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zhannawk [14.2K]
3 years ago
7

Explain what the engineering team should advise in the following scenario.

Engineering
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer: its c

Explanation:

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Match the terms with the correct definitions.
arsen [322]
I think answer should be the first one please give me brainlest let me know if it’s correct or not okay thanks bye
7 0
3 years ago
a coil consists of 200 turns of copper wire and has a cross-sectional area of 0.8mm square . The mean length per turn is 80 cm a
Amanda [17]

Answer:

The picture below with the answer. Hope it helps, have a great day/night and stay safe! Length of the coil,

8 0
3 years ago
Find: factor of safety (n)for point A and B by using both MSS and DE (you can neglect shear stress due to shear force and also n
gladu [14]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Explanation:

Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Attached below is the detailed solution

8 0
3 years ago
have you ever heard the myth that a penny dropped off the empire state building can be dangerous? the penny would be traveling v
Yuri [45]

Answer: yes

Explanation: ontop of a tall building, you drop a small peace of metal covered in zinc. it is possible to be very dangerus because of gravity. some one walking on the side walk who gets hit in the head can get a concusion maybe even a brain injury.

7 0
3 years ago
Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to
7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

    = 380*0.669*0.93

Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0
3 years ago
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