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3 years ago

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A network address of 172.16.0.0 /12 has been given. Which of the following accurately describes this network? (select one or mor

e)
(A) The ending address of this network is 172.255.255.255
(B) The ending address of this network is 172.31.255.255
(C) This is private class A network with 4 bits of sub-netting
(D) This is not a valid network address because the 255.240.0.0 mask is wrong
(E) This is private class B network using a default mask

1 answer:

ludmilkaskok [199]3 years ago

3 0

Answer:

B and E is correct.

Explanation:

Given that network address

172.16.0.0/12

This is class B network type.

The ending of this network will be 172.31.255.255

In IP version 4 there are four following type of classes

1)Class A (0-127)

2)Class B (128-191)

3)Class C (191-223)

4)Class D(224-239)

5)Class E (240-255)

Generally class A,B,C and D are used.

So our options B and E is correct.

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A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur

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Answer:

$Q_{in} = 46.454\,kJ$

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

$Q_{in} = U_{2} - U_{1}$

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

$P = 101.42\,kPa$

$T = 100\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 675.761\,\frac{kJ}{kg}$

$x = 0.123$

State 2 - (Liquid-Vapor Mixture)

$P = 476.16\,kPa$

$T = 150\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 1643.545\,\frac{kJ}{kg}$

$x = 0.525$

The mass stored in the vessel is:

$m = \frac{V}{\nu}$

$m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }$

$m = 0.048\,kg$

The heat transfer require to the process is:

$Q_{in} = m\cdot (u_{2}-u_{1})$

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$Q_{in} = 46.454\,kJ$

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3 years ago

If you log into the admin account on windows 10, will the admin be notified ?

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Answer:

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3 years ago

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1 year ago

. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C

MrRissso [65]

Answer:

The inside temperature, $T_{in}$ is approximately 248 °C.

Explanation:

The parameters given are;

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Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×( $T_s$ - $T_{\infty$ ) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

$P = \dfrac{K \times A \times \Delta T}{L}$

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

$2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}$

$T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C$

The inside temperature, $T_{in}$ = 247.99 °C ≈ 248 °C.

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4 years ago

You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from

Aleksandr [31]

Answer:

a) $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

b) attached below

c) type zero system

d) k > $\frac{g}{200}$

e) The gain K increases above % error as the steady state speed increases

Explanation:

Given data:

Motor voltage = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; $\frac{K1}{1+St}$ be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > $\frac{g}{200}$

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3 years ago