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Rufina [12.5K]
3 years ago
12

If the atoms that share electrons have an unequal attraction for the electrons is called

Physics
1 answer:
Karolina [17]3 years ago
6 0

If the atoms that share electrons have an unequal attraction for electrons, the bond is called a Polar covalent bond.

<h3><u>Explanation:</u></h3>

A covalent chemical bond is formed in case of two different non-metals when one or more electron pairs are shared between bonding atoms. A difference in electronegativity of subsequent atoms of a covalent bond leads to formation of a small net charge around nucleus of each atom, pulling the shared electrons to one side of the bond, to the nucleus which has higher electronegativity.

HCl is an example of polar covalent bond and the HCl bond has Chlorine more electronegative. The bonding electrons are more close to Cl than H and hence Cl is partially negatively charged than H which has partial positive charge (HCl bond : H^{+} - Cl^{-}). When electrons shared in a covalent bond have equal attraction, the bond is a Non-Polar covalent bond.

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The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b
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Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})

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a is the inner radius of the conductor = 0.8 mm

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Substitute in these values in the above equation;

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) =  \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V

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A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
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Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

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      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

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     T_2 = 0.625 * 9.8

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The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

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     \tau  = 0.2505 \ N \cdot m

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