2 years ago

Hi gospelgamer10 lol

Engineering

2 answers:

Katyanochek1 [597]2 years ago

6 0

Answer:

I’m not that person but

Explanation:

thamks for the pointzzz

brainliest too plz??? <3<3<3

Kaylis [27]2 years ago

5 0

Answer:

yo not that person either but points here bye

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Find the value of P(-1.5≤Z≤2)

Vladimir [108]

Answer:

0.9104

Explanation:

Suitable technology can tell you the probability.

P(-1.5≤Z≤2) ≈ 0.9104

A phone app gives the probability as 0.9104426667829628.

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2 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

IrinaVladis [17]

Answer:

The original length of the specimen $l_{o} = 104.7 mm$

Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

Original length

$l_{o} = \frac{105.2}{1.00476}$

$l_{o} = 104.7 mm$

This is the original length of the specimen.

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2 years ago

Andreyy89

Answer:

Explanation:

5 + 2 = 7

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Answer:

Mechanical engineer? Thats my guess I didnt have alot of options sorry if I am wrong

Explanation:

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