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andrezito [222]
2 years ago
9

The National Electrical Code specifies that receptacles in certain areas of a house must have ground fault circuit interrupter p

rotection for personnel. In an older home, existing receptacles in these specified areas may not have GFCI protection. If one of these existing receptacles must be replaced, the replacement receptacle.
A. must be a grounding-type receptacle
B. must be the same type of receptacle as the receptacle being replaced
C. must have ground fault circuit interrupter protection
D. must not be a grounding-type or GFCI receptacle
Engineering
1 answer:
Kaylis [27]2 years ago
6 0

Answer:

b

Explanation:

the NEC has expanded the requirements for ground-fault circuit interrupters (GFCI) to protect anyone who plugs into an electrical system. Initially, it was only required for temporary wiring at construction sites and in dwelling unit bathrooms, but in recent years the Code requirements for GFCI protection have expanded to include many other areas, including commercial occupancies, fountains and swimming pools, and temporary installations, to name a few. (For a complete list of 2002 NEC references, see the sidebar below)

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After a strong storm, a worker does not realize that a power transmission line has fallen on his car and is electrocuted while o
joja [24]

Answer and Explanation:

<u>The correct answer choice is Contacting Overhead Power Lines.</u>

This is because the powerlines are making contact (touching) his car, and the example isn't related to the other answer choices.

<u><em>#teamtrees #PAW (Plant And Water)</em></u>

3 0
2 years ago
A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
In DC electrode positive, how much power is at the work clamp?
Korolek [52]

Answer:

1/3 power

Explanation:

I'm just a smart guy

7 0
2 years ago
What's the difference between accuracy and percision in measuring and gaging?
lidiya [134]

Answer:

The term Accuracy means that how close our result to the original result.

Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.

And the term Precision means how likely we get result like this.

Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.

4 0
3 years ago
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