Question

14 of 18 Constants A proton is traveling horizontally to the right at 4.80×106 m/s . Part A Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50 cm . E = 1.317•10−81.317{\cdot}10^{-8} N/C SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 2 attempts remaining Part B θ = ∘ counterclockwise from the left direction SubmitRequest Answer Part C How much time does it take the proton to stop after entering the field? t = s SubmitRequest Answer Part D What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)? E = N/C SubmitRequest Answer Part E θ =

Answer 1