Question

1. A proton has a speed of 1.5 x 10 m/s at a point where the electrical potential is 1000 V. It moves through a point where the electric potential is 900 V. What is its speed at this second point? (e = 1.60 x 10-19, mproton = 1.67 * 10-27 kg) 3.55 x 10 m/s 0.53 * 10 m/s 2.04 * 1010 m/s 2.04 x 10 m/s 04.16 1010 m/s

Answer 1

giver speed at first point u = 1.5x10 mis potential at first point v = 1000 v Potential at second point I = 900v speed at second point U₂ = 2 We know mu, 2+1,9 = 2 m ² + 2 hmyz mupe= = V, q - V₂ q 2 m ( 4 ² 4; 2) = (11-12) q ų ² _ u,2 2 (Y,-V₂) q 2(1000-900) (1.6% to'9) 1.674 1027 = 1.916101010 Uz² = 4,7+(1.9161x1000) = (1.5x107²+ (1.9161410) = 4.166x100 u - Ju. 16681010 = 12.04xlosmis