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3 years ago

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Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed

Mold c)- Preforming d)- Postforming e)- None of the above f)- All the above

1 answer:

Novay_Z [31]3 years ago

4 0

Answer:

it is f all of the above

Explanation:

let me know if im right

im not positive if im right but i should be right

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A 11-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in

Reika [66]

Answer:

F = 8552.7N

Explanation:

We need first our values, that are,

$V_{jet} = 40m/s\\V_{Plate} = 10m/s \\D = 11cm$

We start to calculate the relative velocity, that is,

$V_r = V_{jet}-V_{plate}\\V_r = (40)-(10)\\V_r = 30m/s$

With the relative velocity we can calculate the mass flow rate, given by,

$\dot{m}_r = \rho A V_r$

$\dot{m}_r = (1000)(30) \frac{\pi (0.11)^2}{4}$

$\dot{m}_r = 285.09kg/s$

We need to define the Force in the direction of the flow,

$\sum\vec{F} = \sum_{out} \beta\dot{m}\vec{V} - \sum_{in} \beta\dot{m} \vec{V}\\$

$F = \dot{m}V_r$

$F = (285.09Kg/s)(30)$

$F = 8552.7N$

8 0

3 years ago

Which of the following are all desirable properties of a hydraulic fluid? a. good heat transfer capability, low viscosity, high

Vinvika [58]

Answer:

e.Fire resistance,Inexpensive,Non-toxic.

Explanation:

Desirable hydraulic property of fluid as follows

1. Good chemical and environment stability

2. Low density

3. Ideal viscosity

4. Fire resistance

5. Better heat dissipation

6. Low flammability

7. Good lubrication capability

8. Low volatility

9. Foam resistance

10. Non-toxic

11. Inexpensive

12. Demulsibility

13. Incompressibility

So our option e is right.

5 0

3 years ago

A structural component in the shape of a flat plate 25.0 mm thick is to be fabricated from a metal alloy for which the yield str

balandron [24]

Answer:

The critical length of surface flaw = 6.176 mm

Explanation:

Given data-

Plane strain fracture toughness Kc = 29.6 MPa-m1/2

Yield Strength = 545 MPa

Design stress. =0.3 × yield strength

= 0.3 × 545

= 163.5 MPa

Dimensionless parameter. Y = 1.3

The critical length of surface flaw is given by

= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2

Now putting values in above equation we get,

= 1/3.14( 29.6 / 1.3 × 163.5)^2

=6.176 × 10^-3 m

=6.176 mm

5 0

3 years ago

Read 2 more answers

Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

40 = X

40 x 100 / 120 = X

4000 / 120 = X

33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

133.333 = X

4 0

3 years ago

Miller Indices:

svetlana [45]

Answer:

A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].

B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.

C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.

Compleated question:

1. Miller Indices:

a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.

b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?

c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.

Explanation:

A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]

2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

<u>In this case, for [1 2 1]:</u>

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=1=1/c_0 \rightarrow c_0=1$

<u>for </u> $[1 2 \overline{4}]$ <u>:</u>

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25$

B) The closest distance between planes with the same Miller indices can be calculated as:

With $\pi:[l m n]$ ,the distance is $d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}}$ with lattice constant a.

<u>In this case, for [1 2 1]:</u>

<u /> $d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a$ <u />

<u>for </u> $[1 2 \overline{4}]$ <u>:</u>

$d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a$

C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:

dir₁=|0 0 1|

dir₂=|0 0.5 1.5|≡|0 1 3|

dir₃=|0 1 1|

dir₄=|0 1.5 0.5|≡|0 3 1|

dir₅=|0 0 1|

5 0

3 years ago