Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA
An average femur bone (Young's modulus 9.4×109 N/m2) has a length of 0.48 m and an approximate cross-sectional area of 1.72×10–3 m2, and reportedly can support a maximum 2.4×104 N of force.[*][**] When this force is applied, the femur would compress by:
(A) 2.1×10–9 m.
(B) 9.1×10–9 m.
(C) 7.2×10–8 m.
(D) 7.1×10–4 m.
[*] "30 times the weight of an adult," orthopaedicsone.com/display/Review/Femur.
[**] "Weight of average adult: 178 lbs," wolfr.am/8dc2Ohi7.
Correct answer (highlight to unhide): (D)
Hooke's law is given by:
(F/A) = Y·(∆L/L),
such that the amount that the femur would be compressed is:
∆L = (F·L)/(A·Y),
∆L = ((2.4×104 N)·(0.48 m))/((1.72×10–3 m2)·(9.4×109 N/m2)),
∆L = 0.0007125185552 m,
or to two significant figures, the femur would compress by 7.1×10–4 m.
(Response (A) is (F·L·A)/Y; response (B) is (F·A)/(L·Y); response (C) is A/F.)
Sections 70854, 70855, 73320
Exam code: quiz06m45S
(A) : 3 students
(B) : 0 students
(C) : 3 students
(D) : 67 students
Success level: 92%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.17
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