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hjlf
3 years ago
13

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Engineering
1 answer:
Andreyy893 years ago
4 0
I don’t know but good luck
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Afina-wow [57]

This would be answer choice D: all of the above. Low back pain can be classified by duration as acute (pain lasting less than 6 weeks), sub-chronic (6 to 12 weeks), or chronic (more than 12 weeks).

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3 years ago
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a
Fudgin [204]

Answer:

5.21e-2mm

Explanation:

Please see attachment

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4 years ago
Technician A says that the enable criteria are the criteria that must be met before the PCM completes a monitor test. Technician
DENIUS [597]

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"A"

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3 years ago
40 = 6(X^2) / 4 (tan(180/6)) solve for X
const2013 [10]

Answer:

x = 3.92

Explanation:

The question is given as ;

40 = 6x² / 4 tan { 180/6}

40 = 6x² / 4 tan {30°}

40 = 6x² / 2.309

40 * 2.309 = 6x²

92.38 = 6 x²

92.38/6 = x²

15.40 = x²

√15.40 = x

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3 0
3 years ago
Shear plane angle and shear strain: In an orthogonal cutting operation, the tool has a rake angle = 16°. The chip thickness befo
Oduvanchick [21]

Answer:

shear plane angle Ф = 26.28°

shear strain 2.20

Explanation:

given data

angle = 16°

chip thickness t1 = 0.32 mm

cut yields chip thickness t2 = 0.72 mm

solution

we get here first chip thickness ratio that is

chip thickness ratio = \frac{t1}{t2}    ................. 1

put here value

chip thickness ratio  = \frac{0.32}{0.72}  

chip thickness ratio r = 0.45

so here shear angle will be Ф

tan Ф = \frac{r*cos\alpha }{1-rsin\alpha}   ............2

tan Ф = \frac{0.45*cos16 }{1-rsin16}  

tan Ф = 0.4938

Ф = 26.28°

and

now we get shear strain that is

shear strain r = cot Ф + tan (Ф - α )   ................3

shear strain r  = cot(26.28) + tan (26.28 - 16 )

shear strain r = 2.20

6 0
4 years ago
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