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Mice21 [21]
3 years ago
11

section 10.1 describes a man in the middle attack on the diffie-hellman key exhange protocal in which the adversary generates tw

o public private key pairs for the attack. could the same attack be accomplished with one pair?
Engineering
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

1. James will attack by generating a random private key XD and a corresponding public key YD.

2. Jane transmit YA to another person called Alex.

3. James intercept YA and transmit YD to jane.

4. Jane receive YD and calculate K1

At this point james and jane thinks they share a secret key but instead james has a secret key k1 to Jane and k2 to alex.

5. Alex transmit another key XA to alex for example.

6. James intercept and calculate k2 and vice versa.

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Q1: The first option should always be to get out safely (RUN)
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Answer:

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6 0
2 years ago
You work for a printing company, and you realize that your colleague sent incorrect price quotes to a client. You begin to write
xxTIMURxx [149]

Answer:

The sentence excerpted from the e-mail uses passive voice.  

Given the  purpose of your message, this voice is appropriate.

Explanation:

Because the objective is to remedy the situation a passive voice is great because it emphasizes the action and the object instead of the subject.

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4 0
3 years ago
Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm
Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0
2 years ago
provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki
Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

percentage of solar power absorbed by refrigerant = 60%

Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0
3 years ago
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