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ivolga24 [154]
3 years ago
5

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.
Engineering
1 answer:
Luda [366]3 years ago
4 0

Answer:

what r u on

Explanation:

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A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so
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What are the main causes of injuries when using forklifts?
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The forklift overturning is a very common way of getting injured from a forklift. Overturning the forklift means it tips over onto it's side due to the operator turning it too fast.

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Consider a drug-eluting balloon catheter deployed into a blood vessel. The balloon is inflated to perfectly adhere to the vessel
GaryK [48]

Answer:

a)  Cr = Co - Fx / D

b)   ΔC / Δx = ( CR - Cr )  / ( xR - xRo )

Explanation:

A) Derive an expression for the profile c(r) inside the tissue

F = DΔC / X  = D ( Co - Cr ) / X   ------ 1

where : F = flux , D = drug diffusion coefficient

            X = radial distance between Ro and R

Hence : Cr = Co - Fx / D

B) Express the diffusive flux at outer surface of the balloon

Diffusive flux at outer surface =  ΔC / Δx = CR - Cr / xR - xRo

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3 years ago
Given the following Grammar and the right sentential form draw a parse tree and show the phrases, simple phrases and handle.S →
Shtirlitz [24]

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Find the answer in the attached in the order

Option a, Option c and Option b

Explanation:

4 0
3 years ago
Read 2 more answers
A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh
sdas [7]

Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C

to kelvin = (5 + 273)K = 278 K

From  the " thermophysical properties of gases at atmospheric pressure" table; At T_f = 278 K ; by interpolation; we have the following

\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}  → v 13.93 (10⁻⁶) m²/s

\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})} → k = 0.0245 W/m.K

\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})} → ∝ = 19.6(10⁻⁶)m²/s

\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720} → Pr = 0.713

\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}

The Rayleigh number for vertical cavity

Ra_L  = \frac{g \beta (T_1-T_2)L^3}{\alpha v}

= \frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}

= 8.38*10^5

\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}

NU_L= \frac{hL}{k}= 4.878

\frac{hL}{k}= 4.878

\frac{h*0.06}{0.0245}= 4.878

h = \frac{4.878*0.0245}{0.06}

h = 1.99 W/m².K

Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

3 0
3 years ago
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