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Anna007 [38]
3 years ago
11

Technician A says the term hot wire refers to the section of the circuit after the load or electric device. Technician says the

ground wire refers to the section of the circuit after the load or electric device. Who is right?
Engineering
1 answer:
zmey [24]3 years ago
4 0

Answer:

  Technician B

Explanation:

In simplistic terms, the "hot wire" connects the load device to the source of electrical energy. The ground wire provides the return path for current from the load device to the energy source. In many circuits, the "ground wire" is at, near, or defined as "ground" potential (the actual potential of the Earth).

Technician A seems to be confused. Technician B is more correct.

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The I26 profile beam is supported and loaded as in the figure. It is required to calculate the vertical movement of the free end
jarptica [38.1K]

Answer:

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Explanation:

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Starting the vehicle's engine and listening to its operation can
miskamm [114]
I pretty sure it’s true because you can start your car and it sounds different that probably means something is wrong
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3 years ago
Which of the following is not an example of heat generation? a)- Exothermic chemical reaction in a solid b)- Endothermic Chemica
Sav [38]

Answer:

b) Endothermic Chemical Reactions in a solid

Explanation:

Endothermic reactions consume energy, which will result in a cooler solid when the reaction finishes.

8 0
3 years ago
The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio
Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

v=23.6m/s

Explanation:

Velocity v_c=18.0km/h

Radius r=21m

initial velocity uu=19=>5.27778

Generally the equation for Angle is mathematically given by

\theta=\frac{v_c}{2r}

\theta=\frac{18}{2*21}

\theta=0.45

\theta=25.7831 \textdegree

Generally

Height of mass

h=\frac{rsin\theta}{\theta}

h=\frac{21sin25.78}{0.45}

h=20.3m

Generally the equation for Work Energy is mathematically given by

0.5mv_0^2+mgh=0.5mv^2

Therefore

v=\sqrt{u^2+2gh}

v=\sqrt{=5.27778^2+2*9.81*20.3}

v=23.6m/s

3 0
3 years ago
In an apartment the interior air temperature is 20°C and exterior air temperatures is 5°C. The wall has inner and outer surface
ELEN [110]

Answer:

20 W/m², 20 W/m², -20 W/m²

Yes, the wall is under steady-state conditions.

Explanation:

Air temperature in room = 20°C

Air temperature outside = 5°C

Wall inner temperature = 16°C

Wall outer temperature = 6°C

Inner heat transfer coefficient = 5 W/m²K

Outer heat transfer coefficient = 20 W/m²K

Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)

q = hΔT

Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)

⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²

Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)

⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²

Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)

⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²

Here the magnitude of the heat flux are same so the wall is under steady-state conditions.

7 0
3 years ago
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