Question

A proton accelerates from rest in a uniform electric field of 650 N/C. At one later moment, its speed is 1.40 Mm/s (nonrelativistic because v is much less than the speed of light).

Answer 1

(a) the acceleration a = 6.23 x 10^10 m/s²

(b) the time interview required to reach that velocity is t = 22.5microseconds

(c) the distance traveled is 15.7m

(d) K.E = 1.64 x 10^-15 J

Step-by-step explanation:

Answer:

(a) the acceleration a = 6.23 x 10^10 m/s²

(b) the time interview required to reach that velocity is t = 22.5microseconds

(c) the distance traveled is 15.7m

(d) K.E = 1.64 x 10^-15 J

Step-by-step explanation:

The detailed step by step solution to this problem can be forced below. The electric force on the charge is equal in magnitude to (Eq) front columb's law. The electric force on the charge gives it an acceleration of magnitude a. The force is also equal in magnitude to (ma) from newton's second law.

The acceleration of the charge in the electric field is constant and as a result the equations for constant acceleration motion applies to its motion.

KE = 1/2(MV²)

The complete solution can be found in the attachment below.