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Sergeeva-Olga [200]
3 years ago
5

Have you ever felt a shock sliding down a plastic slide (like in a McDonald’s Play Place)? Do you think this is related to elect

ricity? How?
Physics
2 answers:
USPshnik [31]3 years ago
7 0

Answer:

The same effect can be observed on other non-conductive, polymer surfaces like the mesh of a trampoline. If it is sitting in the sun for a while, you can feel the static charge on it when you approach it, even if no one has been on it recently (so it's not just from friction of users' clothing).

Explanation:

There are always stray electrons flying around, everywhere. The plastic of the slides (and other, non-conductive materials) tend to build up spare electrons just from the air currents moving over them, so its not really the sun that makes them collect charge. As the air around the plastic heats up, it loses its density and humidity, meaning that electrons that were sitting on or close to the plastic get stuck there. Air flowing over the plastic deposit more electrons. As the air cools after the sun goes down (or if it gets cloudy enough), the air around the plastic becomes more dense and can hold more humidity, which means that the electrons can be carried away by stray ions faster, and the amount of excess charge decreases.

Aleksandr [31]3 years ago
6 0

Answer:

Yes because while you slide down the slide you are obviously going to wear clothes and the cotton on your clothes rubbing up against that rubber from the slides, are gonna react by sending a static electricity shock feeling  to your body  

Explanation:

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Answer:

4.12\times 10^{-5}\ J.

Explanation:

Given that,

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3 0
3 years ago
"If E = 7.50V and r=0.45Ω, find the minimum value of the voltmeter resistance RV for which the voltmeter reading is within 1.0%
Virty [35]

Answer:

The  minimum value is R_V =44.552\  \Omega

Explanation:

From the question we are given that

                  The voltage is E = 7.5V

                  The internal  resistance is r = 0.45

The objective of this solution is to obtain the minimum value of the voltmeter resistance for which the voltmeter reading is within 1.0% of the emf of the battery

  What is means is that we need to obtain voltmeter resistance such that

                                V = (100% -1%) of E

Where E  is the  e.m.f of the battery and V is the voltmeter reading

                          i.e    V = 99% of E = 0.99 E = 7.425  

Generally

                E = V + ir

     where ir is the internal potential difference of the voltmeter and

                V is the voltmeter reading

 Making i the subject of the formula above

            i = \frac{(E-V)}{r}

               =\frac{7.50-7.425}{0.45}

              = 0.1667 A

Now the current is constant through out the circuit so,

                  V = iR_V

Where  R_V is the value of voltmeter resistance

                Hence R_V = \frac{V}{i}  = \frac{7.425}{0.1667}

                                  =44.552\  \Omega

                       

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Answer:

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