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KatRina [158]
3 years ago
8

The purpose of the __________ algorithm is to enable two users to exchange a secret key securely that can then be used for subse

quent encryption of messages.
Engineering
1 answer:
Whitepunk [10]3 years ago
8 0

Answer:

 Diffie Hellman key exchange.                

Explanation:

Diffie Hellman key exchange :

 This is the technique which is used to exchanging cryptographic keys in the public channel.This is also represent as DH keys.This is the first public key protocol.This keys given by Diffie and Hellman.These keys enable two users to exchange a secret key securely .

So the answer is Diffie Hellman key exchange.

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Select the right answer<br>​
Kruka [31]

Answer:

for 1st question the answer is 5th option.

for 2nd question the answer is 2nd option

hope it helps you mate

please mark me as brainliast

5 0
3 years ago
Refers to the capability to keep moving forward on a specified grade.
Mademuasel [1]

Answer:

maneuverability

Explanation:

needless to say, I took the quiz

6 0
3 years ago
(True/False) Unix is written in the C language. *<br> True<br> O False
Katarina [22]

Answer:

false

Explanation:

8 0
2 years ago
Read 2 more answers
The enforcement of OSHA standards is provided by federal and state
Gnoma [55]

Answer:

Explanation:

Enforcing OSHA, Occupational Safety and Health Administration, standards is not a job for electricians, lawmakers or tax collectors. The right answer is safety inspectors.

3 0
2 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
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