What is the acceleration of a proton moving with a speed of 9.5 m/s at right angles to a magnetic field of 1.8 T

Question

What is the acceleration of a proton moving with a speed of 9.5 m/s at right angles to a magnetic field of 1.8 T

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Minh Khuê 1 month 2021-08-15T04:46:07+00:00 1 Answers 5 views 0

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    2021-08-15T04:47:52+00:00

    Explanation:

    F = qvBsin(angle)

    F = 1.6 x 10^-19 x 9.5 x 1.8 x Sin (90)

    F = 273.6 x 10^-20N

    a = F/M

    M of a proton = 1.67 x 10^-27kg

    a = 273.6 x 10^-20/1.67 x 10^-27

    a = 1638323353.29m/s

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