Ch01 ISM Hawkes 2e - hw answers PDF

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Chapter 1—INTRODUCTION TO PHYSICS 1.

c

2.

Answers will vary according to the physicist interviewed or followed. It is unlikely that all of the aspects outlined will appear in one week of any single physicist. There may be additional aspects related to interactions with government or industry, computational experiences, etc.

3.

Physics and mathematics are both quantitative disciplines, and both stress the importance of logic in reasoning. They will use many of the same mathematical and computational techniques and similar formalisms. The essential difference is that math can be defined independent of validation in the physical universe, while ultimately physics must always deal with topics that can at least in principle be validated with experiments.

4.

The key point is whether the idea of a parallel universe that cannot be observed from our universe meets the requirement of being physics. However, some work has suggested indirect indications of the existence of parallel universes, so either view can be supported.

5.

(a) Some entries possible on the list are engineering, medical professions, environmental science, art, philosophy, writing (certain types), journalism, patent law (some aspects), museum or outreach, electronics, geophysics, astrophysics, and planetary science.

6.

We first estimate the mean by the midpoint of the range of values given:

mean =

3.4 s + 4.8 s = 4.1 s 2

Since in a normal distribution about 2/3 of the values lie within ±1 standard deviation of the mean, the standard deviation is given by

σ = 4.8 s − 4.1 s = 0.7 s 7.

Many examples are possible (see http://www.fastcodesign.com/3030529/infographic-of-theday/hilarious-graphs-prove-that-correlation-isnt-causation#5) for some fun ones! For example, the correlation between margarine consumption and divorce rate in some U.S. states is almost perfect! In addition to the ones on that website, one of my favourites is the correlation between the number of Nobel Prizes won by a country and the per capita chocolate consumption.

8.

We first estimate the mean by the midpoint of the range of values given: mean =

5.8 m + 2.2 m = 4.0 m 2

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Since in a normal distribution about 95% of the values lie within ±2 standard deviations of the mean, we can compute the standard deviation as

σ= 9.

5.8 m − 4.0 m = 0.9 m 2

b

10. (c) Note the period after the number indicating that the zeros are significant in this case. 11. a 12. c 13. c 14. W = Fx

 [ x] =

[W ] J kg ⋅ m2 ⋅ s −2 = = [F ] N kg ⋅ m ⋅ s −2

=m

15. (a) The metre is currently defined as the distance travelled by light in vacuum during a time interval of 1/299 792 458 s. (b) 299 792 458 m/s (c) In 1790, the French National Assembly defined a metre as the length of a pendulum with a half-period of 1 s. The next year, this definition was modified to refer to 1 × 10−5 the distance from the equator to the North Pole. In 1795, a provisional metre brass bar was constructed. In 1799, this was replaced by a platinum metre bar. In 1889, the first General Conference on Weights and Measures (Conférence générale des poids et mesures, CGPM) defined the metre as the distance between two lines on a standard bar of an alloy of platinum with 10% iridium, measured at the melting point of ice. This definition was modified in 1927 to refer to the distance, at 0 °C, between the axes of the two central lines marked on the prototype bar of platinum-iridium at standard pressure and supported on two cylinders of at least 1 cm diameter, symmetrically placed in the same horizontal plane at a distance of 571 mm from each other. In 1960, the metre was redefined as equal to 1 650 763.73 wavelengths in vacuum of the radiation corresponding to the transition between the 2p10 and 5d5 quantum levels of a krypton86 atom. In 1983, CGPM defined the metre as equal to the distance travelled by light in vacuum during a time interval of 1∕299 792 458 s. 16. The duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of a cesium 133 atom at rest at 0 K 17. (a) A kilogram is equal to the mass of the International Prototype Kilogram (IPK). The IPK is made of a platinum-iridium alloy and is stored in a vault at the International Bureau of Weights and Measures in France.

