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4 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.

1 answer:

Whitepunk [10]4 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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A flow of 100 mgd is to be developed from a 190-mi^2 watershed. At the flow line the area's reservoir is estimated to cover 3900

yaroslaw [1]

Answer:

13-mi 27 acres

Explanation:

7 0

3 years ago

What is the aim of reviewing a research paper?

qaws [65]

Answer:

Purpose of review papers

They carefully identify and synthesize relevant literature to evaluate a specific research question, substantive domain, theoretical approach, or methodology and thereby provide readers with a state-of-the-art understanding of the research topic.

3 0

3 years ago

Read 2 more answers

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operat

Gwar [14]

Answer:

$\eta _{max} = 0.2413 = 24.13%$

$\eta' _{max} = 0.5061 = 50.61%$

Given:

$T_{1max} = 100^{\circ} = 273 + 100 = 373 K$

operating temperature of heat engine, $T_{2} = 10^{\circ} = 273 + 10 = 283 K$

$T_{3max} = 300^{\circ} = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _{max}$ is given by:

$\eta _{max} = 1 - \frac{T_{2}}{T_{1max}}$

$\eta _{max} = 1 - \frac{283}{373} = 0.24$

$\eta _{max} = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_{3max}$ changes to $300^{\circ}$ , so, the new maximum efficiency, $\eta' _{max}$ is given by:

$\eta' _{max} = 1 - \frac{T_{2}}{T_{3max}}$

$\eta _{max} = 1 - \frac{283}{573} = 0.5061$

$\eta _{max} = 0.5061 = 50.61%$

4 0

4 years ago

A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu

trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

Area of A is = $\frac{\pi}{4}(0.01)^{2} m^{2}$

= $7.85 *10^{-5}m^{2}$

Velocity of liquid through A = 0.2 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $0.2 *7.85*10^{-5} \frac{m^{3}}{s}$

The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1039.8 * 0.2 * 7.85 *10^{-5} Kg/s$

= 0.016324 $\frac{Kg}{s}$

From the question the diameter of B = 2 cm = 0.02 m

Area of B = $\frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}$

Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $3.14*10^{-4} *0.01 \frac{m^{3}}{s}$

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1053 * 3.14*10^{-6} \frac{Kg}{s}$

= 0.00330642 $\frac{Kg}{s}$

From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s

And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $13\frac{g}{L} * 0.5 \frac{L}{s}$

= $\frac{6.5}{1000}\frac{Kg}{s}$ Note (1 Kg = 1000 g)

= 0.0065 kg/s

Considering potassium chloride

Let denote the rate at which liquid flows in terms of mass as as $\frac{dm}{dt}$ i.e change in mass with respect to time hence

Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

(0.016324 + 0.00330642) - 0.0065 = $\frac{dm}{dt}$

$\int\limits {\frac{dm}{dt} } \, dx =\int\limits {0.01313122} \, dx$

=> 0.01313122 t = (m - $m_{o}$ )

From the question (m - $m_{o}$ ) is given as = 19.7 Kg

Hence the time when the sample was taken is given as

0.01313122 t = 19.7 Kg

=> t = 1500.2414 sec

t = .4167337 hours (1 hour = 3600 seconds)

5 0

4 years ago

!!!!!!!!!!!!!!PLEASE ANSWER THIS!!!!!!!!!!!!!!!!

dmitriy555 [2]

Answer:

age = int(input())

ticket = 100

total = 0

passengers = 1

while passengers <= 5:

if age < 3:

continue

else:

total += ticket

passengers += 1

print(total)

Explanation:

UuU

6 0

2 years ago