O = 1.070 g

Process

1.- Find the moles and grams of Carbon

                    44g of CO₂ 12 g of C

                     1.9 g of CO₂  x

                     x = (1.9 x 12) / 44

                     x = 0.52 g

                    12 g of C 1 mol

                    0.52 g     x

                    x = (0.52 x 1) / 12

                    x = 0.043 moles

2.- Find the moles and grams of Hydrogen

                   18 g of H₂O 2 g of H

                   1.07 g of H₂O  x

                   x = (1.07 x 2)/18

                   x = 0.12 g of H

                   1 g of H 1 mol of H

                   0.12 g   x

                   x = (0.12 x 1)/1

                   x = 0.12 moles of H

3.- Calculate the moles and grams of Oxygen

Mass of Oxygen = 0.9827 - 0.12 - 0.52

                            = 0.3427 g

                    16 g of O 1 mol

                     0.3427 g x

                     x = (0.3427 x 1) / 16

                     x = 0.021 moles

4.- Divide by the lowest number of moles

Carbon       0.043 / 0.021 = 2

Hydrogen   0.12 / 0.021 = 5.7 ≈ 6

Oxygen       0.021 / 0.021 = 1

5.- Write the empirical formula

                    C₂H₆O