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Ainat [17]
3 years ago
11

The parents of a young child carefully placed plastic plug protectors in all the electrical outlets in their house. Which statem

ent best describes the role of the plastic protectors?
a. Electricity is transported through wires, which are good conductors of electricity. To prevent the flow of electricity into other objects, a material that will prevent the flow of electricity from escaping the wires needs to be used. Certain types of plastics, which are good conductors, are one of many materials that can be used.

b. Electrical outlets contain metal wires that are good conductors of electricity. In order to promote electrical safety, the parents covered the outlets with plastic, an insulator that prevents electrical current from flowing to another object.

c. Plastic is a good conductor of electricity. Placing plastic covers over the electrical outlets allows the electrical current to be redirected back to the plug and ensures the safety of those around the outlets.

d. Plastics and metals are good conductors. This is why plastics can be used to prevent the flow of electricity from materials, like metal wires, to other objects. By preventing the flow of electricity, these help keep people safe.
Physics
2 answers:
dimulka [17.4K]3 years ago
6 0
The answer is B since A, C and D all say that plastic is a good conductor but it isn't. It is an insulator that is there to stop the electrical current being passed on to other items so nobody gets electricuted so B must be correct.
DanielleElmas [232]3 years ago
4 0
I think the answer is c or b but i p ick c
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A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0
defon

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

I=\frac{9.0}{2.0}=4.5 A

2)

To solve this problem, we can use again the same formula

I=\frac{q}{t}

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I is the current

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In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s

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