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oee [108]
1 year ago
6

Imagine you are charged to manage a project that aims to install wireless access points (aps) throughout the university campus.

what are the steps to a foolproof plan for this project?
Physics
1 answer:
Vitek1552 [10]1 year ago
4 0

If you are charged to manage a project that aims to install wireless access points (aps) throughout the university campus, the steps to a foolproof plan for this project would be -

1. Recognize all of your network’s needs.

The most crucial step in any WiFi installation is probably knowing what your network needs are.

2. Select the appropriate hardware for your wireless network

Finding the ideal access point is much simpler if your needs are clear, but the wide range of options might be difficult.

3. Recognize your devices’ network restrictions.

It’s crucial to keep in mind that other factors besides your Internet connection and network hardware might affect how well your network performs.

4. Take into account the various cables you’ll need to use.

5. Consider how nearby interference may affect the installation of your wireless access point.

6. Decide where to put your wireless access point.

7. Analyze signal strength prior to making a decision.

To know more about access points (aps) visit:

brainly.com/question/14231305

#SPJ4

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3 years ago
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Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area
yaroslaw [1]

Answer:

5.63\cdot 10^{-6} C

Explanation:

The capacitor of a parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

A is the area of each plate

d is the separation between the plates

\epsilon_0 is the vacuum permittivity

The energy stored in a capacitor instead is given by

U=\frac{1}{2}\frac{Q^2}{C}

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}

And re-arranging it

Q=\sqrt{\frac{2U\epsilon_0 A}{d}}

Now if we substitute

d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J

We find the charge stored on the capacitor:

Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C

7 0
3 years ago
When you drop a pebble into a pond, the energy from the pebble acts on the water and causes waves. What is the wave?
Art [367]

Answer:

A, The water moving away

Explanation:

When the pebble hits the water the surface tension breaks causing the water to separate away and make a ripple in the water.

3 0
2 years ago
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This problem is based on the whole idea of pressure but I’m having trouble on when the area circle formula is included.
Mice21 [21]

Answer:

6.23x10^6Pa

Explanation:

Data obtained from the question include:

F (force) = 490N

r (radius) = 0.005m

A (area of the circlular heel) =?

P (pressure) =.?

First, we'll begin by calculating the area of the circlular heel. This is illustrated below:

Area of circle = πr^2

Area = 22/7 x (0.00)^2

Area = 7.86x10^-5m^2

Pressure is simply force per unit area. It represented mathematically as

Pressure = Force /Area

Pressure = 490/7.86x10^-5

Pressure = 6.23x10^6N/m2

Recall: 1N/m2 = 1Pa

Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa

Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor

8 0
3 years ago
An object has traveled 20 meters in 40 seconds what is its average speed
Mice21 [21]

Hello,


To solve we need to know the formula for speed

The formula is D/T=S     (Distance of time=speed)

Now all we have to do is plug in the numbers.

20/40= 1/2 or 0.5

SO the speed is 0.5 m/s


Have a great day!

6 0
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