Physics 205A Quiz 6, fall semester 2012
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 10.3
A "000" AWG-gauge copper wire[*] with Young's modulus of 1.2×1011 Pa is 1.0 m long and has a cross-sectional area of 85.0 mm2. If a weight of 300 N is hung from the wire, it will stretch:
(A) 3.6×10–9 m.
(B) 2.1×10–7 m.
(C) 2.8×10–7 m.
(D) 2.9×10–5 m.
[*] wki.pe/American_wire_gauge.
Correct answer (highlight to unhide): (D)
Hooke's law for elastic materials is given by:
(F/A) = Y·(∆L/L),
where the cross-sectional area of the wire is 85.0 mm2 = 8.50×10–5 m2, such that the length that the amount the wire stretches is:
∆L = (F·L)/(A·Y)
∆L = ((300 N)·(1.0 m))/((8.50×10–5 m2)·(1.2×1011 Pa) = 2.941176471×10–5 m,
or to two significant figures, ∆L = 2.9×10–5 m.
(Response (A) is F·L·(1.2×10–11 Pa); response (B) is F·L·(85.0 m2)/Y; and response (C) is A/F.)
Sections 70854, 70855
Exam code: quiz06How3
(A) : 2 students
(B) : 5 students
(C) : 9 students
(D) : 30 students
Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.52
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