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A proton accelerates from rest in a uniform electric field of 680 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton.

Respuesta :

Answer:

Acceleration, [tex]a=6.51\times 10^{10}\ m/s^2[/tex]                                                

Explanation:

Given that,

Electric field, E = 680 N/C

Speed of the proton, v = 1.3 Mm/s

We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :

[tex]qE=ma[/tex]

a and m are the acceleration and mass of the proton.

[tex]a=\dfrac{qE}{m}[/tex]

[tex]a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}[/tex]

[tex]a=6.51\times 10^{10}\ m/s^2[/tex]

So, the acceleration of the proton is [tex]a=6.51\times 10^{10}\ m/s^2[/tex]. Hence, this is the required solution.

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