Question

Question 1 1 pts nucleation rates as a result of diffusion Small supercooling leads to processes. rapid, lower slow, lower OO
Question 3 1 pts The critical radius of a nucleus determines when the nucleus of the solid phase will be stable enough to con
Question 4 1 pts Select all solid phases that can be transformed into other solid phases during the cooling process within a
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Question 1 1 pts nucleation rates as a result of diffusion Small supercooling leads to processes rapid, lower slow, lower rapThe critical radius of a nucleus determines when the nucleus of the solid phase will be stable enough to continue to grow. Tr

The correct answers are marked with green. I am pretty sure of these answers. Please dear thumpsup for the answer. Thank you.

Add a comment
Know the answer?
Add Answer to:
Question 1 1 pts nucleation rates as a result of diffusion Small supercooling leads to processes....
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • material Question 3: For an Iron-Iron Carbide (Fe-FesC) phase diagram, answer the following for 5.0 wt%...

    material Question 3: For an Iron-Iron Carbide (Fe-FesC) phase diagram, answer the following for 5.0 wt% C 10 Points How many phases are possible at this composition and what will be the melting temperature at this a) composition? b) the alloy is hypocutectic or hypereutectoid at 800°C at this composition ? c) the alloy is hypocutectic or hypereutectoid at 600°C at this composition? d) composition and amount of each phase at 1250°C at this composition e) composition and amount of...

  • (12 Question 4: points: 4 pts each) Consider 3.0 kg of austenite containing 1.6 wt% C...

    (12 Question 4: points: 4 pts each) Consider 3.0 kg of austenite containing 1.6 wt% C at 1100 °C, cooled to below 726 °C. (a) What is the proeutectoid phase? (b) How many kilograms of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid cementite form? Composition (at% C) 10 15 5 20 25 0 1600 1538°C 1493°C L 1400 S 1394°C 7+L 1200 1147°C 2.14 4.30 y, Austenite Temperature (°C) 1000 912°C y...

  • please answer question 2,3,4 ​​​​​​​ A critical feature of steel is that a considerable amount of...

    please answer question 2,3,4 ​​​​​​​ A critical feature of steel is that a considerable amount of carbon can be dissolved in the austenite, phase, (up to 2.14 w/o at 1147"C), whereas carbon is essentially insoluble in ferrite, Cooling from point d to e, just above the eutectoid but still in the an increased fraction of the +y region, will produce phase and a microstructure similar to that shown: the particles precinitates out in the form af an intermetallic.compound called iron-carbide...

  • Question 3 (1) Sketch the microstructures (at temperature just below eutectold temperature) of slow-cooled pure Iron...

    Question 3 (1) Sketch the microstructures (at temperature just below eutectold temperature) of slow-cooled pure Iron (Owt%C) and steels with 0.3wt%C, and 0.76wt%C, and explain how the structures develop on cooling from 1100°C to room temperature (Figure 3.1 for Fe-Fe,c phase diagram, see below). (18marks] 1600 Composition Cat C) 10 is 20 25 1538"C 149346 1400 1394°C 2500 72 1200 1147°C > Austenite 2.14 4.30 Temperature) 2000 1000 Temperature (°F) 912°C y + Fe3c 800 1500 0.76 727°C 600 0.022...

  • Question 1: (16 points: 4 pts each) A 45 wt% Pb - 55 wt% Mg alloy...

    Question 1: (16 points: 4 pts each) A 45 wt% Pb - 55 wt% Mg alloy is slowly cooled from 700°C to 400 °C. (a) At what temperature does the first solid phase form? (b) What is the composition of this solid phase? c) At what temperature does the liquid solidify? (d) What is the composition of this last remaining liquid phase? Composition (at% Pb) 10 20 30 40 70 100 700 L 600 L + MezPo M 500 400...

  • Thermodynamics example information to help solve each question CALCULATIONS Question 1. Write the complete exp...

    Thermodynamics example information to help solve each question CALCULATIONS Question 1. Write the complete expression for the free energy change of the L = V equilibrium reaction for pure water (equation 2) at 100 bars and 300-C using the following data: Gy = -210330 joules/mole GL = -210330 joules/mole Recall that the right side (water vapor) of the reaction is the product side and left side (water liquid) of the reaction is the reactant side, Recall that we always calculate...

  • Question 70 2.5 pts In the unemployment rate, part-time workers are: not included in the labor...

    Question 70 2.5 pts In the unemployment rate, part-time workers are: not included in the labor force. included in the labor force, but counted as unemployed. treated the same way as discouraged workers. included in the labor force and counted as employed. Question 69 2.5 pts Which of the following would be officially classified as unemployed? O a school administrator who has been working as a substitute teacher one day per week while looking for a full-time job in administration...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT