A proton moves through an electric potential created by a number of source charges. Its speed is 2.5X105 m/s at a point where the potential

Question

A proton moves through an electric potential created by a number of source charges. Its speed is 2.5X105 m/s at a point where the potential is 1500 V. What will be the proton

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Neala 3 years 2021-07-26T13:26:38+00:00 1 Answers 53 views 0

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    2021-07-26T13:27:57+00:00

    Answer:

    (6.67 × 10⁵) m/s

    Explanation:

    What will be the proton’s speed a short time later when it reaches a point where the potential is -500V.

    According to the law of comservation of energy, the sum of potential and kinetic energy of the proton at any point in time is always constant.

    P.E₁ + K.E₁ = P.E₂ + K.E₂

    P.E₁ = U₁ = qV₁

    q = charge on the proton = (1.602 × 10⁻¹⁹) C

    V₁ = 1500 V

    P.E₁ = U₁ = (1.602 × 10⁻¹⁹) × (1500) = (2.403 × 10⁻¹⁶) J

    K.E₁ = (1/2)mv₁²

    m = mass of a proton = (1.673 × 10⁻²⁷) kg

    v₁ = (2.5 × 10⁵) m/s

    K.E₁ = (1/2)(1.673×10⁻²⁷)(2.5 × 10⁵)² = (5.23 × 10⁻¹⁷) J

    P.E₂ = U₂ = qV₂

    q = charge on the proton = (1.602 × 10⁻¹⁹) C

    V₂ = -500 V

    P.E₂ = U₂ = (1.602 × 10⁻¹⁹) × (-500) = (-8.01 × 10⁻¹⁷) J

    K.E₂ = ?

    P.E₁ + K.E₁ = P.E₂ + K.E₂

    (2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) = (-8.01 × 10⁻¹⁷) + K.E₂

    K.E₂ = (2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) + (8.01 × 10⁻¹⁷) = (3.727 × 10⁻¹⁶) J

    K.E₂ = (1/2)mv₂²

    m = (1.673 × 10⁻²⁷) kg

    v₂ = ?

    (3.727 × 10⁻¹⁶) = (1/2)(1.673 × 10⁻²⁷)v₂²

    v₂² = (4.455 × 10¹¹)

    v₂ = 667,457.863839 m/s = (6.67 × 10⁵) m/s

    Hope this Helps!!!!

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