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GalinKa [24]
2 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]2 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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Find the magnitude: || 5-3i || ...?
____ [38]
The magnitude would be :

\sqrt{5^2 + 3^2}

= √34

Hope this helps
5 0
2 years ago
To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

L= 1.5*10^{-2}H

R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

i= 0.0133A

i_1 = 0.01052A

3 0
2 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
3 years ago
How can water boil without heat?​
kaheart [24]

Answer:

Put water at room temperature into a vacuum chamber and begin removing the air. Eventually, the boiling temperature will fall below the water temperature and boiling will begin without heating. Or if you want to be easy but messy, add dry ice to a bowl of water and watch how the water starts to boil.

3 0
2 years ago
Một mẫu vật liệu có khối lượng riêng là 2500kg/m3
Leokris [45]

Explanation:

khoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO làkhoâng cách MO là

6 0
2 years ago
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