1. L2-s1,12 Physics 114 – Lecture 2 Chapter 1 Read: §1.1 The Nature of Science §1.2 Physics and Its Relation to Other Fields §1.3 Models, Theories and Laws Models, e.g., the solar system Law – a statement of a relationship between certain quantities

2. l2_2,12 Physics 114 – Lecture 2 Theory: a set of principles which may be used to explain a wide range of phenomena A Physical Theory MUST, in order for it to be of any value, have predictive power – e.g., on solar eclipses It remains an acceptable theory only until ANY accepted gives a result which disagrees with one or more of its predictions, c.f., Einstein’s remark It has often been our experience that a theory is ultimately found to have a limited range of applications and then needs to be modified or changed

3. L2-3,12 Physics 114 – Lecture 2 §1.4 Measurement and Uncertainty: Significant Figures 12.0 13.0 → 12.8

4. L2-s4.12 Physics 114 – Lecture 2 What is the most reliable statement that one can make in such a situation? l = 12.8 ± 0.05 cmsALWAYS GIVE UNITS Equivalent to saying that l lies between 12.75 and 12.85 cms. Note that the result of this measurement is given to three significant figures

5. L2-5,12 Physics 114 – Lecture 2 This result, l = 12.8 cms is interpreted to mean, if the error is not stated, that l is nearer to 12.8 cms than to 12.7 cms and to 12.9 cms. Rounding a number to, for example, 3sf: 12.476→12.5 12.527 → 12.5 12.327 →12.3 12.288 →12.3

6. L2_s6,12 Physics 114 – Lecture 2 Scientific Notation: 36500 → 3.65 x 10 4 0.00723 → 7.23 x 10 -3 Note that this removes any ambiguity regarding the number of significant figures Derived quantities: if, for example, we have to calculate an area, A = l x w, where l = 12.3 m and w = 5.72 m, we will obtain A = 70.356 m 2, but we must give the result to 3sf → A = 70.4 m 2 The number of significant figures (sf) must be chosen to be equal to the least number of sf in the data.

7. L2-s7,12 Physics 114 – Lecture 2 §1.5 Units, Standards and the SI System Length: the choice is arbitrary. Initially the choice was that 1 meter is 10 -7 x the distance from the equator to either of the poles. A standard meter has been kept in Paris. This was cumbersome to use, so the meter was redefined in 1960 in terms of the wavelength of light emitted by 86 Kr (krypton) – 1,650,763.73 wavelengths of the orange line emitted by these atoms.

8. L2-s8,12 Physics 114 – Lecture 2 The meter was redefined in 1983 to be the distance traveled by light in vacuum in 1/299,792,458 = 1/c of a second, where c = the velocity of light in vacuum Time: the standard unit of time is the second, and was originally defined as 1/86,400 of mean solar day. Note that 24 hours = 86,400 s. It is now defined as the frequency - number of oscillations per sec – of radiation emitted by cesium atoms, the time taken for 9,162,631,770 of these oscillations Note that the standards for length and time are now easily reproduced throughout the world

9. L2-9,12 Physics 114 – Lecture 2 Mass: The standard unit of mass is the kilogram, the mass of a standard cylinder of platinum and iridium, which is maintained in Paris. Prefixes: 1 kilometer = 1,000 m, 1 millimeter = 10 -3 m, etc. Systems of Units: In science the SI – Système Internationale – is used. This is also known as the MKS system. Other units cgs system - centimeter, gram, second - and the British Engineering system – foot, pound for force and second

10. L2- 10,12 Physics 114 – Lecture 2 Derived Quantities: For example, average speed = distance traveled/time elapsed = 27.8m/2.00s = 13.6 m/s – the units referred to as meters per second

11. L2-11,12 Physics 114 – Lecture 2 §1.6 Converting Units The trick is to multiply by 1 as often as needed; E.g., A speed of 50.0 km/hr to m/s. 1 km =1000m → 1 km/1000m =1 – dividing both sides by 1000m or → 1 = 1000m/1 km - dividing both sides by 1 km 1 hr = 3600 s → 1hr/3600s = 1 = 3600s/1hr → 50.0 km/hr =

12. L2-12/12 Physics 114 – Lecture 2 §1.7 Order of Magnitude Estimating Study examples 1-6 through 1-9 Another Example: Time taken to mow a lawn of length 50m and width 30m Area, A = l x w =50m x 30m = 1500 m 2 Assume lawn mower travels at a speed of 0.50 m/s and cuts a swath of width 0.50 m → dist traveled by mower = 1500 m 2 /0.50 m = 3000 m → time taken = 3000 m/0.50 m/s = 6000 s = 6000 s x 1 min/60 s = 100 min – actual mowing