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A Proton with a Speed of 2 10510^5 M/s Accelerates Through a Potential Difference and Thereby Increases Its

Question 57

Multiple Choice

A proton with a speed of 2.0 x 10510^5 m/s accelerates through a potential difference and thereby increases its speed to 4.0 x 10510^5 m/s. Through what magnitude potential difference did the proton accelerate?
(e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)


A) 630 V
B) 210 V
C) 840 V
D) 1000 V
E) 100 V

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