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3 years ago

Is it possible that two types of dislocation coexist. a)-True b)- False

Engineering

1 answer:

Svetlanka [38]3 years ago

3 0

Answer:

yes it is possible

Explanation:

dislocation are if two type edge and screw dislocations

edge dislocation is a defect where an extra half plane is inside the lattice.

and screw dislocation is one in which can be assumed as the first half of the crystal slips over another.

These dislocation can coexist together where the line direction and burger vectors are neither parallel nor perpendicular then at that condition both dislocation screw and edge will coexist

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Question Set 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.2.2 In

Gnom [1K]

Answer:

# Program is written in python

# 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.

# initializing string

Stringtocheck = "mississippi"

# using count() to get count of s

counter = Stringtocheck.count('s')

# printing result

print ("Count of s is : " + str(counter))

# 2.2 In the string 'mississippi', replace all occurrences of the substring 'iss' with 'ox

# Here, we'll make use of replace() method

# Prints the string by replacing iss by ox

print(Stringtocheck.replace("iss", "ox"))

#2.3 Find the index of the first occurrence of 'p' in 'mississippi'

# declare substring

substring = 'p'

# Find index

index = Stringtocheck.find(substring)

# Print index

print(index)

# End of program

8 0

3 years ago

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is

Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α = $\frac{1}{\pi} (\frac{K}{\sigma Y})^2$ ....................1

put here value we get

α = $\frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2$

α = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0

3 years ago

Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the

Musya8 [376]

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$$P(h$

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore, $x_1=368 \ veh/h$

$=\frac{368}{1000} = 0.368$

Given, $t_1=2+x_1$

= 2 + 0.368

= 2.368 min

At user equilibrium, $t_2=t_1$

∴ $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

= 1368 veh/h

7 0

3 years ago

PythonA group of statisticians at a local college has asked you to create a set of functionsthat compute the median and mode of

skelet666 [1.2K]

Answer:

def median(l):
if(len(l) == 0):
return 0
else:
l.sort()
if(len(l)%2 == 0):
index = int(len(l)/2)
mid = (l[index-1] + l[index]) / 2
else:
mid = l[len(l)//2]
return mid
def mode(l):
if(len(l)==0):
return 0
mode = max(set(l), key=l.count)
return mode
def mean(l):
if(len(l)==0):
return 0
sum = 0
for x in l:
sum += x
mean = sum / len(l)
return mean
lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
print(mean(lst))
print(median(lst))
print(mode(lst))

Explanation:

Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).

In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.

In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).

In the main program, we test the three functions using a sample list and we shall get

20.5

12.5

3 0

3 years ago

Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t

Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

%This brings an end to the program

4 0

3 years ago