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zmey [24]
2 years ago
6

Why would the shear stress be considered as the momentum flux.

Engineering
1 answer:
oksano4ka [1.4K]2 years ago
3 0

Answer:

A fluid flowing along a flat plate will stick to it at the point of contact

Explanation:

and this is known as the no-slip condition. ... This is the precise reason why shear stress in a fluid can also be interpreted as the flux of momentum.

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Circle because it’s round and we all love round things
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3 years ago
A series of experiments is conducted in which a thin plate is subjected to biaxial tension/compression, σ1, σ2 , the plane surfa
Andrew [12]

Answer:

                                               

Explanation:

3 0
3 years ago
Troy must keep track of the amount of refrigerant he uses from a 50-pound cylinder to ensure that accurate
IgorLugansk [536]

Answer:

Amount of gas still in cylinder = 28 pound

Explanation:

Given:

Amount of gas in cylinder = 50 pound

Amount of gas used in Ms. Jones system = 13 pound

Amount of gas used in client system = 9 pound

Find:

Amount of gas still in cylinder

Computation:

Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system

Amount of gas still in cylinder = 50 - 13 - 9

Amount of gas still in cylinder = 28 pound

7 0
3 years ago
An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

first we will get here side of cross section that is express as

S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = (\frac{0.15 + x}{3})^2    m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = \frac{-k\times A\  dt}{dx}      ...............3

put here value and we get

heat conduction q = -k\times (\frac{0.15 + x}{3})^2 \   \frac{dt}{dx}  

it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

5 0
3 years ago
How can you fix a Nintendo switch if it is a blue screen
MAVERICK [17]

Answer:

This website is for answering and or asking questions, but just buy a new one or put rice in it if it went into something wet. Hope this helps!

Explanation:

be safe, good luck!

4 0
3 years ago
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