Aproton accelerates from rest in a uniform electric field of 610 n/c. at some later time, its speed is 1.4 106 m/s.(a) find the magnitude of the acceleration of the proton.(b) how long does it take the proton to reach this speed? (c) how far has it moved in that interval? (d) what is its kinetic energy at the later time?

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

c)  1.70×10⁻¹¹m

d) 1.661×10⁻²⁷KJ

Explanation:

A proton in the field experience a downward force of magnitude,

F = eE. The force of gravity on the proton will be negligible compared to the electric force

F = eE

a= eE/m

= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷

= 5.851× 10¹⁰m/s²

b)

V = u + at

u= 0

v= 1.4106m/s

v= (0)t + at

t= v/a

= 1.4106m/s/5.851 ×10¹⁰

= 2.411×10⁻¹¹s

c)

S = ut + at²

= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²

= 1.70×10⁻¹¹m

d)

Ke = 1/2mv²

    = (1.67×10⁻²⁷×)(1.4106)²/2

 =  1.661×10⁻²⁷KJ

the first, second, and fifth options

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: