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LekaFEV [45]
3 years ago
11

Investigating how slime molds reproduce is an example of applied research. True False

Engineering
1 answer:
azamat3 years ago
3 0

Answer:

false

Explanation:

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Consider a dip-coating process where a very long (assume infinitely long) wire(solid) with radius, ri, is being pulled verticall
Gekata [30.6K]

Answer:

See explaination and attachment.

Explanation:

Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.

The starting point of the Navier-Stokes equations is the equilibrium equation.

The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.

The second step is to relate the deviatoric stress to viscosity in the fluid.

The final step is to impose any special cases of interest, usually incompressibility.

Please kindly check attachment for step by step solution.

6 0
3 years ago
Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected e
Lady_Fox [76]

Answer:

a) 19 or select the closest answer

b) 5%

Explanation:

a)

from the voltage divide rule

V_{in} = V_0 * \frac{R_2}{R_2 + R_f}

\frac{V_0}{V_{in}} = \frac{R_2 + R_f}{R_2} => 1 + \frac{R_f}{R_2} = 20

\frac{R_f}{R_2} = 19

Select the nearest answer

b)

obtained gain = \frac{V_0}{V_{in}} = 1 + \frac{36}{2} = 19

Expected gain = \frac{V_0}{V_{in}} = 20

∴ error = |\frac{19 - 20}{20}| × 100

            = 1/20 × 100            

            = 5%

6 0
3 years ago
You have measured the force of a hammer hitting a nail by using a new sensor. The result is 60 million dynes is this reasonable
Vesnalui [34]

Answer:

Yes, it is Reasonable.

Explanation:

The problem states that by using a new sensor, the result is 60 million dynes.

Converting the force into Newtons, by using the formula,

1 newton = 10⁵ dynes,

i.e, 1 million dynes = 10 newtons.

we get,

60 million dynes = 60 x 10 newtons = 600 newtons.

The result using a new sensor is reasonable because atmosphere exerts a force of about 10 newtons on us and 600 newtons is just 60 times of it.

Thus 600 N can be applied by the hammer on the nail and it is reasonable.

8 0
3 years ago
Which structures are the horizontal slabs that divide a building into different levels? A. beams B. roofs C. floors D. lintels E
lyudmila [28]

Answer: C. Floors

Explanation:

4 0
3 years ago
A layer of viscous fluid of constant thickness (no velocity perpenducilar to plate) flows steadily down an infinite, inclined pl
mixer [17]

Answer:

q = (ρg/μ)(sin θ)(h³/3)

Explanation:

I've attached an image of a figure showing the coordinate system.

In this system: the velocity components v and w are equal to zero.

From continuity equation, we know that δu/δx = 0

Now,from the x-component of the navier stokes equation, we have;

-δp/δx + ρg(sin θ) + μ(δ²u/δy²) = 0 - - - - - (eq1)

Due to the fact that we have a free surface, it means we will not have a pressure gradient in the x-component and so δp/δx = 0

Then our eq 1 is now;

ρg(sin θ) + μ(δ²u/δy²) = 0

μ(δ²u/δy²) = -ρg(sin θ)

Divide both sides by μ to get;

(δ²u/δy²) = -(ρg/μ)(sin θ)

Integrating both sides gives;

δu/δy = -(ρg/μ)(sin θ)y + b1 - - - - (eq2)

Now, the shear stress is given by the formula;

τ_yx = μ[δu/δy + δv/δx]

From the diagram, at the free surface,τ_yx = 0 and y = h

This means that δu/δy = 0

Thus, putting 0 for δu/δy in eq 2, we have;

0 = -(ρg/μ)(sin θ)h + b1

b1 = h(ρg/μ)(sin θ)

So, eq 2 is now;

δu/δy = -(ρg/μ)(sin θ)y + h(ρg/μ)(sin θ)

Integrating both sides gives;

u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y + b2 - - - eq3

Because u = 0 when y = 0, it means that b2 = 0 also because when we plug 0 for u and y into eq3, we will get b2 = 0.

Thus, we now have:

u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y

Factorizing like terms, we have;

u = (ρg/μ)(sin θ)[hy - y²/2] - - - (eq 4)

The flow rate per unit width is gotten by Integrating eq 4 between the boundaries of h and 0 to give;

∫u = (h,0)∫(ρg/μ)(sin θ)[hy - y²/2]

q = (ρg/μ)(sin θ)[hy²/2 - y³/6] between h and 0

q = (ρg/μ)(sin θ)[h³/2 - h³/6]

q = (ρg/μ)(sin θ)(h³/3)

7 0
3 years ago
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