A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 1.70 × 1013 m/s2 in a machine. If the proton has an initia

Question

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 1.70 × 1013 m/s2 in a machine. If the proton has an initial speed of 4.10 × 105 m/s and travels 2.30 cm, what then is (a) its speed and (b) the increase in its kinetic energy?

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Doris 1 year 2021-08-31T01:36:32+00:00 1 Answers 9 views 0

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    2021-08-31T01:37:47+00:00

    Answer:

    a)9.7 × 10⁻⁵m/s

    b)= 6.46 × 10⁻⁶J

    Explanation:

    a) mass m = 1.67 × 10-27 kg

    a =  1.70 × 1013 m/s²

    distance change Δs = 2.30cm = 2.3 × 10⁻²m

    initial speed v(i) = 4.10 × 10⁵m/s

    [tex]v_f^2 = v_i^2 + 2as[/tex]

    = (4.10 × 10⁵)² + 2(1.70 × 10¹³)(2.3 × 10⁻²)

    [tex]v_f = \sqrt{9.5 * 10^1^1} \\ = 9.7 * 10^5[/tex]

    b) we calculate the initial kinetic of its motion

    [tex]K_i = \frac{1}{2} m_pv_f^2\\= \frac{1}{2} * (1.67 * 10^-^2^7)(4.1 * 10^5)^2\\= 1.4 * 10^-^1^6J[/tex]

    we calculate the final kinetic of its motion

    [tex]K_i = \frac{1}{2} m_pv_f^2\\= \frac{1}{2} * (1.67 * 10^-^2^7)(9.7 * 10^5)^2\\= 7.86 * 10^-^1^6J[/tex]

    Finally we calculate the increase in its kinetic energy

    = (7.86 × 10⁻⁶) – (1.4× 10⁻⁶)

    = 6.46 × 10⁻⁶J

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