A proton moves through an electric potential created by a number of source charges. Its speed is 2.5 x 105 m/s at a point where the potential is 1500 V. What will be the proton’s speed a short time later when it reaches a point where the potential is −500 V? The mass of a proton is 27 m 1.67 10 kg.

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

the answer is kevlar

snell's law

refractve index = sin incidence / sin refraction