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anzhelika [568]
3 years ago
15

A cylindrical bar of steel 10.1 mm (0.3976 in.) in diameter is to be deformed elastically by application of a force along the ba

r axis. Determine the force that will produce an elastic reduction of 2.8 ×10-3 mm (1.102 ×10-4 in.) in the diameter. For steel, values for the elastic modulus (E) and Poisson's ratio (ν) are, respectively, 207 GPa and 0.30.
Engineering
1 answer:
Nina [5.8K]3 years ago
4 0

Answer:

given d= 10.2x10-3m

change in diameter d' =3.4x10-6 m

elastic modulus=207x109pa

1/m=0.30

we know that stress/strain=E

but poissons ratio 1/m = lateral starin/longitudinal starin

0.3= (d'/d)/(L'/L)

L'/L = 3.4x10-6/(0.3x10.2x10-3)

= 1.11*10-3

E= (f/A)/(L'/L)

force f= E*A*(L'/L)

f =(1.11*10-3)*207*109*(3.1415*(10.2*10-3)2)/4

= 18775.2N =18.775KN

Explanation:

The force that produce an elastic reduction is 18.775KN

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astraxan [27]

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<u>Explanation:</u>

<u>First Pass:</u>

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( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

<u>Second Pass:</u>

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

<u>Third Pass:</u>

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

<u>Fourth Pass:</u>

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

<u>Fifth Pass:</u>

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

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3 years ago
2. Write a Java program that generates a new string by concatenating the reversed substrings of even indexes and odd indexes sep
Nana76 [90]

Answer:

  1. public class Main {
  2.    public static void main(String[] args) {
  3.        String testString = "abscacd";
  4.        String evenStr = "";
  5.        String oddStr = "";
  6.        for(int i=testString.length() - 1; i >= 0; i--){
  7.            if(i % 2 == 0){
  8.                evenStr += testString.charAt(i);
  9.            }
  10.            else{
  11.                oddStr += testString.charAt(i);
  12.            }
  13.        }
  14.        System.out.println(evenStr + oddStr);
  15.    }
  16. }

Explanation:

Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).

Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.

Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to  testString.length() - 1  (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).

At last, we print the concatenated evenStr and oddStr (Line 18).  

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Explanation:

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Explanation:

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Step 2:

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E true = Ln(1 + 1 ) = 0.69

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Answer:

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