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Answer:

common fate

Explanation:

The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.

Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.

Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.

Answer:

pascals is this = 101352.972 Pa

Explanation:

given data

Atmosphere pressure at sea level = 14.7 lb/in²

we convert 1 Pa = 1 N/m² to lb/ft²

so we convert here  14.7 lb/in² to pascals

we know that 1 lb/ft² = 47.990172 N/m²

so

1 lb/ft² × ft²/(12in)²  = 47.990172  ×  144 N/m²

it will be simplyfy

1 lb/ft²  = 6894.76 N/m²  

so

14.7 lb/in² = 14.7 × 6894.76 N/m²  

14.7 lb/in² =  101352.972 Pa

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)

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:

[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)

Where:

[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.

[tex]D[/tex] - Diameter, measured in meters.

If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:

[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]

[tex]v \approx 5.093\,\frac{m}{s}[/tex]

The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:

[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)

If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:

[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]

[tex]Re = 211.230[/tex]

A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:

[tex]f = \frac{64}{Re}[/tex]

If we get that  [tex]Re = 211.230[/tex], then the friction factor is:

[tex]f = \frac{64}{211.230}[/tex]

[tex]f = 0.303[/tex]

The friction factor is 0.303.

The image of the load applied to the polystyrene is missing, so i have attached it.

Answer:

a_new = 2.00302 in

b_new = 2.00552

Explanation:

From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.

Now, formula for stress is;

Stress(σ) = Force/Area

We are not given force and area but the load and plate thickness.

Thus, stress = load/thickness

We are given;

Load in x - direction = 500 lb/in.

Load in y - direction = 350 lb/in.

Thickness; t = 0.25 in

Thus;

σ_x = 500/0.25

σ_x = 2000 ksi

σ_y = 350/0.25

σ_y = 1400 ksi

From Hooke's law for 2 dimensions, strain is given by the formula;

ε_x = (1/E)(σ_x - vσ_y)

ε_y = (1/E)(σ_y - vσ_x)

We are given v_p = 0.25 and Ep = 597 × 10³ psi

Thus;

ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)

ε_x = 0.00276

ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)

ε_y = 0.00151

From elongation formula, we know that;

Startin is: ε = ΔL/L

Thus; ΔL = Lε

We are given a = 2 and b = 2

Thus;

ΔL_x = 2 × 0.00276

ΔL_x = 0.00552

ΔL_y = 2 × 0.00151

ΔL_y = 0.00302

New dimensions are;

a_new = 2 + 0.00302

a_new = 2.00302 in

b_new = 2 + 0.00552

b_new = 2.00552

Calculate the normal stress along the x-direction.

[tex]\to \sigma_x =\frac{500}{t}=\frac{500}{0.25}= 2000 \ \frac{lb}{in^2}\\\\[/tex]

Calculate the normal stress along the y-direction.

[tex]\to \sigma_y =\frac{350}{t}=\frac{350}{0.25}= 1400 \ \frac{lb}{in^2}[/tex]

Calculate the strain along the x-direction.

[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_y-\sigma_z)]\\\\[/tex]

        [tex]=\frac{1}{597\times 10^3} [2000-0.25(1400+0)] \\\\= 2.764\times 10^{-3}\\\\[/tex]

Calculate the strain along the y-direction.

[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_x+\sigma_z)][/tex]

        [tex]=\frac{1}{597\times 10^3}[1400 -0.25 (2000+0)]\\\\=1508\times 10^{-3}[/tex]

Calculate the strain along the z-direction.

[tex]\to \varepsilon_x= \frac{1}{E}[\sigma_z-v(\sigma_x +\sigma_y)][/tex]

        [tex]=\frac{1}{597\times 10^3} [0-0.25(2000+1400)] \\\\=-1.424\times 10^{-3}[/tex]

Calculate the new dimensions.

[tex]\to b' =b+ \varepsilon_xb\\\\[/tex]

       [tex]= 2+2.764\times 10^{-3} \times 2\\\\= 2.005528 \ in\\\\[/tex]

[tex]\to a' = a + \varepsilon_y a\\\\[/tex]

        [tex]= 2+1.508\times 10^{-3} \times 2\\\\= 2.003016\ in\\\\[/tex]

[tex]\to c'= c + \varepsilon_x c\\\\[/tex]

        [tex]= 0.25 +(-1.424\times 10^{-3}) \times 0.25\\\\= 0.249644\ in\\\\[/tex]

Find out more about the dimensions here:

brainly.com/question/15406884

This question is incomplete, the complete question is;

Find the magnitude of the steady-state response of the system whose system model is given by

dx(t)/dt + x(t) = f(t)

where f(t) = 2cos8t.  Keep 3 significant figures

Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )

Explanation:

Given that;

dx(t)/dt + x(t) = f(t)  where f(t) = 2cos8t

dx(t)/dt + x(t) = f(t)

we apply Laplace transformation on both sides

SX(s) + x(s) = f(s)

(S + 1)x(s) = f(s)

f(s) / x(s) = S + 1

x(s) / f(s) = 1 / (S + 1)

Therefore

transfer function = H(s) = x(s)/f(s) = 1/(S+1)

f(t) = 2cos8t →   [ 1 / ( S + 1 ) ]   →  x(t) = Acos(8t - ∅ )

A = Magnitude of steady state output

S = jw

S = j8

so

A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )

A = 2/√65 = 0.2481

∅ = tan⁻¹( 1/1) = 45°

therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )

Answer: the discharge observed at a v-notch weir is 66.7 in³/s

Explanation:

Given that;

Notch angle ∅ = 90°

height above the weir is 3 inches { head + head correction factor) h + k = 3 in

Discharge Q = ?

To determine the discharge observed, we us the following expression

Q = 4.28Ctan(∅/2) ( h + k )^5/2

where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor

now we substitute

Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2

Q = 4.28 × 1 × 1 × 15.5884

Q =  66.7 in³/s

Therefore the discharge observed at a v-notch weir is 66.7 in³/s

Answer is given below:

Explanation: