Two protons are released from rest when they are 0.750 {\rm nm} apart. a) What is the maximum speed they will reach?

Question

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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Thái Dương 5 years 2021-09-02T11:44:49+00:00 1 Answers 1102 views 2

Answers ( )

  1. Jezebel
    2
    2021-09-02T11:46:48+00:00
    Reply

    Explanation:

    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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