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1 year ago

The locations and type of electrical device required on an architectural plan are referred to as

2 answers:

5 0

The locations and type of electrical devices required on an architectural plan are referred to as <u>electrical symbols</u>. The use of detailed electrical plan ensures that all electrical equipment and wiring is installed exactly as intended. If the electrical plans are insufficient or hazy, the installation is left to the electrician’s discretion.

Electrical plans are prepared using a CAD layered floor plan. Designers should not rely on electricians to design the electrical system, only to install it. Conversely, designers do not plan the position of every wire, only the position and relationship of all fixtures, devices, switches, and controls. This is done with the use of electrical symbols.

Hundreds of electrical symbols are used on floor plans to describe what and where electrical elements will be installed. On simple plans, electrical symbols are often included as a separate layer on the floor plan. For larger or more complex structures, a separate plan is prepared.

The locations and type of electrical device required on an architectural plan are referred to as <u>electrical symbols</u>.

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son4ous [18]1 year ago

4 0

Explanation:

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In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variab

Aliun [14]

Answer:

The given grammar is :

S = T V ;

V = C X

X = , V | ε

T = float | double

C = z | w

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

First (V) = {z, w}

First (C) = {z, w}

R.H.S.

First (T V ; ) = {float, double }

First ( C X ) = {z, w}

First (, V) = ,

First ( ε ) = ε

First (float) = float

First (double) = double

First (z) = z

First (w) = w

Explanation:

7 0

2 years ago

16 . You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Over [174]

When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

<h3>What is the road about?</h3>

Note that a Yellow centerlines can be seen in roads and it is one that is often used to separate traffic moving in different directions.

Note also that Broken lines can be crossed to allow slower-moving traffic and as such, When a person is turning onto a two-lane road divided by a broken yellow line, you know immediately that you are on a two-way road.

See full question below

You are turning onto a two-lane road divided by a broken yellow line. You know immediately that:

Answers

You are on a two-way road.

You are on a one-way road.

The road is under repair.

You must stay to the left of the broken yellow lines.

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5 0

1 year ago

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A

irina [24]

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.

So; we can say :

$D_{pursuit} =D_{police}$

By using the second equation of motion to find the distance S;

$S= ut + \dfrac{1}{2}at^2$

$D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)$

$D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)$

$D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)$

$D_{police} = ut _P + \dfrac{1}{2}at_p^2$

where ;

u = 0

$D_{police} = \dfrac{1}{2}at_p^2$

$D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2$

$D_{police} = 0.98*(t+12)^2$

$D_{police} = 0.98*(t^2 + 144 + 24t)$

$D_{police} = 0.98t^2 + 141.12 + 23.52t$

Recall that:

$D_{pursuit} =D_{police}$

$(187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t$

$(187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0$

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR t = 3.02

Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s

Total time required for the police car to over take the automobile = 15.02 sec

8 0

2 years ago

The force on a cutting tool are 2600N vertically downward and 2100 horizontal. Determine the resultant force acting on the tool

kolezko [41]

Answer & Explanation:

"The force <em>on a cutting tool</em> are 2600N vertically downward" sounds a little unusual, since most of the time, the tool is above the object to be cut in such a way that the force acting "ON the tool" is upwards. We will accept the statement as it is (downwards).

Since the two forces are acting at right angles to each other, the resultant can be found using Pythagoras theorem, namely

resultant = sqrt(2600^2+2100^2) = 3342 N (approx.)

The angle can be found using the arcTangent function, or

angle = arcTangent(2600/2100) = 51.07 degrees below the horizontal, since the 2600 N force is acting downwards.

3 0

2 years ago

Which of the following distinguishes the importance of engineering to the field of sports biomechanics?

Lisa [10]

The first one and the second i’m glad to help

8 0

3 years ago