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3 years ago

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In RSA Digital Signature, Suppose Bob wants to send a signed message (x = 4) to Alice. The first steps are exactly t eps are exa

ctly the same as it is done for an RSA encryption: Bob computes his RSA parameters and sends the public key to Alice. We know p = 3, q=11, and bob choose e=3. (Hint: We learn RSA algorithm and key generation in Week 7) (1) What is the public key pair Bob sends to Alice? (2) What is the value of signature s? (3) What is the value of verkpubA(x,s)? Show all intermediate steps clearly.

1 answer:

Luda [366]3 years ago

4 0

Answer:

what r u on

Explanation:

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An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions. If the

xenn [34]

Answer:C 0.12 V

Explanation:

Given

Concentration of $Fe^{2+} M_1=0.40 M$

Concentration of $Ni^{2+} M_2=0.002 M$

Standard Potential for Ni and Fe are $V_2=-0.25 V$ and $V_1=-0.44 V$

$\Delta V=V_2-V_1-\frac{0.0592}{2}\log (\frac{M_1}{M_2})$

$\Delta V=-0.25-(-0.44)-\frac{0.0592}{2}\log (\frac{0.4}{0.002})$

$\Delta V=0.12\ V$

7 0

3 years ago

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × $10^{4}$ ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates = 0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = $V_o \times e^{t\RC}$ ...........1

so R will be

R = $\frac{t}{C\times ln(\frac{V_o}{V})}$ ....................2

put here value

so for t min 11.5 μs

R = $\frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}$

R min = 28.173 ohm

and

for t max 6.33 ms

R max = $\frac{6.33}{11.5} \times 28.173$

R max = 1.55 × $10^{4}$ ohm

4 0

3 years ago

Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using

VashaNatasha [74]

Answer:

For any string, we use $s = xyz$

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

Consider any string a i b j c k in the language. If i = 1 or i > 2, we take $x = \epsilon$ and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.

For i = 2, we can take and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
Finally, for the case i = 0, we take $x = \epsilon$ , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.

8 0

3 years ago

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?

lianna [129]

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents a crystal system which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have three axes which have only two of its axes equal in length.

5 0

3 years ago