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Mariana [72]
2 years ago
9

14. You’re setting up a home network for your neighbor, who is a music teacher. She has students visiting her home regularly for

lessons and wants to provide Internet access for their parents while they’re waiting on the children. However, she is concerned about keeping her own data private. What wireless feature can you configure on her AP to meet her requests?
Engineering
1 answer:
nata0808 [166]2 years ago
7 0

Answer:

The feature to configure is Isolation Option which can either be Guest network or Wireless Isolation

Create a guest ssid separate from the Internal Wi-Fi used at home. The guest ssid ensures that the  parents have a separate point to the internet from the teacher.

Guset SSID ensure path isolation of guest traffic from her personal data traffic

Explanation:

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Six years ago, an 80-kW diesel electric set cost $160,000. The cost index for this class of equipment six years ago was 187 and
exis [7]

Answer:

new boiler total cost = $229706.825

new boiler total cost = $127512

Explanation:

given data

power p1 = 80 kW

cost C = $160000

cost index CI 1 = 187

cost index CI 2= 194

cost capacity factor f = 0.6

power p2 = 120 kW

current cost = $18000

to find out

total cost and cost of 40 kW

solution

we consider here CN cost for new boiler and CO cost for old boiler

and x is capacity of new boiler

first we find old boiler current cost that is

current cost CO = C × \frac{CI 1 }{CI 2 }   .............1

put here value

current cost = 160000 × \frac{194 }{187 }

new current cost = $165989.304

and

use here power sizing technique for 124 kW

CN/CO = (\frac{p2}{p1} )^{f}    ...............2

put here value and find CN

CN/CO = (\frac{p2}{p1} )^{f}  

CN / 165989.304 = (\frac{120}{80} )^{0.6}  

CN = 211706.825

so new cost = $211706.825

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 211706.825 + 18000

new boiler total cost = $229706.825

and

for 40 kW new cost will be

use equation 2

CN/CO = (\frac{p2}{p1} )^{f}

CN / 165989.304 = (\frac{40}{80} )^{0.6}  

CN = 109512

so new cost is $109512

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 109512 + 18000

new boiler total cost = $127512

7 0
2 years ago
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a
Fed [463]

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = 2*10^{8}  N/m^{2}

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/(2*10^{8}) = 10^{-5} m^2

recall that area = \pi d^{2} /4

10^{-5} = \frac{3.142*d^{2} }{4} = 0.7855d^{2}

d^{2} = \frac{10^{-5} }{0.7855} = 1.273*10^{-5}

d = \sqrt{1.273*10^{-5}  } = 3.57*10^{-3} m = 3.57 mm

<em>maximum diameter of  the steel linkage d = 3.57 mm</em>

4 0
2 years ago
An electrical utility delivers 6.25E10 kWh of power to its customers in a year. What is the average power required during the ye
Sindrei [870]

Answer:

The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh

However, the average power P, required for a year, t  = ? Kw

Expressing their relationship, we will have

             U = P x t

Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)

          t = 8760 hours

P = \frac{62500000000}{8760}

P = 7134.7Kw

Hence, the average power required during the year is 7,135Kw

Now to calculate the energy used by the power plant in a year (in quads)?

Recall, Efficiency, η = Power Output/Power Input (100)

so, we have

η = P₀/P₁, given

0.45 = \frac{7134.7Kw}{P₁}

P₁ = 15,855Kw

the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules

So to convert this to quads, Note;

1 quads of energy = 10¹⁵ Joules

The total energy used is 0.000000139 quads

Now to find the cubic feet of natural gas required to generate this power?

Note: 0.29Kwh of Power generated  = 1 cubic feet of natural gas used

Since, the power plant generated = 62500000000Kwh

The cubic feet of natural gas used = \frac{62500000000}{0.29}

Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.

8 0
2 years ago
In which of the following branches of engineering is the practice not restricted?
fgiga [73]

Answer:

a) civil engineering.

Explanation:

Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.

Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.

Hence, the correct answer is "a)."

6 0
2 years ago
Read 2 more answers
referring to either the CMS file or code book index, what is the cross reference for reduction of a dislocation?
Oksi-84 [34.3K]

Answer:

reposition

Explanation:

if you go to your icd 10 pcs index located on page 1 then  look up reduction the subterm is of a dislocation which leads you to the answer reposition

4 0
2 years ago
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