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ivann1987 [24]
2 years ago
11

Question 54 (1 point)

Engineering
2 answers:
Rudiy272 years ago
6 0
The answer should be Oc 15
Sergeeva-Olga [200]2 years ago
6 0
Oa 8 should be the answer of the moisture of lumber frame for the house
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Oksi-84 [34.3K]

Answer: g

Explanation:

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____ grinders are used to grind diameters, shoulders, and faces much like the lathe is used for turning, facing, and boring oper
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Answer:

Cylindrical

Explanation:

<em>A cylindrical grinder </em><em>is a tool for shaping the exterior of an item. Although cylindrical grinders may produce a wide range of forms, the item must have a central axis of rotation. Shapes such as cylinders, ellipses, cams, and crankshafts are examples of this.</em><em> Cylindrical grinding</em><em> machines are specialized grinding machines that are used to process cylinders, rods, and similar workpieces. The cylinders revolve in one direction between two centers, while the grinding wheel or wheels are close together and rotate in the other direction.</em>

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1 year ago
Pointssss 100 and brainliest :)
Delvig [45]

Answer:

thank you for the free point have a great rest of your day

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2 years ago
james wants to qualify for icp are and licensure. Which degree would be required in order to qualify for a two year master of ar
ololo11 [35]

Answer:

A degree in architecture with 60 credit hours.

Explanation:

The requirements need for a student to qualify for a two year master of architecture degree are;

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4 0
3 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
2 years ago
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