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Bogdan [553]
2 years ago
6

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

Engineering
2 answers:
Vikki [24]2 years ago
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>.

Learn more:

https://brainly.in/question/40648976

son4ous [18]2 years ago
4 0

Explanation:

i hope it's helpful for you ......

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Technician A says amperage cannot exist without both voltage and resistance. Technician B says if amperage is high, then you kno
Ivan

Answer:

Technician A

Explanation:

Ohms law:  I= E/R so rest resistance must be present along with E/potential difference.  Even if just wire shorted together there is resistance but very little.

Tech B: Again ohms law.  Current flow is directly proportional to the voltage and inversely  proportional to R (resistance or impedance).

8 0
3 years ago
A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above
creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation N_{o} = N_{i} * \frac{E_{f}-E_{i}  }{KT}

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, N_{d}  = 1.65*10^{16}  - 10^{16} \\           = 6.5 * 10^{15} per cm cube

5 0
3 years ago
Read 2 more answers
Side milling cutter is an example of ______ milling cutter.
dusya [7]

Answer:

special type

Explanation:

As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.

6 0
3 years ago
Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters
alex41 [277]

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

 

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

5 0
3 years ago
An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).
Phoenix [80]

Answer:

The right choice would be Option b (2.545).

Explanation:

The given values are:

The aggregate blend will be:

W_A = 55%

G_ A = 2.631

W_ B = 25%

G_B = 2.331

W_C = 20%

G_C = 2.609

Now,

On applying the formula, we get

⇒ G_{BA}=\frac{W_A+W_B+W_C}{\frac{W_A}{G_A} +\frac{W_B}{G_B} +\frac{W_C}{G_C}}

On substituting the values, we get

⇒          =\frac{55+25+20}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          = \frac{100}{\frac{55}{2.631} +\frac{25}{2.331} +\frac{20}{2.609}}

⇒          =2.545

8 0
3 years ago
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