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2 June, 22:01

Two protons are released from rest when they are 0.750 {/rm nm} apart.

a) What is the maximum speed they will reach?

b) What is the max acceleration they will reach?

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Answers (1)
  1. 2 June, 23:11
    0
    Given:

    m = 1.673 * 10^-27 kg

    Q = q = 1.602 * 10^-19 C

    r = 0.75 nm

    = 0.75 * 10^-9 m

    A.

    Energy, U = (kQq) / r

    Ut = 1/2 mv^2 + 1/2 mv^2

    1.673 * 10^-27 * v^2 = (8.99 * 10^9 * (1.602 * 10^-19) ^2) / 0.75 * 10^-9

    v = 1.356 * 10^4 m/s

    B.

    F = (kQq) / r^2

    F = m * a

    1.673 * 10^-27 * a = ((8.99 * 10^9 * (1.602 * 10-19) ^2) / (0.075 * 10^-9) ^2

    a = 2.45 * 10^17 m/s^2.
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