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 <em>u</em>, final velocity <em>v</em>, and distance<em> s</em>, 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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