Answer:
the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Explanation:
Given the data in the question;
yield strength σ
= 690 Mpa
plane strain fracture toughness K
= 32 MPa-
minimum component thickness for which the condition of plane strain is valid = ?
Now, for plane strain conditions, the minimum thickness required is expressed as;
t ≥ 2.5( K / σ )²
so we substitute our values into the formula
t ≥ 2.5( 32 / 690 )²
t ≥ 2.5( 0.0463768 )²
t ≥ 2.5 × 0.0021508
t ≥ 0.005377 m or 5.38 mm
Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Answer:
1) The ethical act Amy should display is to open the Email since it came in through her personal inbox. However, further action should not be taken since ordinarily opening the mail will not disclose or corrupt her system either locally of her system on the server and web space.
2) If such email came in, the first action i will take is to open the mail then take a pause to read through the instruction because most scam or malware comes with an array of further instructions.
After carefully reading through, i will call up Davey to confirm if he sent such mail which answer will obviously be a No and in the case where i am unable to get a call through to Davey, i will immediately delete the mail from my inbox to avoid the mistake of clicking any embedded link within the mail.
Answer:
yay yay
Explanation:
im so excited i cant wait
Answer:
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Explanation:
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
