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agasfer [191]
3 years ago
5

Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1

Gbps. What is the queueing delay of packet number 48 ?(in milliseconds, rounded to one decimal place, e.g. 0.01234 seconds would be entered as "12.3")
Engineering
1 answer:
tia_tia [17]3 years ago
4 0

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is :  (n - 1) × L/R

where L :  packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate =  2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

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An incompressible viscous fluid flows through a pipe with a flow rate of 1 mL/s. The pipe has a uniform diameter D0 and a length
Nataly [62]

Answer:

Q_2 = 32 mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, Q_1 = 1 mL/s

Initial diameter, D_1= D_0

Initial length, L_1=L_0

The initial pressure difference to maintain the flow, P_1=P_0

We know for a viscous flow,

$\Delta P = \frac{32 \mu V L}{D^2}$

$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$

$Q \propto \Delta P \times D^4$

$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$

$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$

$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$

$\frac{1}{Q_2}= \frac{1}{32}$

∴ Q_2 = 32 mL/s

6 0
3 years ago
A cantilever beam is 4000 mm long span and has a u.d.l. of 0.30 kN/m. The flexural stiffness is 60 MNm². Calculate: 1. Slope 2.
Viefleur [7K]

Answer:

1. Slope = 53.3 x 10⁻⁶

2. Deflection = -0.00016m

Explanation:

given:

let L = 4 m (span of cantilever beam)

let w = 300 N/m (distributed load)

let EI =60 MNm² (flexural stiffness)

                 dy      w * L³        300 x 4³

1. slope = ------- = --------- =  ------------------- =  53.3 x 10⁻⁶

                  dx        6EI           6 x 60x10⁶

                                  wL⁴               300 x 4⁴

2. Deflection = y = - ----------- =  - ------------------ =   -0.00016m

                                    8EI                8 x 60x10⁶

therefore the deflection is 0.16mm downwards.

3 0
3 years ago
2.18 The net potential energy between two adjacent ions, EN, may be represented by the following equation: (1) Calculate the bon
goblinko [34]

Answer:

2.18

Explanation:

because ya correc

3 0
3 years ago
Ice dams are most directly related to:
musickatia [10]

Answer:

(B) heated air in an uninsulated, unvented attic.

Explanation:

-Ice dams are formed by an interaction between snow cover, outside temperatures, and heat lost through the roof – more specifically, from heat loss through the poorly insulated ceiling of the house into the attic area and under the roof.

-Hot, humid air trapped in attics can also cause moisture to accumulate, resulting in mold or mildew.

7 0
3 years ago
The coefficient of performance of a reversible refrigeration cycle is always (a) greater than, (b) less than, (c) equal to the c
bezimeni [28]

Answer:

Option B is correct

Explanation:

Let

Higher temperature = T_H

Lower temperature = T_L

We know that COP is given by

COP = \frac{T_L}{T_H-T_L}

We see that COP is depends only on the temperature difference & Temperature difference is maximum for the Carnot cycle.

Therefore the COP of reversible refrigeration cycle is always less then the COP of  an irreversible refrigeration cycle when each operates between the same two thermal reservoirs.

Therefore option B is correct

5 0
3 years ago
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