1) [tex]6.11\cdot 10^{10} m/s^2[/tex]
The force experienced by the proton is
[tex]F=qE[/tex]
where
[tex]q=1.6\cdot 10^{-19}C[/tex] is the proton charge
[tex]E=635 N/C[/tex] is the strength of the electric field
Substituting into the equation,
[tex]F=(1.6\cdot 10^{-19} C)(635 N/C)=1.02\cdot 10^{-16}N[/tex]
The acceleration of the proton is given by Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
[tex]F=1.02\cdot 10^{-16}N[/tex] is the force exerted on the proton
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the proton's mass
Substituting,
[tex]a=\frac{1.02\cdot 10^{-16}N}{1.67\cdot 10^{-27}kg}=6.11\cdot 10^{10} m/s^2[/tex]
2) [tex]2.13\cdot 10^{-5} s[/tex]
We can use the following equation:
[tex]a=\frac{v-u}{t}[/tex]
where
[tex]a=6.11\cdot 10^{10} m/s^2[/tex] is the acceleration of the proton
[tex]v=1.30\cdot 10^6 m/s[/tex] is the final velocity
u = 0 is the initial velocity
t is the time
Solving the equation for t, we find
[tex]t=\frac{v-u}{a}=\frac{1.30\cdot 10^6 m/s -0}{6.11\cdot 10^{10} m/s^2}=2.13\cdot 10^{-5} s[/tex]
3) 13.86 m
The distance travelled by the proton is given by the equation
[tex]d=ut + \frac{1}{2}at^2[/tex]
where
u = 0 is the initial velocity
[tex]t=2.13\cdot 10^{-5} s[/tex] s the time
[tex]a=6.11\cdot 10^{10} m/s^2[/tex] is the acceleration of the proton
Substituting,
[tex]d=0 + \frac{1}{2}(6.11\cdot 10^{10}m/s^2)(2.13\cdot 10^{-5} s)^2=13.86 m[/tex]
4) [tex]1.41\cdot 10^{-15} J[/tex]
The final kinetic energy of the proton is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where we have
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the proton's mass
[tex]v=1.30\cdot 10^6 m/s[/tex] is the final velocity
Substituting into the formula,
[tex]K=\frac{1}{2}(1.67\cdot 10^{-27}kg)(1.30\cdot 10^6 m/s)^2=1.41\cdot 10^{-15} J[/tex]
