7.) A proton accelerates from rest in a uniform electricfield of 670 N/C. At som

Question

7.) A proton accelerates from rest in a uniform electricfield of 670 N/C. At some later instant,its speed is 1.60 X106 m/s(nonrelativistic, because v is much less than the speed oflight). (a) Find the acceleration of the proton.
m/s2
(b) After what time interval does the proton reach this speed?
s
(c) How far does the proton move in this time interval?
m
(d) What is its kinetic energy at the end of this timeinterval?
J (a) Find the acceleration of the proton.
m/s2
(b) After what time interval does the proton reach this speed?
s
(c) How far does the proton move in this time interval?
m
(d) What is its kinetic energy at the end of this timeinterval?
J

Explanation / Answer

a) let acceleration be a, since its charge is q =1.6*10-19 C, mass m = 1.67*10-27 kg we have: F=qE = 1.6*10-19 C*670N/C = 1.072*10-16N = ma a = 1.072*10-16 N / 1.67*10-27 kg =6.42*1011 m/s2 acceleration is:  6.42*1011m/s2 b) let time be t v=at =>t = v/a = 1.60 X106m/s / 6.42*1011 m/s2 =2.49*10-6 s c) distance d = (1/2)at2 = 0.5*6.42*1011m/s2 *(2.49*10-6 s)2 = 1.99m d) Kinetic energy KE = (1/2)mv2 =0.5* 1.67*10-27 kg*(1.60X106 m/s)2 = 2.14*10-15 J b) let time be t v=at =>t = v/a = 1.60 X106m/s / 6.42*1011 m/s2 =2.49*10-6 s c) distance d = (1/2)at2 = 0.5*6.42*1011m/s2 *(2.49*10-6 s)2 = 1.99m d) Kinetic energy KE = (1/2)mv2 =0.5* 1.67*10-27 kg*(1.60X106 m/s)2 = 2.14*10-15 J

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.