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2 years 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]2 years ago

4 0

Explanation:

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masha68 [24]

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3 years ago

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The spring has a stiffness k = 200 N>m and an unstretched length of 0.5 m. If it is attached to the 3-kg smooth collar and th

ruslelena [56]

Answer:

15.467 m/s

Explanation:

See attached picture.

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4 years ago

A 24-tooth gear has AGMA standard full-depth involute teeth with diametral pitch of 12. Calculate the pitch diameter, circular p

torisob [31]

Answer:

Explanation:

Given:

Tooth Number, N = 24

Diametral pitch pd = 12

pitch diameter, d = N/pd = 24/12 = 2in

circular pitch, pc = π/pd = 3.142/12 = 0.2618in

Addendum, a = 1/pd = 1/12 =0.08333in

Dedendum, b = 1.25/pd = 0.10417in

Tooth thickness, t = 0.5pc = 0,5 * 0.2618 = 0.1309in

Clearance, c = 0.25/pd = 0.25/12 = 0.02083in

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3 years ago

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Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 20 µm, J = 0

Naily [24]

Question:

Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.

(a) Find the total current crossing the surface z = 0.1 m in the az direction.

(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.

Answer:

a. -39.8μA

b. -15.81mC/m³

c. 29.05m/s

Explanation:

Given

Density = J = −10^6z^1.5az A/m²

Region: 0 ≤ ρ ≤ 20 µm

ρ ≥ 20 µm

J = 0.

a. Total current is calculated by.

J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.

Where J = Density = -10^6 * z^1.5

ρ1 = Upper bound of ρ = 20

ρ0 = Lower bound of ρ = 0

π = 22/7

φdza = 10^-6

z = 0.1

Total current

= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= −39.7543477278310

= -39.8μA

b. Calculating velocity charge density at (ρv)

Density (J) = ρv * V

Where J = Density = -10^6 * z^1.5

V = 2 * 10^6

z = 0.1

Substitute the above values

-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6

ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)

ρv = -0.1^1.5/(2)

ρv = -0.015811388300841

ρv = -0.01581 --------- Approximated

ρv = -15.81mC/m³

c. Calculating Velocity

Velocity = J/V

Where Velocity Charge Density = -2000 C/m3

Where J = -10^6 * z^1.5

z = 0.15

J = -10^6 * 0.15^1.5

J = -58094.75019311125

Velocity = -58094.75019311125/-2000

Velocity = 29.047375096555625m/s

Velocity = 29.05m/s

8 0

3 years ago

Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and

ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, $Q=16 m^{3}/min$

Diameter d= 8cm= 0.08 m

Assumptions

The flow is jet flow hence momentum-flux correction factor is unity
Gravitational force is not considered
The flow is steady, frictionless and incompressible
Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

$V=\frac {Q}{A}$ where V is the exit velocity and A is area

Area, $A=\frac {\pi d^{2}{4}$ where d is the diameter

By substitution

$V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min$

To convert v to m/s from m/s, we simply divide it by 60 hence

$V=\frac {3183.098862 m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s$

3 0

3 years ago