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5 June, 11:09

An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases to 6.80x10^5 m/s in a distance of 4.80 cm. Assume its acceleration is constant.

(a) Determine the magnitude of the force exerted on the electron.

(b) Compare this force (F) with the weight of the electron (Fg), which we ignored. F/Fg

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  1. 5 June, 11:34
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    Answer: A) Force = 3.841*10^-18 N.

    B) force (f) is 4.30 * 10^12 times greater than the weight (Fg).

    Explanation: mass of electronic charge = 9.11*10^-31kg

    v = final velocity = 6.80*10^5 m/s

    u = initial velocity = 2.40 * 10^5 m/s

    S = distance covered = 4.8cm = 0.048m

    a = acceleration

    Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

    Thus, recall that

    v² = u² + 2aS

    (6.80*10^5) ² = (2.40*10^5) ² + 2*a (0.048)

    46.24 * 10^10 = 5.76 * 10^10 + 0.096a

    46.24 * 10^10 - 5.76 * 10^10 = 0.096a

    40.48 * 10^10 = 0.096a

    a = 40.48 * 10^10/0.096

    a = 4.2167*10^12m/s².

    Force = mass * acceleration

    Force = 9.11*10^-31 * 4.2167*10^12

    Force = 3.841*10^-18 N.

    Weight = Fg = mg where g = acceleration due gravity = 9.8m/s²

    Fg = 9.11*10^-31 * 9.8

    Fg = 8.9278 * 10^-30 N

    By comparing the force and the weight, we have that

    F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30 * 10^12.

    This implies that the force (f) is 4.30 * 10^12 times greater than the weight (Fg).
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