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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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AleksandrR [38]

Explanation:

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I total = Vt/Rt

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Ix= 600 mA

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6 0
3 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

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solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
The error in the trapezoidal rule is proportional to h^2 True b. False
Lynna [10]

Answer:

b) False

Explanation:

in trapezoidal rule the error is proportional to h^3 and the order of accuracy is proportional to h^3 .

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4 0
3 years ago
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Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

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mass of the car = 2000 kg

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F c = 2666. 67 N

F f= μ N

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μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

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Evgesh-ka [11]

Answer:

Explanation:

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For the step by step solution to the question you asked, go through the attached documents.

4 0
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