The two basic categories of electrical switches are manual and ??

Answers

Answer 1

Explanation:

Single Pole Single Throw Switch

Single Pole Double Throw Switch

Double Pole Single Throw Switch

Double Pole Double Throw Switch

Push Button Switch.

Toggle Switch.

Limit Switch.

Float Switches.

In the simplest case, a switch has two conductive pieces, often metal, called contacts, connected to an external circuit, that touch to complete (make) the circuit, and separate to open (break) the circuit

Answer 2

Two basic categories of electrical switches are manual and automatic.

What are electrical switches?

A switch that opens and closes an electric circuit is typically electromechanical in nature. This, correspondingly, stops and starts the flow of electric current.

A mechanical handle, lever, or rocking mechanism is used to manually activate (or push up or down) a toggle switch. These are frequently employed as switches for lighting.

An electrically controlled switch is a relay. It is made up of a set of working contact terminals and a set of input terminals for one or more control signals. Any number of connections in different contact configurations, such as making contacts, breaking contacts, or combinations of both, may be included on the switch.

Therefore, electrical switches can be divided into two fundamental categories: manual and automatic.

To learn more about electrical switches, refer to the link:

https://brainly.com/question/29045327

#SPJ2


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Answers

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Answers

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Answers

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Solve the Truss. Your end result should be
a diagram with all member and reaction
forces labeled with the amount of force and
either the direction indicated with an arrow
or whether it is under Tension(positive) or
compression(negative).

Answers

Answer:

Please find attached the diagram end result of the truss with the member reaction forces labelled with the amount of forces as required, created with Microsoft Visio

Explanation:

To answer the question, we note that the sum of the forces in the Truss is zero

Therefore, we have, ∑Fₓ = [tex]\Sigma F_y[/tex] = 0 N

The forces on the members CD and AD are therefore as follows;

Horizontal forces;

[tex]F_{CD}[/tex] × cos(45°) - [tex]F_{AD}[/tex] × cos(30.964°) = 100.0 N...(1)

[tex]F_{CD}[/tex] × sin(45°) + [tex]F_{AD}[/tex] × sin(30.964°) = 750.0 N...(2)

Solving the above system of equations using a graphing calculator gives;

[tex]F_{CD}[/tex] ≈ 715.94 N (Tension)

[tex]F_{AD}[/tex] ≈ 473.76 N (Compression)

[tex]F_{Cy}[/tex] + [tex]F_{Ay}[/tex]  = 750.0 N

[tex]F_{Ax}[/tex] = 100.0 N

From ∑M[tex]_c[/tex] = 0, we have;

[tex]F_{Ay}[/tex] × 8 = 750 × 3 + 100 × 3

[tex]F_{Ay}[/tex] = 2550/8 = 318.75 N

∴ [tex]F_{Cy}[/tex]  = 750.0 N - [tex]F_{Ay}[/tex] = 750.0 N - 318.75 N = 431.25 N  

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