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BARSIC [14]
3 years ago
14

Someone help me please.

Physics
1 answer:
Alekssandra [29.7K]3 years ago
7 0

Answer:

ionic

ionic

covalent

covalent (i think)

Explanation:

ionic= donate electrons (gain/lose)... (results in ions that have +/- charge, are no longer neutral)

covalent= share electrons

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
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Answer:

The value is  u  =  3.23 *10^{7} \  m/s

Explanation:

From the question we are told that

   The diameter of the nucleus is d =  5.50 \ fm = 5.50 *10^{-15} \  c

   The charge of the proton that makes up the nucleus is  Q_2 = \frac{12}{2}  * 1.60 *10^{-19} =9.6*10^{-19} \ C

    The energy to be impacted is  KE_f  =  2.30 \  MeV  =  2.30 *10^{6} \  eV =   2.30 *10^{6}  *  1.60 *10^{-19} = 3.68*10^{-13} \  J

Generally the radius of the nucleus is mathematically represented as

         r  =  \frac{d}{2}

=>      r  =  \frac{5.50 *10^{-15}}{2}

=>      r  =  2.75 *10^{-15} \  m

Generally from the law energy conservation we have that

     Initial \  total  \  energy \ of the \  proton =  final \  total  \  energy \ of the \  proton

i.e

    T_i  =  T_f

Here

   T_i  =  KE_i + PE_i

Here KE_i is the initial kinetic energy which is mathematically represented as

       KE_I  =  \frac{1}{2}  *  m * u ^2

Here  PE_i is the initial potential energy of the proton and the value is  0 J given that the proton is moving

Also  T_f is mathematically represented as

         T_f  =  KE_f + PE_f

Here  

        PE_f  is the final potential energy which is mathematically represented as

         PE_f  = \frac{k * Q_1 * Q_2}{r}

Here Q_1 is the charge on the proton with a value of Q_1 =  1.60 *10^{-19} \  C

So

        PE_f  = \frac{9*10^{9} *(1.60 *10^{-19} ) * ( 9.6 *10^{-19})}{ 2.75 *10^{-15}}

=>     PE_f  = 5.027 *10^{-13 } \  J

So  

         KE_i + PE_i  =  KE_f + PE_f

=>       \frac{1}{2}  *  m * u ^2 +0  =  3.68*10^{-13} + 5.027 *10^{-13 }

Here m is the mass of the moving proton with value m =  1.67*10^{-27} \  kg

So

       \frac{1}{2}  *  1.67*10^{-27}  * u ^2 +0  =  3.68*10^{-13} + 5.027 *10^{-13 }

=>      u  =  \sqrt{\frac{3.68*10^{-13} + 5.027 *10^{-13 }}{0.5  *  1.67*10^{-27}} }

=>       u  =  3.23 *10^{7} \  m/s

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