Question

A proton is projected in the positive x direction into a region of a uniform electric field →E =(-6.00 × 10⁵) i^ N/C at t=0 . The proton travels 7.00 cm as it comes to rest. Determine (b) its initial speed, and

Answer 1

Answer:

The initial speed of the proton that is projected in the positive x direction into a region of a uniform electric field is 2.84×〖10〗^6  m/s

Explanation:

The equation of motion expresses the speed of an object with initial velocity u, final velocity v, and distance s, as;

v^2= u^2+2 a.s

Making u the subject of the formula,

u^2= -2 a.s

Given that:

a=5.76×〖10〗^13 m/s^2

s =7.00cm or 0.07m

Inserting values,

u^2= -2 ×(-5.76×〖10〗^13  m/s^2)×0.07m

u^2=8.06×〖10〗^12 m^2/s^2

Therefore,

u= √8.06×〖10〗^12 m^2/s^2

u=2.84×〖10〗^6  m/s

Therefore, the initial speed of the proton is 2.84×〖10〗^6  m/s

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