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Shkiper50 [21]
4 years ago
15

Assume that an attacker knows that a user’s password is either p1 = abcd or p2 = bedg. Say the user encrypts his password using

the Vigen´ere cipher, and the attacker sees the resulting ciphertext c. Show how the attacker can determine the user’s password, or explain why this is not possible, when the period t used by cipher is 1, 2, 3, or 4 respectively.
Engineering
1 answer:
denis23 [38]4 years ago
8 0

Answer: I will Show how the attacker can determine the user’s password

Explanation:

If we assume an attacker knows that a user’s password is either abcd or bedg. Say the user encrypts his password using the shift cipher, and the attacker sees the resulting ciphertext. Show how the attacker can determine the user’s password, or explain why this is not possible.  

The alphabet{A, B, . . . , Z}is identified  with the set Σ = {0, 1, . . . 25} and all additions are implicitly taken mod26.  

Then, the possible passwords are p0 = abcd = (0, 1, 2, 3) and p1 = bedg = (1, 4, 3, 6).  

All possible encryptions of p0 are C0 = {(k, k + 1, k + 2, k + 3) | k ∈ Σ} and the ones of p1 are C1 = {(k + 1, k + 4, k + 3, k + 6) | k ∈ Σ}.  

These two sets are disjoint and so in order to obtain the password  it is necessary to check in which set the ciphertext lies.

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When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing
Alexandra [31]

Answer:

Pressure = P2=73.13 lbf/ft^2

Explanation:

The pressure is calculated by Bernoulli's equation.

P1+\frac{1}{2} dVi^2=P2+\frac{1}{2} dVf^2\\

Solving it for P2

P2=P1 +\frac{1}{2} d(Vi^2-Vf^2)

Now inserting values

P1 = 0 as it is given that initial point is at atmospheric point.

d= density = 2.05 sl/ft3

Vi = 200 mph = 293.33 ft/s

Vf = 273 mph = 400.4 ft/s

Putting these values

P2 = 0 + 0.5* 2.05(293.33^2-400.4^2)\\P2=-76sl/ft.s^2

Changing the units to pounds per square foot

P2=73.13 lbf/ft^2

7 0
4 years ago
Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrus
saw5 [17]

Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = \frac{5}{18}\times50 = 13.89 m/s

Now,

We have the relation

\tan\theta=\frac{v^2}{gR}

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

\tan\theta=\frac{13.89^2}{9.81\times100}

or

\tan\theta=0.1966

or

θ = 11.125°

3 0
3 years ago
A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at
Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, T_{s}=280^{o}C

Temperature of air, T_{\infinity}=20^{o}C

Velocity of air flow V=5.5m/s

The film temperature is determined as:

T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C

from the table, properties of air at 1 atm pressure

At T_{f}=150^{o}C

Thermal conductivity, K = 0.03443 W/m^oC; kinematic viscosity v=2.860 \times 10^{-5} m^2/s; Prandtl number Pr=0.70275

The reynolds number for the flow is determined as:

Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92

sice the obtained reynolds number is less than 2\times10^5, the flow is said to be laminar.

The nusselt number is determined from the relation given by:

Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}

Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11

The covective heat transfer coefficient is given by:

Nu_{cyl}=\frac{hD}{k}

Rewrite and solve for h

h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m

The rate of heat transfer from the wire to the air per meter length is Q=340.42W/m

6 0
3 years ago
What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent leng
skad [1K]

Answer:

0.067wc

Explanation:

The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction

We substitute values

actual static pressure = 0.1

Total equivalent length = 150 ft

0.1 = (150ft/100) multiplied by Rate of friction

Friction rate at 100ft = 0.067

So we have that the required friction needed is 0.067wc

6 0
3 years ago
Nancy ate a 500 Cal lunch. Neglecting efficiency issues (i.e., assuming 100% conversion of energy to work), to what height could
VladimirAG [237]

Answer:

4265.04\ \text{m}

2.38\times 10^{10}\ \text{W}

Explanation:

PE = Energy of food = 500 cal = 500\times4184=2.092\times10^6\ \text{J}

m = Mass of object = 50 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Potential energy of food is given by

PE=mgh\\\Rightarrow h=\dfrac{PE}{mg}\\\Rightarrow h=\dfrac{2.092\times 10^6}{50\times 9.81}\\\Rightarrow h=4265.04\ \text{m}

Nancy could raise the weight to a maximum height of 4265.04\ \text{m}.

Mass of H_2 used per year = 25\times 10^{9}\ \text{kg/year}

Energy of H_2 = \dfrac{30\times10^9}{1000}=30\times 10^6\ \text{J/kg}

Power

P=25\times 10^{9}\ \text{kg/year}\times 30\times 10^6\ \text{J/kg}\\\Rightarrow P=7.5\times 10^{17}\ \text{J/year}\\\Rightarrow P=\dfrac{7.5\times 10^{17}}{365.25\times 24\times 60\times 60}\\\Rightarrow P=2.38\times 10^{10}\ \text{W}

The power requirement is 2.38\times 10^{10}\ \text{W}.

6 0
3 years ago
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