contestada

A proton accelerates from rest in a uniform electric field of 700 N/C. At one later moment, its speed is 1.10 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. m/s2 (b) Over what time interval does the proton reach this speed? s (c) How far does it move in this time interval?

Respuesta :

jorgezarate

Answer:

[tex]a=6.75*10^{10}m/s^{2}[/tex]

[tex]t=1.6*10^{-5}s[/tex]

[tex]x=8.64m[/tex]

Explanation:

[tex]F=Eq=ma[/tex]

[tex]a=Eq/m[/tex]

charge and masse of proton

[tex]q=1.61*10^{-19}C[/tex]

[tex]m=1.67*10^{-27}kg[/tex]

so: [tex]a=700*1.61*10^{-19}/(1.67*10^{-27})=67485029940m/s^{2}[/tex]

Speed:

[tex]v_{f}=at[/tex]

[tex]t=v_{f}/a=1.10*10^{6}/67485029940=1.6*10^{-5}s[/tex]

Distance:

[tex]x= 1/2*at^{2}=1/2*67485029940*(1.6*10^{-5})^{2}=8.64m[/tex]

Otras preguntas