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anzhelika [568]
3 years ago
10

Suppose we use radix sort to sort the English-language strings below using standard lexicographic ordering (i.e. sort in alphabe

tical order). We sort least-to-greatest and consider the numbers in top-to-bottom order when assigning them to bins. Assume the empty string" comes before all letters in lexicographic order. PART TRIP
TARP
ART
TRAP
CHIP

a) (1 point) How many passes are required to sort the strings?
b) (1 point) How many buckets would radix sort allocate to sort the strings?
c) (5 points) For each of the following pairs of words, fill in the circle next to the word that would appear earlier in the list after two passes of radix sort.
i) TRIP or TARP
CHIP or TRIP
iii) ART or PART
iv) PART or TARP
v) TARP or TRAP

d) State the runtime of radix sort on each of the following inputs set as precisely as you can. Include any known constant factors. i) (1 pt) Runtime on English-language strings of length d: ii) (1 pt) Runtime on decimal integers of length d:

Engineering
1 answer:
nlexa [21]3 years ago
7 0

Answer:

a) 4 passes are required to sort the string.

b) 4

c)   i) TARP

    ii) CHIP

    iii) PART

    iv) TARP

    v) TARP

d) O(k+n), n is no. of strings, k is largest no. of character in among the string

   O(d*(n+10)), n is no. of integers

Explanation:

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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
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Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

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