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irina [24]
3 years ago
8

When economist determine that a nations GDP has declined, they can point to this as a sign of what?

Engineering
1 answer:
Alex73 [517]3 years ago
3 0
When economists determine that a nation's GDP has declined, they can point to this as a sign of economic shrinkage.
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1. In a series circuit the sum of all voltage drops produced by the loads in the circuit
GarryVolchara [31]

Answer:

ik the answer but I'm busy rn

5 0
3 years ago
Are there engineering students here?​
leva [86]
Uh, I’d assume so because Brainly has a whole section of questions for them.
7 0
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Design a PLC ladder logic program to control the operation of a conveyor-storage system using the following sequence: - 1. Progr
Phantasy [73]

Answer:

See explaination

Explanation:

Kindly check attachment for the step by step solution of the given problem.

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3 years ago
The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is
nexus9112 [7]

Answer:

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa.

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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