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Answer:
LAOD = 6669.86 N
Explanation:
Given data:
width
thickness 
crack length 2c = 0.5 mm at centre of specimen
stress intensity factor = k will be
we know that
[c =0.5/2 = 2.5*10^{-4}]
K = 0.1724 Mpa m^{1/2} for 1000 load
if
then load will be
LAOD = 6669.86 N
Answer:
The solution code is written in Python:
- def convertCSV(number_list):
- str_list = []
- for num in number_list:
- str_list.append(str(num))
-
- return ",".join(str_list)
- result = convertCSV([22,33,44])
- print(result)
Explanation:
Firstly, create a function "convertCSV" with one parameter "number_list". (Line 1)
Next, create an empty list and assign it to a new variable <em>str_list</em>. (Line 2)
Use for-loop to iterate through all the number in the <em>number_list</em>.(Line 4). Within the loop, each number is converted to a string using the Python built-in function <em>str() </em>and then use the list append method to add the string version of the number to <em>str_list</em>.
Use Python string<em> join() </em>method to join all the elements in the str_list as a single string. The "," is used as a separator between the elements (Line 7) . At the end return the string as an output.
We can test the function by calling the function and passing [22,33,34] as an argument and we shall see "22,33,44" is printed as an output. (Line 9 - 10)
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Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= 
By putting values, we get
= 
= 
= 
hence,
The average force will be:
= 
= 
= 
Or,
= 
