Aproton accelerates from rest in a uniform electric field of 700 n/c. at one later moment, its speed is 1.10 mm/s (nonrelativistic because v is much less than the speed of light). (a) find the acceleration of the proton. m/s^2 (b) over what time interval does the proton reach this s (c) how far does it move in this time m (d) what is its kinetic energy at the end of this j

Explanation:

a ) Force = a charge q in an electric field E = q x E

Acceleration = Force / mass

Acceleration of proton

= q E / m

= 6.7 x 10¹⁰ m /s²

b ) Initial speed u = 0 , Final speed v = 1.10 x 10⁶ m/s , acceleration a = 6.7 x 10¹⁰ m /s²

v = u + at

1.1 x 10⁶ = 0 +6.7 x 10¹⁰ t

t = 16.4 x 10⁻⁶ s

c ) s = ut + 1/2 at²

s = 0 +.5 x 6.7 x 10¹⁰ x (16.4x 10⁻⁶ )²

= 9.01 m .

d ) K E = 1/2 m v²

= .5 x 1.67 x 10⁻²⁷x (1.1 x 10⁶ )²

= 10⁻¹⁵ J

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer