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Anna35 [415]
1 year ago
5

which of the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick

sheet of fiber-reinforced plastic (two best answers): (a) abrasive jet machining, (b) chemical milling, (c) edm, (d) laser beam machining, (e) oxyfuel cutting, (f) water jet cutting, and (g) wire edm
Engineering
1 answer:
Elena-2011 [213]1 year ago
3 0

Laser beam machining and water jet cutting are the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick sheet of fiber-reinforced plastic. Hence option d and f is correct.

<h3>What is fiber reinforced plastic?</h3>

Fiber reinforced plastic is defined as a polymer-based composite material that is strengthened by fiber support. Fiber-reinforced plastics (FRP) are composite materials that use glass or carbon fibers as reinforcement and polymer resins as a matrix.

Because laser cutting is a far less aggressive and abrasive process than water jet cutting, it is much more accurate. A laser can carefully and safely cut materials as thin as 0.006 inches, whereas water jet cutters can't handle cutting through surfaces smaller than 0.02 inches.

Thus, laser beam machining and water jet cutting are the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick sheet of fiber-reinforced plastic. Hence option d and f is correct.

To learn more about fiber-reinforced plastic, refer to the link below:

brainly.com/question/11941367

#SPJ1

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The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had
kipiarov [429]

Answer:

  • Moisture/ water content w = 26%
  • Void ratio , e =  0.73

Explanation:

  • Initial mass of saturated soil w1 = mass of soil - weight of container

                                                 = 113.27 g - 49.31 g = 63.96 g

  • Final mass of soil after oven w2 = mass of soil - weight of container

                                                  = 100.06 g - 49.31 g = 50.75

Moisture /water content, w =   \frac{w1-w2}{w2} = \frac{63.96-50.75}{50.75} = 0.26 = 26%

Void ratio =  water content X specific gravity of solid

                  = 0.26 X 2.80 =0.728

5 0
3 years ago
Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass
snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0
3 years ago
A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s
Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})

where

'H' is the initial depth of the layer

C_c is the Compression index

e_o is the inital void ratio

\bar{\sigma_o} is the initial effective stress at the depth

\Delta \bar{\sigma_o} is the change in the effective stress at the given depth

Applying the given values we get

\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04

3 0
3 years ago
The __________________ refers to the main screen of the computer.
Sedbober [7]

Answer:

<em>D</em><em>e</em><em>s</em><em>k</em><em>t</em><em>o</em><em>p</em>

Explain:

<em>Desktop refers to the main screen of the computer. It is the first screen you see after logging in. The desktop’s appearance can vary widely because it is highly customizable, but generally desktops will feature a large image, icons, and a taskbar(covered later on this page).</em>

4 0
3 years ago
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