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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 electrical symbols. 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 electrical symbols.

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son4ous [18]2 years ago

4 0

Explanation:

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

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You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0

3 years ago

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resis

Rufina [12.5K]

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

Therefore, the relation IV > I²R holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

6 0

3 years ago

A water tank is completely filled with liquid waterat 20°C.The tank material is such that it can withstand tensioncaused by a vo

Xelga [282]

Answer:

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Explanation:

From literature, the coefficient of volume expansion of water between 20°C and 50°C = β = (0.377 × 10⁻³) K⁻¹

Volume expansivity is given by

ΔV = V β ΔT

ΔV = Change in volume

V = initial volume

β = Coefficient of volume expansion = (0.377 × 10⁻³) K⁻¹ = 0.000377 K⁻¹

ΔT = Change in temperature = ?

It is given in the question that maximum volume increase the tank can withstand is

(ΔV/V) × 100% = 0.8%

(ΔV/V) = 0.008

V β ΔT = ΔV

β ΔT = (ΔV/V)

β ΔT = 0.008

ΔT = (0.008/β)

ΔT = (0.008/0.000377)

ΔT = 21.22°C

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Hope this Helps

5 0

4 years ago

A sports car has a drag coefficient of 0.29 and a frontal area of 20 ft2, and is travelling at a speed of 120 mi/hour. How much

Andrej [43]

Answer:

Power required to overcome aerodynamic drag is 50.971 KW

Explanation:

For explanation see the picture attached

4 0

3 years ago

A coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is placed in a magnetic field of 0.56 T.

SVEN [57.7K]

Answer:

Explanation:

it is given that diameter = 8.6 cm

$radius =\frac{8.6}{2}=4.3\ cm=4.3\times 10^{-2}\ m$

current =2.7 ampere

number of turns = 15

$area =\pi r^2=3.14\times \left ( 4.3\times 10^{-2} \right )^{2}=0.005806 m^{2}$

magnetic field =0.56 T

maximum torque= BINASINΘ for maximum torque sinΘ=1

so maximum torque==0.56×2.7×0.005806×15=0.13174 Nm

4 0

3 years ago