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Fantom [35]
3 years ago
10

A metal spehere hangs by a thread. when the north of a bar magnet is brought near the sphere is strongly attracted to the magnet

. Then the magnet is reversed and its south pole is borught near the sphere. How does the sphere respond
Physics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

It gets attracted

Explanation:

Materials that attracts magnet gets attracted to the magnet at both the North and South Pole. This can be compared to how neutral objects also gets attracted to the positively and negatively charged rod through the force of polarisation.

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Explanation:

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Mrs. Davis has asked Kyle to retrieve the compound mixture from the chemical cabinet. Which of the following should Kyle retriev
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Why isn't iron used to make coin
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Read 2 more answers
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodie
vivado [14]

Answer:

The length of the rod for the condition on the question to be met is L  =  1.5077 \ m

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is M  =  3.41 \ kg

           The mass of each small bodies is  m =  0.249 \ kg

           The moment of inertia of the three-body system with respect to the described axis is   I  =  0.929 \ kg \cdot  m^2

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        I =  I_r + 2 I_m

Where  I_r is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        I_r  =  \frac{ML^2 }{12}

And   I_m the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           I_m  =   m * [\frac{L} {2} ]^2 =  m*  \frac{L^2}{4}

Thus  2 *  I_m  =  2 *  m  \frac{L^2}{4}  = m  *  \frac{L^2}{2}

Hence

       I  =  M  *   \frac{L^2}{12}  +  m  * \frac{L^2}{2}

=>   I  =    [\frac{M}{12}  + \frac{m}{2}] L^2

substituting vales  we have  

        0.929   =    [\frac{3.41}{12}  + \frac{0.249}{2}] L^2

       L  =  \sqrt{\frac{0.929}{0.40867} }

      L  =  1.5077 \ m

     

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