An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached to this spri

Question

An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.022 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

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Tài Đức 3 months 2021-08-22T19:07:37+00:00 1 Answers 2 views 0

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    2021-08-22T19:08:54+00:00

    Answer:

    A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

    B) Magnitude of charges = 1.206 × 10^(-6) C

    Explanation:

    We are given;

    Spring constant;k = 350 N/m

    Spring length;L = 0.39 m

    Stretched length of spring;x = 0.022 m

    A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

    B) Force (F) required to stretch spring is given by the formula;

    F = kx

    Thus:

    F = (350 × 0.022)

    F = 7.7 N

    Now, if we assume point charges, then the distance (r) between them will be given as:

    r = (0.39 + 0.022) = 0.412 m

    Coulomb’s Law has a formula:

    F = k(q1×q2)/r²

    where k is coulomb’s constant = 8.99 × 10^(9) Nm²/C²

    Making q1 × q2 the subject, we have;

    (q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

    (q1 × q2) = 14.54 × 10^(-11) C

    We are told that both charges are equal, thus; |q1| = |q2|

    So;

    q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

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