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A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 3V0, what speed would it gain? Group of answer choices

Sagot :

Manetho

Answer:

[tex]v_{0,new} = v0\sqrt{}2[/tex]

Explanation:

Initial work done on the proton is given by, [tex]\DeltaW0 = q V_o[/tex]

we know that, [tex]\DeltaW = \DeltaK.E[/tex]

[tex]qV0 = (1/2) m v_0^2[/tex]

[tex]v_0 = \sqrt{}2 q V_0 / m[/tex]                                                        { eq.1 }

If it were accelerated instead through a potential difference of 2V0, then it would gain a speed will be given as :

using the above formula, we have

[tex]v_{0,new} = \sqrt{}2 q (2V0) / m[/tex]    

[tex]v_{0,new} = \sqrt{}4 q V0 / m[/tex]    

[tex]v_{0,new} = v0\sqrt{}2[/tex]

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