Solucionario Fisica De Zemnaski

1.1: [pic]
Although rounded to three figures, this conversion is exact because the given conversion from inches to centimeters defines the inch.

1.2: [pic]

1.3: The time required for light to travel any distance in a vacuum is the distance divided by the speed of light;
[pic]

1.4: [pic]

1.5: [pic]

1.6: [pic]
[pic]
The daily consumption must then be
[pic]

1.7: [pic]
[pic]1.8: [pic]

1.9: [pic]

1.10: a) [pic]

b) [pic]
c)[pic]

1.11: The density is mass per unit volume, so the volume is mass divided by density. [pic]
Use the formula for the volume of a sphere, [pic]
to calculate [pic]

1.12: [pic]

1.13: a) [pic]
b) Since the distance was given as 890 km, the total distance should be 890,000 meters.
To report the total distance as 890,010meters, the distance should be given as 890.01 km.

1.14: a) [pic] (two significant figures).
b)[pic] = 0.50 (also two significant figures).
c) 36 mm (to the nearest millimeter).
d) 6 mm.
e) 2.0.

1.15: a) If a meter stick can measure to the nearest millimeter, the error will be about [pic]b) If the chemical balance can measure to the nearest milligram, the error will beabout [pic] c) If a handheld stopwatch (as opposed to electric timing devices) can measure to the nearest tenth of a second, the error will be about [pic]

1.16: The area is 9.69 ( 0.07 cm2, where the extreme values in the piece’s length and width are used to find the uncertainty in the area. The fractional uncertainty in the area is [pic] = 0.72%, and the fractional uncertainties in the lengthand width are [pic] = 0.20% and [pic] = 0.53%.

1.17: a) The average volume is
[pic]
(two significant figures) and the uncertainty in the volume, found from the extreme values of the diameter and thickness, is about [pic], and so the volume of a
cookie is [pic] (This method does not use the usual form for progation of errors, which is not addressed in the text. The fractional uncertainty inthe thickness is so much greater than the fractional uncertainty in the diameter that the fractional uncertainty in the volume is [pic], reflected in the above answer.)
b) [pic]

1.18: (Number of cars [pic] miles/car.day)/mi/gal = gallons/day
(2 [pic]108 cars [pic] 10000 mi/yr/car [pic] 1 yr/365 days)/(20 mi/gal) = 2.75 [pic] 108 gal/day

1.19: Ten thousand; if it were to contain ten million,each sheet would be on the order of a millionth of an inch thick.

1.20: If it takes about four kernels to fill 1 cm3, a 2-L bottle will hold about 8000 kernels.

1.21: Assuming the two-volume edition, there are approximately a thousand pages, and each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercise and problems), so anestimate for the number of words is about [pic].

1.22: Assuming about 10 breaths per minutes, [pic] minutes per day, 365 days per year, and a lifespan of fourscore (80) years, the total volume of air breathed in a lifetime is about [pic]. This is the volume of a room [pic], which is kind of tight for a major-league baseball game, but it’s the same order of magnitude as the volume of theAstrodome.

1.23: This will vary from person to person, but should be of the order of [pic].

1.24: With a pulse rate of a bit more than one beat per second, a heart will beat 105 times per day. With 365 days in a year and the above lifespan of 80 years, the number of beats in a lifetime is about [pic]. With [pic] L (50 cm3) per beat, and about [pic] gallon per liter, this comes to about [pic]gallons.

1.25: The shape of the pile is not given, but gold coins stacked in a pile might well be in the shape of a pyramid, say with a height of [pic] and a base [pic]. The volume of such a pile is [pic], and the calculations of Example 1-4 indicate that the value of this volume is [pic]

1.26: The surface area of the earth is about [pic], where R is the radius of the earth, about [pic], so...