Chapter 1, problem 7 (sig figs: 2; technically 1 by problem statement): Hydraulic engineers in the United States often often use, as a unit of volume of water, the acre-foot, defined as the volume of water that will cover 1 acre of land to a depth of 1 foot. A 2 severe thunderstorm dumped dumped 2.0 inches in 30 minutes on a town town of area 26 km . What volume of water, in acre-feet, fell on the
Google!), 247.11 acr (thanks, 247.11 247.1 1 acr 1 ft 2 2 V A 26 km 2.0 in 26 km 2.0 in 1070.81 acr ft 2 1 km 12 in 2
2
town? You will need to convert “acre” to “km ”; we have 1 km
Technically, Technically, “ 26 km
2
(1.1)
volume, just in weird units, so it’s worth it to box it. 2.0 in ” is the correct volume,
Additional problem: Discuss the “ampere“ampere -hour”, in the same spirit, and talk about why it means total charge, and has nothing to do with “current”. Chapter 1, problem 12 (sig figs: 2): the fastest growing plant on record is the hesperoyucca whipplei that grew 3.7 m in 14 days. What was its growth rate in micro meters per millisecond? 6 m 10 m y 3.7 m m m 0 . 0 0 3 0 6 3 . 0 6 ms t 14 days 86400 103 day m s s
(1.2)
Fun fact: this problem is exactly identical to finding the vertical average velocity of a point-projectile. point-projectile. That universality is one of the beauties of physics. Important: It is important to distinguish three parts of any physics problem (1) the setting up of the equations and putting things in the right place, (2) the algebra, and sometimes (3) the calculator work (if you choose to plug in numbers). Surprisingly, (1) is usually the harder part! In fact, the steps are in order of descending difficulty! Chapter 1, problem 14 (sig figs: 3): a lecture period (50min) is close to 1 microcentury. a.) How long is a microcentury in minutes?
Ce 106 100 yr 104 yr
365.25 days
yr
24 60 mi min day
52.6 min
(1.3)
b) Use the formula below to find the percentage difference from the approximation.
%
Ac App Ac
100%
52.6 50 52.6
100% 4.94%
(1.4)
Remembering Remembering the pesky “percent difference” formula : In the numerator, you can write either Ac App or write App Ac , which is just the negative of the abovementioned. But, you must always divide by the “Accepted”, not the not the “approximation”! “approximation”! Can anyone tell me why this is? This is because an accepted value will never be 0, otherwise it won’t be of interest. Things are always theoretically 0. You actually can never measure something that is 0, by definition, because…there’s because…there’s literally nothing (zero!) to measure! Chapter 1, problem 14 (sig figs: n/a): Five clocks are being being tested in a laboratory. Exactly Exactly at noon, as determined determined by WWV timesignal, on successive days of a week. The clocks read as in the following table. Rank the five clocks according to their relative value as good timekeepers, best to worst. Justify your answers.
Clock: A B C D E
Sun. 12:36:40 11:59:59 15:50:45 12:03:59 12:03:59
Mon. 12:36:56 12:00:02 15:51:43 12:02:52 12:02:49
Tues. 12:37:12 11:59:57 15:52:41 12:01:45 12:01:54
Wed. 12:37:27 12:00:07 15:53:39 12:00:38 12:01:52
Thurs. 12:37:44 12:00:02 15:54:37 11:59:31 12:01:32
Fri. Sat. 12:37:59 12:38:14 11:59:56 12:00:03 15:55:35 15:56:33 11:58:24 11:57:17 12:01:22 12:01:12
Modification of problem statement: This would require Microsoft EXCEL to do the most painlessly. None of the clocks advance advance by exactly 24 h in a 24-h period but this is not the most most important criterion criterion for judging their quality quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to
another, it cannot be corrected since it would impossible to tell what the correction should be. The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning. CLOCK
Sun. -Mon. 16 3 58 +67 +70
A B C D E
Mon. -Tues. 16 +5 58 +67 +55
Tues. -Wed. 15 10 58 +67 +2
Wed. -Thurs. 17 +5 58 +67 +20
Thurs. -Fri. 15 +6 58 +67 +10
Fri. -Sat. 15 7 58 +67 +10
Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correcti on for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to +10 s, for clock E it is in the range from -70 s to -2 s. After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E. Chapter 1, problem 19 (sig figs: 3): Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height h = 1.70 m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t = 11.1 s, what is the radius 'r' of the Earth? TYPO IN BOOK: the H is supposed to be lower-case. This problem teaches us about drawing schematics, a very important skill in physics.
When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B. In the schematic, label d r tan .
Compute “r” using the pythagorean formula,
d 2 r 2 (r h)2 r 2 2rh h2 r 2 tan 2 r 2
solve for r
r 2 tan2 2rh h2 0
(1.5)
Taylor expansion: a skill you will need to learn is series-expansion. The theory behind it involves derivatives, but the upshot is that various functions are equal to simpler (often linear forms). To see this, divide (1.5) through by r , and NOTE that the 1 quantity h/r is a pure number, without dimensions ! 2
tan 2 2 hr hr tan 2 2 hr 0 0 r 2 tanh2 2h cot 2
(1.6)
How do we compute ? We use what we know about the length of a day,
360
t 24 h
(360 )(11.1 s) (24 h)(60 min/h)(60 s/min)
0 .0 46 25 ;
(1.7)
Now we’re home free,
r 2(1.7) cot 2 (0.04625) 5.2 10 6 m
1
Only quantities without dimension may be said to be “small”, unlike what the book says.
(1.8)