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expeople1 [14]
3 years ago
9

Consider a 6-bit cyclic redundancy check (CRC) generator, G = 100101, and suppose that D = 1000100100. 1. What is the value of R

? (10 pts)
Engineering
1 answer:
dmitriy555 [2]3 years ago
3 0

Answer:

The value of R is 10101

Explanation:

As per the given data

D = 1000100100

G = 100101

Redundant bit = 6-bits - 1-bit = 5-bits

No add fice zero to D

D = 100010010000000

Now calculate R as follow

R = D / G

R = 100010010000000 / 100101

R = 10101

Workings are attached with this question

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Explanation:

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When water precipitates from the sky, runs off downhill along the ground, or infiltrates down into the soil, its gravitational p
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Geothermal energy.

Explanation:

Geothermal energy is called a renewable energy source because the water is replenished by rainfall, and the heat is continuously produced by the earth.

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3 years ago
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
saveliy_v [14]

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

4 0
3 years ago
Who here is a genius?
Pachacha [2.7K]

Answer:

im kinda smart  whyy?

Explanation:

3 0
3 years ago
Read 2 more answers
Write a method called letterCount that takes two String arguments, one containing some text and the other containing a single le
jenyasd209 [6]

Answer:

I am writing a Python program.  Here is the function letterCount. This function takes two string arguments text and letter and return count of all occurrences of a letter in the text.

def letterCount(text, letter):  

 count = 0  # to count occurrences of letter in the text string

 for char in text:  # loop moves through each character in the text

   if letter == char: # if given letter matches with the value in char

     count += 1  # keeps counting occurrence of a letter in text

 return count # returns how many times a letter occurred in text

   

Explanation:

In order to see if this function works you can check by calling this function and passing a text and a letter as following:

print(letterCount('apples are tasty','a'))

Output:

3

Now lets see how this function works using the above text and letter values.

text = apples are tasty

letter = a

So the function has to compute the occurrences of 'a' in the given text 'apples are tasty'.

The loop has a variable char that moves through each character given in the text (from a of apples to y of tasty) so it is used as an index variable.

char checks each character of the text string for the occurrence of letter a.

The if condition checks if the char is positioned at a character which matches the given letter i.e. a. If it is true e.g if char is at character a of apple so the if condition evaluates to true.

When the if condition evaluates to true this means one occurrence is found and this count variable counts this occurrence. So count increments every time the occurrence of letter a is found in apples are tasty text.

The loop breaks when every character in text is traversed and finally the count variable returns all of the occurrences of that letter (a) in the given text (apples are tasty). As a occurs 3 times in text so 3 is returned in output.

The screen shot of program along with output is attached.

4 0
3 years ago
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