A proton initially moves left to right along the x axis at a speed of 7.0 × 10³ m/s.It moves into an electric field,which points in the negative x direction,and travels a distance of 0.70 m before coming to rest.What acceleration magnitude does the proton experience? A)7.0E+3 m/s² B)3.5E+7 m/s² C)1.8E+9 m/s² D)1.4E+12m/s² E)2.5E+9m/s²

Question

Quizplus · Accepted Answer

The correct answer is B because the acceleration of the proton is given by the equation a = (v^2 - u^2) / 2s, where v is the final velocity, u is the initial velocity, and s is the distance traveled. In this case, v = 0 m/s, u = 7.0 × 10^3 m/s, and s = 0.70 m. Plugging these values into the equation, we get a = (0^2 - (7.0 × 10^3)^2) / 2(0.70) = 3.5 × 10^7 m/s^2.

A Proton Initially Moves Left to Right Along the X