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(b) On April 7, 1795, the gram was defined as equal to “the absolute weight of a volume of water equal to the cube of the hundredth part of the metre, at the temperature of melting ice.” Later on, a metallic artifact was used to define a kilogram. In 1799, an all-platinum kilogram prototype was fabricated, which was subsequently ratified as the Kilogram of the Archives, and the kilogram was defined as being equal to its mass. 18. While the ampere is operationally equal to a current of 1 C passing a point in 1 s, it is derived formally in the following manner (https://physics.nist.gov/cuu/Units/ampere.html): “The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10−7 newton per metre of length.” 19. “The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 × 1012 Hz and that has a radiant intensity in that direction of 1∕683 watts per steradian.” This definition is from “Base unit definitions: Candela” in the NIST Reference on Constants, Units, and Uncertainty (http://physics.nist.gov/cuu/Units/candela.html). Previous definitions: 1946: The value of the new candle is such that the brightness of the full radiator at the temperature of solidification of platinum is 60 new candles per square centimetre. Reference: Taylor, Barry N. (1992). The Metric System: The International System of Units (SI). U.S. Department of Commerce, p. 18. (NIST Special Publication 330, 1991 ed.). ISBN 0-941375-74-9. 1967: The candela is the luminous intensity, in the perpendicular direction, of a surface of 1/600 000 of a square metre of a black body at the temperature of freezing platinum under a pressure of 101 325 newtons per square metre. Reference: 13th CGPM Resolution 5, CR, 104 (1967), and Terrien, J. (1968). ”News from the International Bureau of Weights and Measures.” Metrologia 4: 43–44. doi:10.1088/0026-1394/4/1/006.

20. d 21.

σA NA

=

σB NB

σ A = 2σ B 2σ A σA NA

=

1 = NA

NB

2 NB

NB = 2 NA NB = 4 NA = 4× 16 = 64

Thus, 64 data points are needed.

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22. d 23. Answers will vary extensively according to the journal and article chosen. However, in most cases, there will be evidence of the collaborative, quantitative, model-driven nature of the discipline and use of simplifying concepts and equations. Answers may indicate the theoretical or applied aspects of physics, and possibly there will be direct evidence of creativity and interface with other subjects. The paper of course will represent one form of communication, a key part of science. 24. Answers will vary according to the science news source chosen and the particular story. Note that most research is done by collaborative teams, often with participants from multiple countries. There will frequently be evidence of development of new techniques or novel experiments In most cases, a model will play a central role to be validated (or not) with experimental results. Almost every story will make the quantitative nature of physics obvious. While research may seem complex, urge students to look for examples of simplifying approaches. Finally, the role of communication is obvious in any story. 25. The answers to this exercise will vary extensively. Among topics likely to be listed in the category of unanswered questions in physics are dark energy, dark matter, the nature of gravity, the presence of higher dimensions, life in the universe (only partly a physics topic), the nature of the early universe, and various particle physics questions. Scientific American speculated on a possible top five in 2013: http://blogs.scientificamerican.com/observations/2013/10/25/physics-biggest-unansweredquestions/. In terms of which are likely to be answered in the next decade, one might argue the most likely to be answered are ones in particle physics (ongoing experiments from the LHC have already answered some of these) and possibly the nature of dark matter. We have a wealth of new evidence on exoplanets but seem far from a direct detection of life elsewhere. 26. The interview should demonstrate characteristics of good journalism, such as clearly answering essential questions in a balanced manner and telling a story in an interesting way. If possible, ask a journalist or someone from communications/media relations at your university to speak to the class prior to completion of this exercise. 27. The choice of a subject for a video may be current (e.g., browse physics news sites for inspiration) or historical. Many women physicists have made overlooked contributions; among those that might be considered are Lise Meitner, Irène Curie-Joliot, Rosalind Franklin, Jocelyn Bell Burnell, Cecilia Payne-Gaposchkin, and Vera Rubin. Many physicists who have won Nobel Prizes are not well known in their own country and would be good choices (see problem 75, which develops a list). Consulting lists of award winners by the Canadian Association of Physicists and NSERC will suggest current physicists who do important research. 28. The precision of the ultrasound range finder is better (in that it has a measurement resolution of 0.001 m versus 0.01 m for the metre stick). However, the information indicates that it has a much worse accuracy. The metre stick is accurate to a few hundredths of a metre, while the ultrasonic range finder is probably accurate to a few tenths of a metre. It is perhaps not

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conclusive, but there are some indications that there is a systematic bias in the ultrasound ranger and it reads high. 29. Various answers are possible. For example, a kitchen scale might have a precision of 0.1 g but an accuracy of 0.5 g, or a digital outdoor thermometer might have a precision of 0.1°C but an accuracy of 1°C.

