A proton with mass 1.67×1027 kg and charge 1.60×1019 C accelerates from rest in

Question

A proton with mass 1.67×1027 kg and charge 1.60×1019 C accelerates from rest in a uniform electric field of strength 500 N/C. (a) What is the magnitude of the acceleration of the proton? (b) How long does it take the proton to reach a speed of 35,000 m/s? (c) What distance has the proton traveled when it reaches this speed? (d) What is the kinetic energy of the proton at 35,000 m/s?

Explanation / Answer

mp = 1.67x10-27 Qp = 1.6x10-19 , E= 500N/C (a) F = ma = qE => 1.67x10-27 * a =1.6x10-19* 500 => a = 4.79x1010m/s2(b) t=? with speed Vf = 35000m/s,Vi = 0m/s Use Vf = Vi + at 35000 = 0 + 4.79x1010 * t t = 7.3 * 10-7 s (c) Distance: Use distance = Vit+(1/2)at2             = 0 + (1/2)x4.79x1010x(7.3x10-7)2               = 0.0128m (d) kinetic energy =(1/2)mv2(1/2)x1.67x10-27x(35000)2= 1.02x10-18 Jouls

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