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goblinko [34]
3 years ago
6

Q9. A cylindrical specimen of a metal alloy 54.8 mm long and 10.8 mm in diameter is stressed in tension. A true stress of 365 MP

a causes the specimen to plastically elongate to a length of 61.8 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 54.8 mm to a length of 64.7 mm. (10 points)
Engineering
1 answer:
wolverine [178]3 years ago
7 0

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

ε = true strain = Change in Length/Original Length

ε = (61.8 - 54.8)/54.8 = 0.128

n = strain hardening exponent = 0.2

Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

k = 550.62 MPa

Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

ε = true strain = Change in Length/Original Length

ε = (64.7 - 54.8)/54.8 = 0.181

n = strain hardening exponent = 0.2

Therefore,

σ = (550.62 MPa)(0.181)^0.2

<u>σ = 391.2 MPa</u>

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vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

4 0
3 years ago
13. Write a function which is passed two strings. The function creates a new string from the two original strings by copying one
attashe74 [19]

Answer:

I am writing the code in C++. Let me know if you want the program in some other programming language.

#include <iostream>  // includes header file for input output functions

using namespace std;     //to identify objects like cin cout

string CopyStrings(string string1, string string2)  

{   string newString = "";    

   for (int loop = 0; loop < string1.length() ||  

                   loop < string2.length(); loop++)      {      

       if (loop < string1.length())  

           newString += string1[loop];          

       if (loop < string2.length())  

           newString += string2[loop];      }  

   return newString;   }  

int main()  

{   string stringA = "ace";  

   string stringB = "bdf";  

   cout << CopyStrings(stringA, stringB);   }

Output:

abcdef

Explanation:

The function CopyStrings() function takes two strings i.e. string1 and string2 as parameters to copy characters from both the string one character from each.

The newString variable stores the new string after copying characters from both strings string1 and string2.

Then the for loop starts which has a variable loop which is an index variable that traverses through both the strings stored in string1 and string2. The loop continues to execute until it moves through entire length of string1 and string2 which means it copies all the characters from both string1 and string2. length() is used here which returns length of the string1 and string2.

If statement in the for loop checks the character that loop (index) variable is pointing to is less than the string1 length which means it checks each character stored in string1. For example if string1 contains "ace" and loop variable is moving through the string and is currently at "a" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character a into the newString variable.

Next If statement checks character that loop variable is pointing to is less than the string2 length which means it checks each character stored in string2. For example if string2 contains "bdf" and loop variable is moving through the string and is currently at "b" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character b into the newString variable.

Then the second iteration starts which again first stores the next character i.e. c from string1 into newString and then stores next character i.e d from string2 into newString.

Then the third iteration starts which again first stores the next character i.e. e from string1 into newString and then stores next character i.e f from string2 into newString.

Then the loop breaks as the loop variable reaches end of both the string1 and string2.

return newString will return the copied string into the output screen which is abcdef.

The screenshot of code along with output is attached.

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Answer:

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Explanation: Check the attachment

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At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut
professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

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Therefore,

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Average Arrival rate, λ = 30 vehicles per hour

Now,

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thus,

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or

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Average time spent by the vehicle = 2 min/veh

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Option (d) 2 min/veh

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Where are the statements then bbs lol
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