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PHY 320 Intermediate Physics Laboratory II. Important items of business. 1. Go to course web site to print copy of syllabus; review syllabus 2. All PHY 320 students are required to conduct the experiment that measures the half-life of a radionuclide; safety matters discussed today. - PowerPoint PPT Presentation
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PHY 320 Intermediate Physics Laboratory II
Important items of business
1. Go to course web site to print copy of syllabus; review syllabus
2. All PHY 320 students are required to conduct the experiment that measures the half-life of a radionuclide; safety matters discussed today.
3. Instructor will provide oral critiques of drafts of reports.
4. Note due dates and times for reports in syllabus:
Caution: You should insure the functionality of the apparatus for any experiment long prior to conducting it.
11 Apr Individual Presentations1: Brown, Frost, Gallis, Larkin, Lawton 18 Apr Individual Presentations1: LeGalle, Lechner, Massaro, Mayer, McCauley 25 Apr Individual Presentations1: Savoy,
Sheriff, Snyder, Sutton, Zeits
2 May Group Presentations2
2 Group presentations are assigned as follows:
1. Zeeman Effect: Brown, Frost, Gallis, Larkin, Lawton
2. Nuclear Spectroscopy: LeGalle. Lechner, Massaro, Mayer, McCauley
3. Hall Effect: Savoy, Sheriff, Snyder, Sutton, Zeits
21 Mar Panel discussion on Shroud of Turin data, 3:00 p.m.
Format: Oral arguments before the Supreme Court
Class members to be divided into two groups:
1. Defend paper in Nature2. Challenge paper in Nature
Note: Paper in Nature restricts itself to dating the artifact.
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Faulty Assumption: Scientists and the research papers they publish are free of errors and biases.
… a different example from Nature
1 Sverdrup (Sv) = 106 m3/s
A measure of volume of water transport. Benchmark: The entire global input of fresh water from rivers to the oceans is equal to about 1 Sv.
Q: Is 23 ± 6 significantly different from or inconsistent with 15 ± 6?
How did the authors obtain the “about 30 percent” quoted in the abstract?
(22.9 – 14.8)/22.9 = 0.353
(23 – 15)/23 = 0.35
Why did the authors not compute the change as
(23 – 15)/15 = 0.53 ?