Physics quiz question: pressure difference from climbing one floor

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A wearable fitness tracker has an air pressure sensor[*]:
Your tracker is using an altimeter, which measures when air pressure decreases slightly. It adds one floor every time you increase your elevation by 10 feet [3.0 meters] while moving, which is the average distance between two floors.
(The density of air is ρair = 1.3 kg/m3.) An increase in elevation of 3.0 m experienced by this fitness tracker would correspond to a change in air pressure of:
(A) 4.2 Pa.
(B) 13 Pa.
(C) 38 Pa.
(D) 2.9×104 Pa.

[*] Logan Strain, "How Accurate Is Fitbit? Here's What The Research Says About Fitbit Accuracy" (March 29, 2017), wearablezone.com/news/how-accurate-is-fitbit/.

Correct answer (highlight to unhide): (C)

For static fluids, the energy density relation between pressure difference ∆P and elevation change ∆y is given by:

0 = ∆P + ρair·g·∆y.

The difference in pressure ∆P for climbing one floor is then:

P = –ρair·g·∆y,

P = –ρair·g·(yfy0),

P = –(1.3 kg/m3)·(9.80 m/s2)·((+3.0 m) – (0 m)),

P = –(1.3 kg/m3)·(9.80 m/s2)·(+3.0 m),

P = –38.22 Pa,

such that the difference in pressure for climbing one floor (to two significant figures) is 38 Pa (the negative sign in the above calculation denotes that the air pressure decreased with increasing elevation).

(Response (A) is ρair·g/∆y; response (B) is ρair·g; and response (D) is ρwater·g·∆y.)

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 6 students
(B) : 6 students
(C) : 34 students
(D) : 6 students

Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.80

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