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3 years ago

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Georgia Tech is committed to creating solutions to some of the world’s most pressing challenges. Tell us how you have improved o

r hope to improve the human condition in your community.

1 answer:

Kaylis [27]3 years ago

6 0

Answer:

Georgia Tech is committed to WGAR 53566 THE ANSWER IS JELLY IS KING AND THE JELLY IS KING AND hope to improve the human condition in your community.

Explanation:

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The observations of 124.53, 124.55, 142.51, and 124.52 are obtained when taping the length of a line. What should the observer c

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A flat site is being considered for a new school that will have a steel frame and brick façade. The steel columns will have a ma

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3 years ago

g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultima

nikdorinn [45]

Answer:

A) - Yield strength before operation = 32 kpsi

- Ultimate Strength before operation = 49.5 kpsi

- Yield strength after operation = 61.854 kpsi

- Ultimate Strength after operation = 61.875 kpsi

- Percentage increase of yield strength = 93.29%

- Percentage increase of ultimate strength = 25%

B) ratio before operation = 1.55

Ratio after operation = 1

Explanation:

From online values of the properties of this material, we have;

Yield strength; S_y = 32 kpsi

Ultimate Strength; S_u = 49.5 kpsi

Modulus; m = 0.25

Percentage of cold work; W_c = 0.2

S_o = 90 kpsi

A) Let's calculate the strain(ε) from the formula;

A_o/A = 1/(1 - W_c)

A_o/A = 1/(1 - 0.2)

A_o/A = 1.25

Thus, strain is;

ε = In(A_o/A)

ε = In(1.25)

ε = 0.2231

Yield strength after the cold work operation is;

S'_y = S_o(ε)^(m)

Plugging in the relevant values;

S'_y = 90(0.2231)^(0.25)

S'_y = 61.854 kpsi

Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%

Ultimate strength after the cold work operation is;

S'_u = S_u/(1 - W_c)

S'_u = 49.5/(1 - 0.2)

S'u = 61.875 kpsi

Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%

B) Ratio of ultimate strength and yield strength before cold work operations is;

S_u/S_y = 49.5/32

S_u/S_y = 1.547

Ratio of ultimate strength and yield strength after cold work operations is;

S'_u/S'_y = 61.875/61.854 = 1

The ratio after the operation is less than before the operation, thus the ductility reduced.

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3 years ago

The combustion in a gasoline engine may be approximated by a constant volume heat addition process. There exists the air-fuel mi

ddd [48]

Answer:

the pressure at the end of the combustion is 2.68 MPa

Solution:

As per the question:

Initial Pressure, P = 0.95\ MPa

Temperature before combustion, $T = 425^{\circ}C$ = 273 + 425 = 698 K

Temperature after combustion, $T' = 1700^{\circ}C$ = 1973 K

Now,

To calculate the pressure at the end of combustion, P':

By using the Pressure-Temperature relation from Gay- Lussac's law:

$\frac{P}{T} = \frac{P'}{T'}$

$P' = \frac{P}{T}\times T'$

$P' = \frac{0.95\times 10^{6}}{698}\times 1973 = 2.68\times 10^{6}\ Pa = 2.68\ MPa$

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3 years ago

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mote1985 [20]

Answer:

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Explanation:

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