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14 June, 05:11

Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant hor-izontal force on a charged object. A 15 g charged projectile is launched with a speed of 6.0 m/s at an angle 35° above the hori-zontal. It lands 2.9 m in front of the launcher. What is the magni-tude of the electric force on the projectile?

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  1. 14 June, 08:09
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    Magnitude of electric force = 0.03345 N

    Explanation:

    We are given;

    Mass; m = 15g = 0.015kg

    Angle above horizontal; θ = 35°

    Speed; v = 6 m/s

    Horizontal displacement; d = 2.9m

    Now formula for time of flight is given as;

    time of flight; t = (2Vsinθ) / g

    Thus, plugging in values, we have

    t = (2 x 6.0 x sin35) / 9.8

    t = (12 x 0.5736) / 9.8

    t = 0.7024 s

    Now, let's find the acceleration

    The formula for horizontal displacement is given by;

    d = (Vcosθ) t + (1/2) at²

    Plugging in the relevant values;

    2.9 = [6 (cos35) x 0.7024] + (1/2) a (0.7024) ²

    2.9 = (4.2144 x 0.8192) + (0.2467) a

    2.9 = 3.45 + (0.2467) a

    (0.2467) a = 2.9 - 3.45

    a = - 0.55/0.2467

    a = - 2.23 m/s²

    Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²

    We know that F = ma

    Thus, Force = 0.015kg x 2.23m/s² =

    = 0.03345 N
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