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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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The correct option is;

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This system consists of a marble rolling down a ramp. Assume the mass of the marble is 5.0 g. Decide what would be needed to mea
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U₁ = m. g. h

As we know m, and g is a constant equal to 9.8 m/s², we will need to measure  the height h, either directly, or in an indirect way from the value of the angle that the ramp does with the horizontal, and the measured value of  the distance travelled along the ramp, x.

So, we could write U₁ as follows:

U₁ = m . g. x. sin θ

Now, at the bottom of the ramp, neglecting fricition, all this potential energy must become kinetic energy, as follows:

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Simplifying and solving for v₂ (the speed of the marble at the top of the bottom), we have:

v₂ = √2.g.h

Once the marble has reached to the bottom of the ramp, it has no more net  external forces acting on it (neglecting friction), it must continue moving at constant speed, equal to v₂.

This value can be measured easily, measuring the displacement between 2 points, and the time used to pass between those points, and computing v₂ as follows:

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5 0
3 years ago
An object is at rest on top of a smooth sphere with a radius of ???? = 15.3 m that is buried exactly halfway under the ground. I
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Answer:

10.2 m .

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mgcosθ - R is net force acting, which provides centripetal force

mgcosθ - R = mv² / r

But v² = 2g r( 1-cosθ )   [ object falls by height ( r - r cosθ ).

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When the object is no longer in touch with sphere,

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1/3 r

1/3 x 15.3

5.1 m

Height from the ground

15.3 - 5.1

10.2 m .

4 0
3 years ago
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