σ

30. SDOM =

=

1.2

N 16 mean = 4.5 ± 0.3

= 0.3

Note that we write the mean to only two significant digits, since the uncertainty in the mean supports that precision. 31. N =

σ2 SDOM 2

=

2.42 =4 1.22

32. The mean is the sum divided by the number of measurements:

x=

x

i

N

=

30.11 = 3.35 9

We calculate the standard deviation from the following relationship (we have rounded the final answer to three significant digits):

σ=

 i =1.. N

( xi − x ) 2 N −1

=



( xi − 3.35) 2

i =1..9

9 −1

=

6.42 = 0.896 8

The SDOM is obtained from

SDOM =

σ N

=

0.90 = 0.299 9

With this information, we would write the mean to an appropriate number of significant digits as 3.4 ± 0.3, since if we are uncertain by this amount it does not make sense to write the mean to more than one digit to the right of the decimal point. 33. The journal article specified 19 times out of 20, or about 95%, so we know that the range given must correspond to two standard deviations on each side of the mean. Therefore, we have the following results for the mean (which is given) and the standard deviation: mean = 10.5 0.5 σ= = 0.25 2

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34. As a technique, have your friend measure the object repeatedly and have someone else use a different and validated instrument to also repeatedly measure the length of the object. For each set of measurements, determine the mean and standard deviation. If your friend’s repeated measurements of the length of the object have one (or more) measurements that are outside the mean by more than the standard deviation would suggest, that would flag possible true errors. Calculate the SDOM for both sets of measurements. If the two means vary by more than that suggested by these SDOM values, a systematic error of some sort is suggested. The standard deviation of your friend’s repeated measurements is an indication of the random errors associated with those measurements. 35. If we use the techniques of this chapter, we obtain a mean of 8.58, a standard deviation of 1.51, and an SDOM of 0.50. We would expect in 95% of the cases that the value will be within two standard deviations. This suggests that the value of 4.56 is probably the result of a true error. If we accept this and recalculate using only the other eight values, we obtain a mean of 9.08, a standard deviation of 0.16, and an SDOM of 0.06 m/s2. This suggests that the random errors are typically 0.16 m/s2. Since the mean is less than the “true” value, it would suggest that our methodology has some sort of systematic error that makes the accelerations low. 36. (a) We have plotted the points and error bars on the graph below, along with an approximate best fit plot line.

(b) Yes, it is possible to have a best fit line, as shown, that passes through all data points within the uncertainty expressed by their error bars. (c) The circle is the new point. The amount that its positive-pointing y-axis error bar would need to be at a minimum is drawn with the dashed line. It implies a y-axis error bar of at least +3.5 (no constraint is placed on the negative-pointing error bar). 37. (a) The Anscombe’s quartet is a collection of four datasets that have the same statistical measures (e.g., same mean, same standard deviation) but very different natures, which are only visible when the data are actually plotted. The idea was developed and popularized by the U.K. statistician Francis John “Frank” Anscombe (1918–2001). (b) This is the commonly reproduced Anscombe’s set from Wikipedia. Of course this is not the only solution possible.

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Source: By Anscombe.svg: Schutz Derivative works of this file:(label using subscripts): Avenue (Anscombe.svg) [GPL (http://www.gnu.org/licenses/gpl.html), CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons

(c) The answer in part (b) shows the importance of plotting, at least roughly, any data being analyzed and not depending only on the mean and the standard deviation to characterize the data. 38. (a) 4 (b) 2 (c) 4 (d) 2 (e) 3 (f) 4 39. 17.39 ± 0.04 40. (a) The subtraction yields 0.556 but should only be expressed to one significant digit, so the final answer is 0.6. (b) 6.551 to the correct number of significant digits 41. In the numerator, the first term has three significant digits and the second has four significant digits. Therefore, the product will have three significant digits. The denominator has four significant digits; therefore, the final answer will have the lesser of three and four significant digits and is written as 25.8. 42. The numbers add to 25.95 m, but according to the rule for addition we should keep the number relative to the decimal point of the least significant input (the 12 and 9 in this case). Therefore, it should be expressed as 26 m.

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28.3 m − 27.5 m = 0.8 m/s. After the distance subtraction, we only have one significant 1.00 s − 0.00 s digit, so the final answer will have one significant digit.

43. v =

44.

45.

