A proton accelerates from rest in a uniform electric field of 660 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because

Question

A proton accelerates from rest in a uniform electric field of 660 N/C. At one later moment, its speed is 1.30 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? m (d) What is its kinetic energy at the end of this interval? J

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bonexptip 4 months 2021-08-01T14:03:46+00:00 1 Answers 2 views 0

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    2021-08-01T14:04:56+00:00

    Answer:

    a)  a = 6.31 10¹⁰ m / s²
    , b) t = 2.06 10⁻⁵ s
    , c)  x = 13.39 m
    , d)  ΔK = 1.41 10⁻¹⁵ J

    Explanation:

    a) Since they indicate that the speeds are non-relativistic, we can use the kinematics relations and Newton’s second law

                      F = m a

                     

    The force in electrical is

                     F = qE

                    qE = m a

                    a = qE / m

    we calculate

                    a = 1.6 10⁻¹⁹ 660 / 1.673 10⁻²⁷

                    a = 6.31 10¹⁰ m / s²

    b) Let’s use the one-dimensional kinematics relation

                     v = v₀ + a t

    as part of rest its initial velocity is zero

                     v = a t

                     t = v / a

                     t = 1.30 10⁶ / 6.31 10¹⁰

                     t = 2.06 10⁻⁵ s

    c) We use the kinematics displacement equation

                     x = v₀ t + ½ a t²

    initial velocity is zero

                    x = ½ a t²

                    x = ½ 6.31 10¹⁰ (2.06 10⁻⁵)²

                    x = 1,339 10¹ m

                    x = 13.39 m

    d) the kinetic energy is

                ΔK = Kf -K₀

                ΔK = ½ m v² – 0

                ΔK = ½ 1.673 10⁻²⁷ (1.30 10⁶) 2

                ΔK = 1.41 10⁻¹⁵ J

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