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erastovalidia [21]
3 years ago
5

Question in image. Question from OSHA.

Engineering
2 answers:
Maru [420]3 years ago
4 0

Answer:

welding screens

Explanation:

WITCHER [35]3 years ago
4 0
Welding screens and protective clothing seems to be the most logical answers.
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Find the rate of heat transfer through a 6 mm thick glass window with a cross-sectional area of 0.8 m2 if the inside temperature
kiruha [24]

Answer:

6.9

Explanation:

I had the same question lol your welcomr if itd not right in sorry

3 0
2 years ago
Motor oil is responsible for
Lelechka [254]

Answer:

lubricating all moving parts in the engine

Explanation:

like the pistons, pushrods, and the crank

5 0
2 years ago
Select four examples of fluid or pneumatic power systems.
Dennis_Churaev [7]

Answer:

Students learn about the fundamental concepts important to fluid power, which includes both pneumatic (gas) and hydraulic (liquid) systems. Both systems contain four basic components: reservoir/receiver, pump/compressor, valve, cylinder.

Explanation:

8 0
3 years ago
Read 2 more answers
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using
kow [346]

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

4 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40MPa. It has been dete
Nataly [62]

Answer:

Yes, fracture will occur

Explanation:

Half length of internal crack will be 4mm/2=2mm=0.002m

To find the dimensionless parameter, we use critical stress crack propagation equation

\sigma_c=\frac {K}{Y \sqrt {a\pi}} and making Y the subject

Y=\frac {K}{\sigma_c \sqrt {a\pi}}

Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, \sigma_c  is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain

Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682

When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m

\sigma_c=\frac {K}{Y \sqrt {a\pi}}  and making K the subject

K=\sigma_c Y \sqrt {a\pi}  and substituting 260 MPa for \sigma_c  while a is taken as 0.003m and Y is already known

K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa

Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material

6 0
3 years ago
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