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4 June, 01:18

A particle (q = 3.0 mC, m = 20 g) has a speed of 20 m/s when it enters a region where the electric field has a constant magnitude of 80 N/C and a direction which is the same as the velocity of the particle. What is the speed of the particle 3.0 s after it enters this region?

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  1. 4 June, 04:54
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    56 m/s

    Explanation:

    The electirc force applied on the particle by the field will be

    F = q * E

    F = 3*10^-3 * 80 = 0.24 N

    This force will cause an acceleration:

    F = m * a

    a = F/m

    a = 0.24 / 0.02 = 12 m/s^2

    The equation for speed under constant acceleration is:

    V (t) = V0 + a*t

    V (3) = 20 + 12 * t = 56 m/s

    The final speed will be 56 m/s.
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