Aproton is projected in the positive x direction into a region of uniform electric field e (-6.60 x 105) i n/c at t 0. the proton travels 7.20 cm as it comes to rest. (a) determine the acceleration of the proton. magnitude direction elect m/s2 (b) determine the initial speed of the proton. magnitude directionselect. m/s (c) determine the time interval over which the proton comes to rest.

a = - 6.23 10 13 m / s²   ,   Vo = 3.0 10⁶ m / s    and   t = 4.8 10⁻⁸ s

Explanation:

a) To find the acceleration we use Newton's second law

     F = ma

 Where the strength is

    F = q E

    q E = ma

    a = qE / m

    a = 1.6 10-19 (-6.6 105) / 1.67 10-27

    a = - 6.23 10 13 m / s²

b) Let's use the kinematic equations to find the speed

     X = 7.20 cm (1m / 100cm) = 7.2 10⁻² m

     Vf² = Vo² - 2 to x

     0 = Vo² - 2ax

     Vo = √ 2ax

     Vo = √ (2 6.23 1013 7.2 10⁻²

     Vo = 3.0 10⁶ m / s

The direction is in the direction posita of the field, so that the repulsive force between the field that goes to the left brakes the particle

C) Let's calculate the time for the proto to stop

       Vf = Vo - at

        0 = Vo - at

        t = Vo / a

        t = 3.0 10⁶ /6.23 10¹³

        t = 4.8 10⁻⁸ s

pe = weightxheight = 2000x1000=2,000,000joules

temp rises, and expansion