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4 years ago

A circuit-switching scenario in whichNcs users, each requiring a bandwidth of 25 Mbps, must share a link of capacity 150 Mbps.

Engineering

1 answer:

Monica [59]4 years ago

3 0

Answer:

0.09

Explanation:

Packet switching involves breaking a message into packets and sending them independently. Since the user only needs to transmit 10 percent of the time, the probability that a given (specific) user is transmitting = 10% = 0.1

The probability that a user is not transmitting = 100% - 10% = 90% = 0.9

Therefore, the probability that a given (specific) user is transmitting, and the remaining users are not transmitting = 0.1 * 0.9 = 0.09

You might be interested in

Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display

Citrus2011 [14]

Algorithm of the Nios II assembly program.

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

<h3>The Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally, the program will be written using a cpulator simulator in order to attain best result.

We are to

Attain data for simulation from the SW11-0, on the DE2-115 Simulator
The data will be read from the switches in loop.
The decimal output is displayed using the seven-segment displays and done using the loop.
The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

DE2-115
DE2-115_SW11
DE2-115_HEX3
DE2-115_HEX4
DE2-115_HEX5
DE2-115_HEX6
DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

6 0

2 years ago

g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,

irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0

3 years ago

Locate the centroid y¯ of the composite area. Express your answer to three significant figures and include the appropriate units

german

Answer:

Please see the attached Picture for the complete answer.

Explanation:

4 0

3 years ago

When Ontario

Marat540 [252]

Answer:

Explanation:

This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.

4 0

3 years ago

A 2 m3 rigid tank initially contains air at 100 kPa and 22 degrees C. The tank is connected to a supply line through a valve. Ai

Vaselesa [24]

Answer:

a. 9.58kgs b. 340.32KJ

Explanation:

Volume of tank= 2m³

Initial Pressure Pi= 100KPa

Initial Temperature Ti= 22 C= 295K

Line Pressure P₁= 600 KPa

Line Temperature T= 22 C= 295K

Final Pressure P2= 600 KPa

Final Temperature T2= 77 C= 350K

Use Ideal Gas Equation

PV= mRT

P₁V₁= m₁RT₁

m₁= (100 x 2)/(0.287 x 295) = 2.3622kg

P₂V₂= m₂RT₂

m₂= (600 x 2)/(0.287 x 350) = 11.946 kg

Since valve is closed and no mass leave

m₁ + mi = m₂ + me

as per above condition me= 0

mi= m₂ - m₁ = 11.946 - 2.3622 = 9.5838kg

Applying energy equation

m₁u₁ + mihi + Q = m₂u₂ + mehe + W

me and W=0

m₁u₁ + mihi + Q = m₂u₂

m₁CvT₁ + miCpTi + Q = m₂CvT₂

Q = m₂CvT₂- m₁CvT₁ - miCpTi

Q = (11.946 x 0.717 350) - (2.3622 x 0.717 x 295) - (9.5868 x 1.004 x 295)

Q = -340.321 KJ (Negative sign doesn't matter as energy is not a vector quantity)

4 0

3 years ago