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NeX [460]
3 years ago
13

For a charged hollow metal sphere with total charge Q and radius R centered on the origin: (Determine whether each of the follow

ing statements is correct or incorrect.) The field inside the shell is zero. The field for r > R will be the same as the field of a point charge, Q, at the origin The charge is on the inside surface. Only + charges can be on the outside surface. The E-field on the outside is perpendicular to the surface. Inside the metal the field is strongest.
Physics
1 answer:
Sloan [31]3 years ago
5 0

Answer:

1. The field inside the shell is 0; TRUE

2. The field for r>R will be the same as a field for the point charge Q, at the origin: TRUE

3. only positive charges can be on the outside surface: FALSE (Negative can too!!!)

4. The field on the outside is perpendicular to the surface: TRUE

Explanation:

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Answer:

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Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

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a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

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Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

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Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

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1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

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b) Now, the kinetic energy of each nucleus is:

For He-4:

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E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

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