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GalinKa [24]
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.

Physics
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

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Given the speed and the distance, to find time you can use the formula speed is equal to distance over time. From there you can manipulate the equation for time to equal the distance divided by speed. Time is equal to 18.4 meters divided by 35m/s which equals 0.526 seconds.

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3 years ago
What is a crystal as applied in physics ​
Scilla [17]

Answer:

The correct answer is - A matter that has an ordered arrangement of atoms, molecules, or ions.

Explanation:

In physics, a crystal is a type of solid matter in which a highly arranged molecule or atoms present to form a lattice that extended in all directions. It is a lightweight clear solid which is normally is colorless.

It can be cubic, hexagonal, triclinic, monoclinic, orthorhombic, tetragonal, and trigonal that are ordered arrangments. Its internal symmetry is visible to its surface.

4 0
3 years ago
Which particles do not affect the stability of the atom
lutik1710 [3]
Electrons


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5 0
3 years ago
If one bullet is dropped from a certain height and another is fired from a gun horizontally from the same height which one will
Zinaida [17]

If the ground is flat, and both bullets are released at the same time from the same height, then they both hit the ground at the same time.

The horizontal motion of the one from the gun has no effect on its vertical motion.

7 0
4 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
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