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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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Engineering Careers Scavenger Hunt
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Answer:

c

Explanation:

it's the only engineering career

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3 years ago
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What type of bridge is the sunshine skyway bridge?
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That's either a cable-stayed bridge or a cantilever bridge

6 0
3 years ago
A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with specific gravity of 1.7. The ma
MArishka [77]

Answer:

5320.6 Pascal

Explanation:

Manometer is a pressure measuring device use to measure gas pressure .

Pressure difference in Manometer is a function of density,gravity and the height difference of the liquid.

Pressure difference = density x acceleration due to gravity x difference in height of liquid

Density of liquid = specific gravity of object x density of water.

Density of water = 997 kg/m^3

Specific gravity of liquid = 1.7

Density of liquid = 997 x 1.7 =1694.9kg/m^3

g= 9.81 m/s^2

h =320mm = 0.320m

Pressure difference = 1694.9 x 9.81 x 0.320 = 5320.6 Pascal

3 0
3 years ago
Read 2 more answers
Tech A says that failure of a battery hold down can cause battery plate damage. Tech B says that a bungee
Pavel [41]

Answer:

Technician A

Explanation:

1- It's was easy to understand.

2- Holding down a battery could actually cause battery plate damage. Most of the batteries I interact with only use a plastic cover and a small spring to hold the battery in place.

-Hope this helped :) -

7 0
2 years ago
A heating system must maintain the interior of a building at 20°C during a period when the outside air temperature is 5°C and th
Anettt [7]

Answer:

a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Explanation:

(COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }

W=Q_{1} \frac{T_{1}-T_{2}  }{T_{1} }

a. the ground at 15°C.

T_{1}=20°C = 273 K + 20 = 293 K

T_{2}=15°C = 273 K + 15 = 288 K

Q_{1}=3x10^{6} kJ

W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}

W = 0.051195x10^{6} kJ

W = 51,194.54 kJ

b. a pond at 10°C.

T_{2}=10°C = 273 K + 10 = 283 K

W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}

W = 0.102390x10^{6} kJ

W = 102,390 kJ

c. the outside air at 5°C.

T_{2}=5°C = 273 K + 5 = 278 K

W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}

W = 0.153585x10^{6} kJ

W = 153,585 kJ

Hope this helps!

3 0
3 years ago
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