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kvasek [131]
3 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]3 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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Answer:

I know ship has stream line body

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U 4. Find 2 bridges in the US and answer the following:
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Im guessing this is for CEA for PLTW, if so look up the exact assignment number and look at online examples of the exact same assignment.

Explanation:

6 0
3 years ago
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest
sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

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Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

3 0
3 years ago
How would I get this python code to run correctly? it's not working.​
Elanso [62]

Answer:

See Explanation

Explanation:

This question requires more information because you didn't state what the program is expected to do.

Reading through the program, I can assume that you want to first print all movie titles then prompt the user for a number and then print the move title in that location.

So, the correction is as follows:

Rewrite the first line of the program i.e. movie_titles = ["The grinch......."]

for each_movie_titles in movie_titles:

   print(each_movie_titles)

   

usernum = int(input("Number between 0 and 9 [inclusive]: "))

if usernum<10 and usernum>=0:

   print(movie_titles[usernum])

Line by line explanation

This iterates through the movie_titles list

for each_movie_titles in movie_titles:

This prints each movie title

   print(each_movie_titles)

This prompts the user for a number between 0 and 9    

usernum = int(input("Number between 0 and 9 [inclusive]: "))

This checks for valid range

if usernum<10 and usernum>=0:

If number is valid, this prints the movie title in that location

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6 0
3 years ago
A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
Marrrta [24]

Answer:

a) Ef = 0.755

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c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0
2 years ago
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