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ehidna [41]
3 years ago
11

How does a rubber rod become negativley charged trough friction/

Physics
2 answers:
iren2701 [21]3 years ago
7 0

Answer:

Hey

here's ur answer ⤵️

Explanation:

It touches a negatively charged object, and protons move off of the rod it is rubbed with another object and electrons move onto the rod.

It is rubbed with another object and protons move off of the rod.

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps </em>

natali 33 [55]3 years ago
3 0

Answer: Answer below hope it helps

Explanation: How does a rubber rod become negatively charged through friction? It touches a negatively charged object, and protons move off of the rod. ... It is rubbed with another object, and electrons move onto the rod. It is rubbed with another object, and protons move off of the rod.

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Answer:

The value is  B =  3.33 *10^{-6} \  T

Explanation:

From the question we are told that

  The distance of separation is  d = 0.6 \  m

  The current on the one wire is I_1 =  9 \  A

  The current on the second wire is I_2 =  4 \ A

Generally the magnitude of the field exerted between the current carrying wire is

        B  =  B_1 - B_2

Here B_1 is the magnetic field due to the first wire which is mathematically represented as

         B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}

Here d_1 is the distance to the half way point of the separation and the value is  

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B_2 is the magnetic field due to the first wire which is mathematically represented as

         B_2  = \frac{\mu_o * I_2 }{2 \pi * d_2}

Here d_2 is the distance to the half way point of the separation and the value is  

    d_2 =  0.3 \  m  

This means that d_1 = d_2 = a =  0.3

So

     B =  \frac{\mu_o * I_1 }{2 \pi * d_1}  -  \frac{\mu_o * I_2 }{2 \pi * d_2}

=>  B =  \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }

=>  B =  \frac{  4\pi * 10^{-7}  * (9- 4)}{2 * 3.142  *0.3 }

=>  B =  3.33 *10^{-6} \  T

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