Ask question

Ask question

All categories

3 years ago

14

7. Plasma from blood (density = 1025 kg/m3) flows along a vertical channel in a steady, incompressible, fully developed laminar

film of thickness h. (i) Simplify the continuity and Navier-Stokes equations to model this flow field. (ii) Obtain expressions for the velocity profile, (iii) the shear stress distribution, (iv) the volume flow rate, and (v) the average velocity. (vi) Relate the plasma film thickness to the volume flow rate per unit depth of surface normal to the flow. (vii) Calculate the volume flow rate for h = 0.5 mm, flowing down a surface b = 2 cm wide.

1 answer:

Goryan [66]3 years ago

6 0

Answer: Q = 8.37×10^-9L/s

Explanation: since the depth of the surface is normal to the flow, ø = 90

Find the attached file for the solution

You might be interested in

Describe two sources of earth's energy that are not produced

Anvisha [2.4K]

Here is the answer. Two sources of Earth's energy that are not produced would be Cosmic rays and Tidal Energy. Cosmic rays <span>are high-energy protons and atomic nuclei that come from outside the solar system. Whereas, tidal energy is the energy produced by both the moon (2/3) and the sun (1/3). Hope this answers your question.</span>

6 0

3 years ago

How does friction make it possible for you to walk across the floor?

Tema [17]

if we are walking on a perfectly smooth ground which has no friction our force would simply cancel out the force reverted by the ground and we would fall.

We need it to help push out feet off the ground

Hope those helps :)

5 0

3 years ago

A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

jeka57 [31]

Answer:

a. $3.1 m/s^2$

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

$mg sin \theta - \mu_k R = ma$

where :

m = 6.0 kg is the mass of the box

g = 9.8 m/s^2 the acceleration of gravity

$\theta=39^{\circ}$ is the angle of the slope

$\mu_k = 0.40$ is the coefficient of friction

R is the normal reaction

a is the acceleration

- Along the perpendicular direction:

$R-mg cos \theta =0$

From the 2nd equation, we get an expression for the reaction force:

$R=mg cos \theta$

And substituting into the 1st equation, we can find the acceleration:

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2$

6 0

3 years ago

A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi

Volgvan

Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

$y=V_{o}sin\theta t-\frac{gt^{2}}{2}$ (1)

$V_{y}=V_{oy}-gt$ (2)

Where:

$y$ is the height of the cat

$V_{oy}=V_{o}sin\theta$ is the cat's initial velocity

$\theta=60\°$

$g=9.8m/s^{2}$ is the acceleration due gravity

$t$ is the time

$V_{y}$ is the y-component of the velocity

Now the cat will have its maximum height $y_{max}$ when $V_{y}=0$ . So equation (2) is rewritten as:

$0=V_{oy}-gt$ (3)

Finding $t$ :

$t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}$ (4)

$t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}$ (5)

$t=0.24 s$ (6)

Substituting (6) in (1):

$y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}$ (7)

Finally:

$y_{max}=0.287 m \approx 0.29 m$ (8)

3 0

3 years ago

A small bar of pure gold whose density is 19.3g/cm. Displaces 80 cm

nasty-shy [4]

Answer:

The mass of the gold bar is 1,544 g

Explanation:

3 0

3 years ago