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kvasek [131]
4 years ago
15

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

utectoid or hypereutectoid alloy? Why?
Engineering
1 answer:
WINSTONCH [101]4 years ago
4 0

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

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bagirrra123 [75]

Answer:

Explanation:

Step by step solved solution is given in the attached document.

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3 years ago
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Vanyuwa [196]
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5 0
3 years ago
Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
3 years ago
Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at
makkiz [27]

Answer:

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

Explanation:

The Young's module is:

E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }

E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}

Let assume that both specimens have the same geometry and load rate. Then:

E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}

The displacement rate for steel is:

\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}

\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

7 0
3 years ago
Read 2 more answers
A 30 mm thick AISI 1020 steel plate is sandwiched between two 10 mm thick 2024-T3 aluminum plates and compressed with a bolt and
denis-greek [22]

Answer:

275 MPa

Explanation:

Regardless of what it is holding, the stiffness of a bolt depends on its own material properties and geometry.

The stiffness is:

k = E * \frac{A}{l}

I assume this one is made of steel, because regular bolts are steel.

The Young's modulus for steel is E = 210 GPa

The longitude is given. (But note that in a real application you have to consider the length up to the nut.)

The section is (using the nominal diameter of 10 mm)

A = \frac{\pi * d^2}{4} = \frac{\pi * 0.01^2}{4} = 7.85e-5 m^2

Then:

k  = 2.1e11 * \frac{7.85e-5}{0.06} = 275e6 Pa = 275 MPa

5 0
3 years ago
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