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AVprozaik [17]
2 years ago
9

Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch ca

bles to connect them; however, he is unable to reach the new network segment from his workstation. He can only connect to it from a workstation within that segment. Which of the following is MOST likely the problem?
A. One of the hubs is defective.
B. The new hub is powered down.
C. The patch cable needs to be a CAT6 patch cable.
D. The technician used a straight-through cable.
Engineering
1 answer:
mart [117]2 years ago
3 0

Answer:

Option D. is correct

Explanation:

Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment,  he is not able to reach the new network segment from his workstation.

The most problem is that the technician used a straight-through cable.

Option D. is correct.

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A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s
grandymaker [24]

Answer:

the endurance strength  S_e = 421.24  MPa

Explanation:

From the given information; The objective is to estimate the endurance strength, Se, in MPa .

To do that; let's for see the expression that shows the relationship between the ultimate tensile strength and Brinell hardness number .

It is expressed as:

200 \leq H_B \leq 450

S_{ut} = 3.41 H_B

where;

H_B = Brinell hardness number

S_{ut} =  Ultimate tensile strength

From ;

S_{ut} = 3.41 H_B; replace 290 for H_B ; we have

S_{ut} = 3.41 (290)

S_{ut} = 988.9 MPa

We can see that the derived value for the ultimate tensile strength when the Brinell harness number = 290 is less than 1400 MPa ( i.e it is 988.9 MPa)

So; we can say

S_{ut} < 1400

The Endurance limit can be represented by the formula:

S_e ' = 0.5 S_{ut}

S_e ' = 0.5 (988.9)

S_e ' = 494.45 MPa

Using Table 6.2 for parameter for Marin Surface modification factor. The value for a and b are derived; which are :

a = 1.58

b =  -0.085

The value of the surface factor can be calculate by using the equation

k_a = aS^b_{ut}

K_a = 1.58 (988.9)^{-0.085

K_a = 0.8792

The formula that is used to determine the value of  k_b for the rotating shaft of size factor d = 10 mm is as follows:

k_b = 1.24d^{-0.107}

k_b = 1.24(10)^{-0.107}

k_b = 0.969

Finally; the the endurance strength, Se, in MPa if the rod is used in rotating bending is determined by using the expression;

S_e =k_ak_b S' _e

S_e= 0.8792×0.969×494.45

S_e = 421.24  MPa

Thus; the endurance strength  S_e = 421.24  MPa

8 0
3 years ago
Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/
kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k

Obtain the following properties at 10kPa from the table "saturated water"

h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K

Calculate the enthalpy at exit of the turbine using the energy balance equation.

\frac{dE}{dt}=Q-W+m(h_1-h_2)

Since, the process is isentropic process Q=0

0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

a.)

Calculate the isentropic turbine efficiency.

\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

b.)

Find the quality of the water at state 2

since h_f at 10KPa <h_2<h_g at 10KPa

Therefore, state 2 is in two-phase region.

h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

Calculate the entropy at state 2.

s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K

Calculate the rate of entropy production.

S=\frac{Q}{T}+m(s_2-s_1)

since, Q = 0

S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

6 0
3 years ago
For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.
horrorfan [7]

Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

5 0
3 years ago
Match the military operation to the category of satellite that would perform it.
SOVA2 [1]

Answer:

1. Location of enemy ground troops  - EARTH OBSERVING.

Using earth observing satellite imagery, the military can observe vast expanses of land and in so doing, find the location of enemy ground troops.

2. Routine reconnaissance of an unfamiliar climate  - WEATHER

In other to find out more about the climate of an area, a weather satellite can be used to observe the areas and its changing weather patterns.

3. Analyze waterways in an unfamiliar location  - NAVIGATION

Using navigation satellites, navigation conduits such as roads and waterways can be observed.

4. Provide warning of an attack - COMMUNICATION.

Communications satellites enable people to communicate over great distances and so can be used by the military to warn of an impending attack.

5 0
3 years ago
What major advancement in machine tools occurred in the 1970s and what benefits did it provide? describe in your own words.
mixer [17]

Answer:

I'm just a seventh grader

4 0
3 years ago
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