A proton of mass m = 1.67 x 10-27 kg and charge q = 1.60 x 10-19 C is accelerated from rest through a potential difference of 4000 V. What final velocity does it achieve? A proton of mass...

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Answer 1

Answer #1

Kinetic energy = q* $\Delta$ V

0.5mv^2 = q* $\Delta$ V

0.5*1.67*10^-27*v^2 = 1.6*10^-19*4000

v = 8.75*10^5 m/s

so the final velocity is 8.75*10^5 m/s

Answer 2

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