A proton has a speed of 3.5 x 10^5 m/s at a point where the electrical potential is 600 V. It moves through a point where the electric potential is 1000 V. What is its speed at this second point?
(will mark you brainliest)

4.46 * 10^5 m/s

Explanation:

proton speed = 3.5 * 10 ^ 5 m/s

initial electrical potential = 600 V

second point electrical potential = 1000 V

Applying the law of conservation of energy to determine the speed at the second point

q* ( V2 - V1 ) = 1/2 M ( v2^2 - v1^2 )

making v2( speed at second point) subject of the formula( v2 ) =

V2 = 1000

V1 = 600

M = 1.67262178 * 10 ^-27 kg

q = 1.6 * 10 ^ -19 c

inserting the given values into the given equation

speed at the second point ( v2) = 4.46 * 10 ^5 m/s

note : V = POTENTIAL DIFFERENCE , v = speed