Answered

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This question is divided into two parts. This is part (a) of the question. A proton accelerates from rest in a uniform electric field of 580 N/C. At some later time, its speed is 1.00 x 106 m/s. (a) Find the magnitude of the acceleration of the proton. (Mass of the proton is 1.67 x 10-27 kg and charge is 1.60 x 10-19 C) (in the following options 10^10 m/s^2 is 1010 m/s2)

Sagot :

Numenius

Answer:

The acceleration of proton is 5.56 x 10^10 m/s^2 .

Explanation:

initial velocity, u = 0

Electric field, E = 580 N/C

final speed, v = 10^6 m/s

(a) Let the acceleration is a.

According to the Newton's second law

F = m a = q E

where, q is the charge of proton and m is the mass.

[tex]a= \frac{q E}{m}\\\\a = \frac{1.6\times10^{-19}\times 580}{1.67\times 10^{-27}}\\\\a= 5.56\times 10^{10} m/s^2[/tex]

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