Aproton is traveling horizontally to the right at 4.20×10^6m/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 3.50cm. part b: counterclockwise from the left directionpart c: how much time does it take the proton to stop after entering the field? part d: what minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

2630250 N/C, horizontally left

0

1434.825 N/C, horizontally left

Explanation:

m = Mass of particle

u = Initial velocity =

v = Final velocity = 0

t = Time taken

s = Displacement = 3.5 cm

q = Charge of particle =

Force is given by

Acceleration is given by

Magnitude of electric field is 2630250 N/C

Direction is horizontally to the left

The angle counterclockwise from left is zero.

The time taken is

Acceleration is given by

Magnitude of electric field is 1434.825 N/C

Direction is horizontally to the left

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

uranium fissions/breaks up giving out energy and neutrons