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igor_vitrenko [27]
3 years ago
11

A copper wire of original diameter .80 m exhibits a maximum tensile load/ strength at an engineering stress= 248.2 mpa. its duct

ility is measured as 75 reduction of area. Determine the true strain at failure.
Engineering
1 answer:
ddd [48]3 years ago
4 0

Answer:

True Strain at failure = 1.386

Explanation:

For the question, ductility is given in reduction In the area to be 0.75

Let the initial Cross sectional Area of wire be Ao

And the final cross sectional Area of wire of cross sectional Area of wire at fracture be A

(Ao - A)/Ao = ductility = 0.75

Ao - A = 0.75Ao

A = Ao - 0.75Ao

A = 0.25 Ao

True Strain = In (Lf/Lo)

To obtain the ratio of the lengths of wire,

The volume of the wire stays constant, that is, Vo = Vf; Vo = Ao × Lo and Vf = A × Lf

AoLo = ALf

Lf/Lo = Ao/A

In (Lf/Lo) = In (Ao/A) = In (Ao/0.25Ao) = In 4 = 1.386

True Strain = 1.386

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creativ13 [48]

Answer:

Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.

Explanation: lol

8 0
4 years ago
1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-
Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

  = 0.0214 * 3.738

  = 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0
3 years ago
The importance of reading a circuit diagram to interpret a wiring diagram?
Nataly [62]

Answer:

The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.

Explanation:

7 0
3 years ago
Which of the following is true about the n-way analysis of variance (ANOVA)?
Ne4ueva [31]

Answer:

It is a type of ANOVA that can analyze several independent variables at the same time.

Explanation:

This is the statement that correctly describes the n-way analysis of variance (ANOVA). ANOVA is a type of analysis of variance that can analyze several independent variables at the same time. In this type of analysis, a dependent variable is measured by different levels of independent variables. When the results are obtained, these are assumed to be the consequence of the different levels of the independent variables, plus random error. The computation necessary for this analysis can be done in most types of statistical software.

7 0
3 years ago
The mass fractions of total ferrite and total cementite in an iron–carbon alloy are 0.91 and 0.09, respectively. Is this a hypoe
WINSTONCH [101]

Answer:

hypoeutectoid

Explanation:

ferrite:  pure form of iron

cementite: It is iron carbide with 93.3% iron and 6.67% carbon

hypoeutectoid: Eutectoid steel with carbon fraction less than 0.8%

hypereutectoid: Eutectoid Steel with carbon content more than 0.8%

For the mentioned iron-carbide alloy,

% of carbon in iron-carbide alloy= percentage of cementite × percentage of carbon in cementite

% of carbon in iron-carbide alloy= 0.09× 0.0667

                                                     = 0.6%

so the alloy is hypoeutectoid

4 0
4 years ago
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