I have the 1st two answers, I just don't know how to calculate the other two. Everything I have tried comes out with a wrong answer.
(a) 479.04e+8 m/s2
(b) 7.30627923e-7 s
(c) m
(d) J
A proton with mass 1.67×10−27 kg and charge 1.60×10−19 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?
5 answers
F = q E = m a
so
a = (q/m)E
= (1.6*10^-19/1.67*10^-27)5*10^2
= 4.79 * 10^10 m/s^2
v = a t
3.5*10^4 = 4.79 * 10^10 t
t = .731 * 10^-6 = 7.31*10^-7 seconds
d = (1/2) a t^2
Ke = (1/2) m v^2
I think you can do c and d
so
a = (q/m)E
= (1.6*10^-19/1.67*10^-27)5*10^2
= 4.79 * 10^10 m/s^2
v = a t
3.5*10^4 = 4.79 * 10^10 t
t = .731 * 10^-6 = 7.31*10^-7 seconds
d = (1/2) a t^2
Ke = (1/2) m v^2
I think you can do c and d
I don't understand what the "^2" means or how to use it in the equation.
The ^2 means squared. v^2 is velocity squared.
what would be the mass of the proton?
0 answers