# Answers/Solutions: Giancoli Chapter 1

Q10: We'd need to make an estimate the number of car repairs per day in the city, and how many repairs per day each mechanic can make. This would tell us how many mechanics we need.

To start then: let's suppose that there are 1 million people in San Francisco, that about half of them own cars, and that each one needs (on average) a mechanic two times per year. This is 1x10^{6} visits per year, and dividing by (let's say) 250 work days per year, we obtain 4000 visits per day. If each mechanic can service (say) three cars per day, we would need approximately 1300 mechanics.

This number is an estimate. Maybe there are only 500 or maybe there are 2500. We could safely guess "about a thousand."

P21: (a) lightspeed is 2.998x10^{8} meters/second. To get meters/year we must multiply this number by the number of seconds per year:

2.998x10^{8} meters/second x (3600 seconds/hour) x (24 hours/day) x (365 days/year) = 9.45x10^{15} meters/year

(b) We are given kilometers/AU. To get AU/lightyear, we can divide kilometers/lightyear by kilometers/AU:

N (AU/ly) = (9.45x10^{12} km/ly)/(1.50x10^{8} km/AU) = 6.3x10^{4} AU/ly

(c) lightspeed = 2.998x10^{8} m/s x (1/1.5x10^{11} AU/m) x (3600 s/hour) = 7.2 AU/hour

P32: Method 1: $1000/day x 30 days = $30,000

Method 2: (the long way): Day 1 = 0.01; Day 2 = 0.02; Day 3 = 0.04; Day 4 = 0.08; Day 5 = 0.16; Day 6 = 0.32; Day 7 = 0.64; Day 8 = 1.28; Day 9 = 2.56; Day 10 = 5.12; Day 11 = 10.24; Day 12 = 20.48; Day 13 = 40.96; Day 14 = 81.96; Day 15 = 163.84; Day 16 = 327.68; Day 17 = 655.36; Day 18 = 1310.72; Day 19 = 2621.44; Day 20 = 5242.88; Day 21 = 10485.76; Day 22 = 20971.52; Day 23 = 41943.04; Day 24 = 83886.08; Day 25 = 167772.16; Day 26 = 335544.32; Day 27 = 671088.64; Day 28 = 1342177.28; Day 29 = 2684354.56; Day 30 = 5368709.12 dollars

So on day 30 you would get over five million dollars!

(the short way): To estimate this (instead of going through the addition), note that each day's pay is double that of the previous day, so Day 1 is one cent or 2^{0} cents, Day 2 is then 2^{1} cents, Day 3 is 2^{2} cents, Day 4 is 2^{3} cents, and so Day N will be 2^{N-1} cents. The pay for Day 30 then will be 2^{29} cents, and my calculator says this equals the 5.3 million listed above!

(if you really care): Notice that the SUM TOTAL of pay you have received on (say) Day 4 is almost the same (one cent less) than the NEXT DAY'S pay. That is, on Day 4 you receive eight cents, and the total you have received for the first four days is 15 cents, which is one cent less than what you will receive on Day 5.

SO: The TOTAL PAY through 30 days would be just one cent less than that you would have received (if you worked that day) on Day 31, or about 10.6 million dollars. I'd suggest choosing to be paid by Method 2.

P37: (a) Cannot possibly be correct. "x" is in meters, but the term vt^{2} would have dimensions of (meters/sec)*(sec)^{2} which gives (meters-sec); likewise the term 2at would have dimensions of (meters/sec^{2})*(sec) which gives meters/sec. This equation is similar to saying "5 apples = 3 bananas + 2 oranges." It makes no sense.

(b) and (c) are possible. Each term in the equations have units of meters.

P43: We can estimate the volume of a lung at about 1 liter = 0.001 cubic meter. If there are 300 million alveoli, the volume of each one should be about:

V = (.001 m^{3}/lung) x (1 lung/3x10^{8} alevoli) = 3.3 x 10^{-12} m^{3}/alveoli.

Given this volume of an alveoli, to estimate the diameter let's assume the alveoli are cubical. Then if the length of one side is a, the volume would be V = a^{3}, and since we have a number for V, we can simply take its cube-root to get a. This would give a = 1.6x10^{-4} meters.

Try it assuming that each alveoli is a sphere, whose volume is (4/3)R^{3}, and whose diameter would then be 2R.

P48: I count roughly 11 balls vertically and about 10 horizontally. Assuming the chamber has a "depth" (i.e., the dimension into the page) equal to the horizontal dimension, there should be roughly 11x10x10 = 1100 balls. Or just say about one thousand.