Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 1.29
The U.S. standard kilogram "K4" was set to be exactly one kilogram in 1889, but by 1999 had lost 4.1×10–5 kg when compared to the international prototype kilogram in Sèvres, France.[*] The "K4" standard is made of Pt‑10Ir, which has a density[**] of 21.5 g/cm3.
The volume lost by the "K4" standard over 110 years compared to the international prototype kilogram is:
(A) 1.9×10–9 cm3.
(B) 8.9×10–7 cm3.
(C) 1.9×10–3 cm3.
(D) 0.89 cm3.
[*] wki.pe/Kilogram.
[**] smt.sandvik.com/en/materials-center/material-datasheets/wire/sandvik-bioline-ptir-alloys/ .
Correct answer (highlight to unhide): (C)
The only relevant conversion factor is 1 kg = 1,000 g. To find volume would require that unwanted units cancel (kg and g) while the desired unit remains (cm3), the only unique way to combine the two given factors (4.1×10–5 kg, 21.5 g/cm3) to do this would then be:
(4.1×10–5 kg)/(21.5 g/cm3) = 1.90697674419×10–6 cm3·(kg/g),
which results in units of cm3 in the numerator, and to eliminate the (kg/g) term:
1.90697674419×10–6 cm3·(
or to two significant figures is 1.9×10–3 cm3.
Sections 70854, 70855, 73320
Exam code: quiz01w34K
(A) : 9 students
(B) : 17 students
(C) : 32 students
(D) : 16 students
Success level: 43%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.53
No comments:
Post a Comment