Arquitectura

Evaluate: The conversion 1 y = 3.156x 107 s assumes 1 y = 365.24 d, which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year.
1.8. Identify: Apply the given conversion factors.
Set Up: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.
Execute: (180,000 furlongs/fortnight)) ¥ i^T^f = 67 mi/h
v 6/6 ^ 1 furlong)[14 days Jl 24 h J
Evaluate: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number.
1.9. Identify: Convert miles/gallon to km/L. Set Up: 1 mi = 1.609 km. 1 gallon = 3.788 L.
Execute: (a) 55.0 miles/gallon = (55.0 miles/gallon)^^j09 ^T&TL) = 23 4 km/L .
,, „ , „ , ■ 1500 km ^ , T 64.1 L , , ,
(b) The volume of gas required is= 64.1 L . = 1.4 tanks .
23.4 km/L 45 L/tank
Evaluate: 1 mi/gal = 0.425 km/L . A km is very roughly half a mile and there are roughly 4 liters in a gallon, so 1 mi/gal ~ km/L , which is roughly our result. 1.10. Identify: Convert units.
Set Up: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g = 1 kg .
, , Ln mi 1 f 1h 1 f 5280 ft 1 „„ft Execute: (a) I 60— J(b) 132ft 1 I 3048cm
3
(c) I1.0
Execute: V = mcntical /density = | ^ kg31|1022_S | = 3080 cm
19.5 g/cm3 1.0 kg
3V / 3
r = 3 — = .3 — (3080 cm3) = 9.0 cm . V4n \4n
Evaluate: The density is very large, so the 130 pound sphere is small in size.
1.12. Identify: Use your calculator to display n x 107. Compare that number to the number of seconds in a year. Set Up: 1 yr = 365.24days, 1 day = 24 h, and 1 h = 3600 s.
Execute: (365.24 days/1 yr) ^ 14Jl 3 15567... x107 s; nx 107 s = 3.14159... x 107 s
The approximate expression is accurate to two significant figures. Evaluate: The close agreement is a numerical accident.
1.13. Identify: The percent error is the error divided by the quantity.
Set Up: The distance from Berlin to Paris is given to the nearest 10 km.Execute: (a) —^^— = 1.1x10-3%. 890x103 m
(b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total distance to only three significant figures.
Evaluate: In this case a very small percentage error has disastrous consequences.
I 1h 1 | I 5280 ft |
l3600s) | l 1mi , |
1 m 1 | m |
| | = 9.8—2 |
100 cm J | s2 |
I 1 kg ' | 1 = 1034 |
11000g| J m3 |
1ft
100 cm 1 m
Evaluate: The relations 60 mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The relation 32 ft/s2 = 9.8 m/s2 is accurate to only two significant figures. 1.11. Identify: We know the density and mass; thus we can find the volume using the relation
density = mass/volume = m/V . The radius is then found from the volume equation for a sphere and the result for the volume.Set Up: Density = 19.5 g/cm3 and mcrilical = 60.0 kg. For a sphere V = jnr3.
1.14. Identify: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of the decimal that matters.

Set Up: 12 mm has two significant figures and 5.98 mmhas three significant figures. Execute: (a) (12 mm) x (5.98 mm) = 72 mm2 (two significant figures)
(b) 5 98 mm = 0.50 (also two significant figures)
12 mm
(c) 36 mm (to the nearest millimeter)
(d) 6 mm
(e) 2.0 (two significant figures)
Evaluate: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to thenearest mm.
1.15. Identify and Set Up: In each case, estimate the precision of the measurement.
Execute: (a) If a meter stick can measure to the nearest millimeter, the error will be about 0.13%.
(f) If the chemical balance can measure to the nearest milligram, the error will be about 8.3 x 10-3%.
(g) If a handheld stopwatch (as opposed to electric timing devices) can measure...