34.56 − 29.35 5.21 = = 0.944. Here we have three significant digits in both numerator and 8.75 − 3.23 5.52 denominator after the subtractions, so the final answer should be expressed to three significant digits. 5.16 + 9.12 14.28 = = 3.586. Here the original values in the numerator were three 2.750 + 1.232 3.982 significant digits, but when added according to the rules they yield a numerator with four significant digits. The denominator addition has four significant digits. Therefore, by the division rule we should express the answer to four significant digits as well.

46. (a) When we take 10% of 5.00 m, we obtain 0.5 m. Therefore, it is not written correctly and should be written as 5.0 m. (b) In the standard way of expressing the result, it would be 5.0 ± 0.5 m. 47. (a) When we take 1% of 8.0 m, we obtain 0.08 m. Therefore, it is not written correctly and should be written as 8.00 m. (b) In the standard way of expressing the result, it would be 8.00 ± 0.08 m. 48. (a) 2.452 × 103 (b) 5.92 × 10−1 (c) 1.2 × 104 (d) 4.5 × 10−5 49. (a) 8.900 × 103 (b) 8.9 × 103; yes, it is different since there are now just two significant digits. 50. 7.5 × 105; the numerator product has two significant digits, so the overall answer also only has two significant digits. 51. (a) 3670 (b) 0.000 000 000 002 25 (c) 240 000 (d) 1200.

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52. kg/m3 53. (a)

(b)

N m2 kg ⋅ m kg = 2 2 m ⋅s m ⋅ s2

54. N ⋅ m =

kg ⋅ m ⋅ m kg ⋅ m2 = s2 s2

55. [RC] = Ω∙F = m2∙kg∙s−3∙A−2∙m−2∙kg−1∙s4∙A2 = s

kg ⋅ m N kg ⋅ m kg ⋅ m V m 2 ⋅ kg 56. , = = 3 = 2 = 3 3 C s ⋅s ⋅ A s ⋅ A m m⋅ s ⋅ A s ⋅ A 57. We know that [U] = kg∙m2∙s−2, [m] = kg, [g] = m∙s−2, and [h] = m, so U ~ mgh. 58.

[ R] =

[PV ] = ⋅ 2 ⋅ −2 ⋅ −1 ⋅ −1 = ⋅ −1 ⋅ −1 kg m s K mol J K mol [ nT ]

59. We see from Table 1-3 that the units of V (in terms of base SI units) are m2∙kg∙s−3∙A−1, the units for I are simply A, and the units of R are m2∙kg∙s−3∙A−2. We can see by inspection that the relationship V ~ IR makes the units on both sides the same. 60. [ c] =

[F ] kg ⋅ m ⋅ s−2 = kg ⋅ m−1 2 = 2 −2 [v ] m ⋅s

61. [ b] =

[F ] kg ⋅ m ⋅ s −2 = = kg ⋅ s −1 −1 [ v] m ⋅s

62.

75.0 km 1000 m 1h × × = 20.8 m/s h 1 km 3600 s

63. 1

cm 1m 109 nm 1 yr × × × = 0.32 nm/s yr 100 cm 1m 365.25 days/yr× 24 h/day× 60 min/h× 60 s/min

Therefore, 1 cm/year to 2 cm/year will be 0.3 nm/s to 0.6 nm/s, to one significant digit. 2

2

1000 m   1000 mm  1.00 1012 mm2 64. 1.00 km2 ×  ×  ×  =  1 km   1 m 

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65. We will approximate the number by dividing the total volume by the volume of a single cell of yeast. The radius of one yeast cell is r = 2.0 × 10−6 m, so the volume of one yeast cell is

4 3 π r = 3.35× 10−17 m3 3 =18 mL = 18 cm3 = 1.8 × 10−5 m3

Vcell = Vtotal

If we assume that the space between the cells is negligible, then N =

Vtotal = 5.4 ×1011 cells Vcell

The technique overestimates the true number, since we have not accounted for space between cells. In more advanced courses, you will encounter something called the packing ratio to provide a more precise answer. Recognizing the approximation, we should only state the answer to a single significant digit as about 500 billion cells. 2

 10 −6 m  = 1.19 ×10 −11 m 2 66. 2.50 µ m × 4.75 µ m ×   µ  1 m  67. When we express the angle 4.00° in radians...