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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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It will results in mechanical hardening.

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In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.
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Where is Eq.(28) ?? You should show it to find the result
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3 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

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3 years ago
In which type of shoot is continuous lighting used?
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A type of shoot in which continuous lighting used is: 1) studio.

<h3>What is a photoshoot?</h3>

A photoshoot simply refers to a photography session which typically involves the use of digital media equipment such as a camera, to take series of pictures (photographs) of models, group, things or places, etc., especially by a professional photographer.

<h3>The types of shoot.</h3>

Basically, there are four main type of shoot and these include the following:

  • Studio
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In photography, a type of shoot in which continuous lighting used is studio because it enhances the photographs.

Read more on photography here: brainly.com/question/24582274

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The question “How do plants convert sunlight to energy?” best represents which of the following?
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D

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I would say this awnser because its the only one that makes sence to me

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