0/5 points SerPSE10 22.6.OP019. [4040922] A proton is projected in the positive x direction into a region of uniform ele

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0 5 Points Serpse10 22 6 Op019 4040922 A Proton Is Projected In The Positive X Direction Into A Region Of Uniform Ele 1 (18.02 KiB) Viewed 10 times

0 5 Points Serpse10 22 6 Op019 4040922 A Proton Is Projected In The Positive X Direction Into A Region Of Uniform Ele 2 (35.7 KiB) Viewed 10 times

0/5 points SerPSE10 22.6.OP019. [4040922] A proton is projected in the positive x direction into a region of uniform electric field E=(-6.90 x 105) I N/C at t = 0. The proton travels 6.90 cm as it comes to rest. (a) Determine the acceleration of the proton. magnitude direction -Select-X (b) Determine the initial speed of the proton. x 3.02e+06 m/s magnitude direction -Select- x 6.620-13 m/s² X (c) Determine the time interval over which the proton comes to rest. x 4.57e-08 Need Help? Read It Watch It
0/2 points SerPSE10 22.6.OP.020. [4098718] Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 564 N/C. If the particles are free to move, what are their speeds (in m/s) after 47.6 ns? electron X 4.720+06 m/s 2570 m/s proton Solution or Explanation We can use the particle under constant acceleration model for each particle, with a =. is the fundamental electric charge. For the electron, m m = m = 9.11 x 10-31 kg for the proton, m = m₂ = 1.67 x 10-27 kg Now, apply v, - v,+ at, where v, - 0, t= tx 10-9 s, and a = - For the electron, Vfe = Vie + at = 0 + and for the proton, Need Help? at - 0 + Read It eE m eE + (²) = ((1.61 ((1.60 x 10-19 C)(564 N/C)) (4.76 x 10- s) = 4.72 x 10 m/s 9.11 x kg (1.60 x 10-19 C)(564 N/C) 1.67 x 10-27 kg In both cases, F= QE=eE, where e (4.76 x 10-8 s) = 2,570 m/s