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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.

1 answer:

Luda [366]3 years ago

4 0

Answer:

what r u on

Explanation:

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Determine the following for a south facing surface at 30� slope in

lesantik [10]

Answer:

z=60.32°, i=0.32°, Beam Radiation = 1097.2 W/m², Id = 94.2 W/m², Ir=14.1W/m², total radiation = 1205.4 W/m², Local time=1:21PM

Explanation:

A. Zenith Angle:

As we know that,

Zenith angle=z=90⁰-α=L(latitude)=29.68⁰

Another way to do it is to find α first,

At solar time hour angle is 0⁰. So, solar altitude becomes equal to latitude which could be written as

sinα=cosL

α=sin⁻¹(cosL)=sin⁻¹(cos29.68⁰)=60.32°

B. Angle of incidence:

angle of incidence= cosi=sin(α+β)=sin(60.32°+32°)=sin92.32°

i=cos⁻¹(sin92.32°)=0.32°

C. Beam Radiation:

First we need to calculate extra terrestrial radiations

Iext.=1353[1+0.034cos(360n/365)]

where n=264

=1345 W/m²

Now,

Beam Radiation=CIext⁻ⁿ

where n=0.1/sin60.32°

Beam Radiation = 1097.2 W/m²

D. Diffude Radiation:

difuse radiation = Id = 0.0921ₙcos²(β/2)

where β=30°

Id = 94.2 W/m²

E. Reflected Radiations:

Ir=pIn(sinα+0.092)sin²(β/2)

= (0.2)(1097.1)(sin60.32+0.092)sin²(30/2)

= 14.1W/m²

F. Total Radiation:

total radiation = beam radiation + diffuse radiation + reflected raddiation

= 1205.4 W/m²

G. Local Time:

LST= ST-ET-(lₓ-l(local))4min/₀

= 12:00-7.9min-(75°-82.27°)4min/₀

=12:21PM

Local time

LDT=LST+=12:21+1:00=1:21PM

5 0

3 years ago

The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are

IrinaK [193]

Answer:

the density of the electrum is 14.30 g/cm³

Explanation:

Given that:

The equilibrium fraction of lattice sites that are vacant in electrum = $9.93*10^{-8}$

Number of vacant atoms = $5.854 * 10^{15} \ vacancies/cm^3$

the atomic mass of the electrum = 146.08 g/mol

Avogadro's number = $6.022*10^{23$

The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)

$5.854*10^{15}$ = $9.93*10^{-8}$ × Total number of sites(N)

Total number of sites (N) = $\dfrac{5.854*10^{15}}{9.93*10^{-8}}$

Total number of sites (N) = $5.895*10^{22}$

From the expression of the total number of sites; we can determine the density of the electrum;

$N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}$

where ;

$N_A$ = Avogadro's Number

$\rho_{electrum} =$ density of the electrum

$A_{electrum}$ = Atomic mass

$5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}$

$5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}$

$8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}$

$\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}$

$\mathbf{ \rho _{electrum}=14.30 \ g/cm^3}$

Thus; the density of the electrum is 14.30 g/cm³

7 0

3 years ago

In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at

Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

$V = 690V$

$P_{real} = 2.3MW$

Real power in 3 phase

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Now the Phase Voltage is,

$V_{ph} = \frac{V}{\sqrt{3}}$

$V_{ph} = \frac{690}{\sqrt{3}}$

$V_{ph} = 398.37V$

The current phase would be,

$P_{real} = 3V_{ph}I_{ph} Cos\theta$

Rearranging,

$I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}$

Replacing,

$I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}$

$I_{ph}= 2.26kA/phase$

Therefore the current per phase is 2.26kA

6 0

3 years ago

Help I will brainliest

ValentinkaMS [17]

Answer:

hacking?

Explanation:

jsd

3 0

3 years ago

What is the present value of the future receipts of $2,000, 5 years from now at 10% compounded annually?

Elden [556K]

Answer:

P = $ 766.28

Explanation:

present value = ?

Future value = $ 2000

time = 5 years

compounded annually at the rate of = 10 %

$A = P + P(1+\dfrac{r}{100})^t$

$2000 = P + P(1+\dfrac{10}{100})^5$

2000 = P + 1.61 P

2.61 P = 2000

P = $ 766.28

hence, the present value of amount invested to get the future value of $2000 is equal to P = $ 766.28

3 0

3 years ago