Fisica Universitaria Zemansky (Solucionario)

UNITS, PHYSICAL QUANTITIES AND VECTORS

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1.1.

IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in. = 2.54 cm , 1 km = 1000 m , 12 in. = 1 ft , 1 mi = 5280 ft . ⎛ 5280 ft ⎞⎛ 12 in. ⎞⎛ 2.54 cm ⎞⎛ 1 m ⎞⎛ 1 km ⎞ EXECUTE: (a) 1.00 mi = (1.00 mi) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ 2 ⎟⎜ 3 ⎟ = 1.61 km ⎝ 1 mi ⎠⎝ 1 ft ⎠⎝ 1 in. ⎠⎝ 10 cm ⎠⎝ 10 m ⎠

1.2.

⎛ 103 m ⎞⎛ 102 cm ⎞ ⎛ 1 in. ⎞⎛ 1 ft ⎞ 3 (b)1.00 km = (1.00 km) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 3.28 × 10 ft 1 km ⎠⎝ 1 m ⎠ ⎝ 2.54 cm ⎠⎝ 12 in. ⎠ ⎝ EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. IDENTIFY: Convert volume units from L to in.3 . SET UP: 1 L = 1000 cm3 . 1 in. = 2.54 cm
⎛ 1000 cm3 ⎞ ⎛ 1 in. ⎞ 3 0.473 L × ⎜ ⎟×⎜ ⎟ = 28.9 in. . ⎝ 1 L ⎠ ⎝ 2.54 cm ⎠ EVALUATE: 1 in.3 is greater than 1 cm3 ,so the volume in in.3 is a smaller number than the volume in cm3 , which is 473 cm3 . IDENTIFY: We know the speed of light in m/s. t = d / v . Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is v = 3.00 × 108 m/s . 1 ft = 0.3048 m . 1 s = 109 ns . 0.3048 m EXECUTE: t = = 1.02 × 10−9 s = 1.02 ns 3.00 × 108 m/s EVALUATE: In 1.00 s light travels 3.00 × 108 m = 3.00 × 105 km =1.86 × 105 mi . IDENTIFY: Convert the units from g to kg and from cm3 to m3 . SET UP: 1 kg = 1000 g . 1 m = 1000 cm .
EXECUTE: EXECUTE:
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1.3.

1.4.

g ⎛ 1 kg ⎞ ⎛ 100 cm ⎞ 4 kg × 11.3 ⎟×⎜ ⎟ = 1.13 × 10 3 ⎜ cm ⎝ 1000 g ⎠ ⎝ 1 m ⎠ m3

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1.5.

EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3 . IDENTIFY: Convert volume units from in.3 to L. SET UP: 1L = 1000 cm3 . 1 in. = 2.54 cm . EXECUTE:

( 327 in. ) × ( 2.54 cm in.) × (1 L 1000 cm ) = 5.36 L
3 3 3

1.6.

EVALUATE: The volume is 5360 cm3 . 1 cm3 is less than 1 in.3 , so the volume in cm3 is a larger number than the volume in in.3 . IDENTIFY: Convert ft 2 to m 2 and then to hectares. SET UP: 1.00 hectare = 1.00 × 104 m 2 . 1 ft = 0.3048 m . EXECUTE:

⎛ 43,600 ft 2 ⎞ ⎛ 0.3048 m ⎞The area is (12.0 acres) ⎜ ⎟⎜ ⎟ ⎝ 1 acre ⎠ ⎝ 1.00 ft ⎠

2

⎛ 1.00 hectare ⎞ = 4.86 hectares . ⎜ 4 2 ⎟ ⎝ 1.00 × 10 m ⎠

1.7.

EVALUATE: Since 1 ft = 0.3048 m , 1 ft 2 = (0.3048) 2 m 2 . IDENTIFY: Convert seconds to years. SET UP: 1 billion seconds = 1 × 109 s . 1 day = 24 h . 1 h = 3600 s . EXECUTE:

⎛ 1 h ⎞⎛ 1 day ⎞ ⎛ 1 y ⎞ 1.00 billion seconds = (1.00 × 109 s ) ⎜ ⎟ = 31.7 y . ⎟⎜ ⎟⎜ ⎝3600 s ⎠⎝ 24 h ⎠ ⎝ 365 days ⎠
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Chapter 1

1.8.

EVALUATE: The conversion 1 y = 3.156 × 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. IDENTIFY: Apply the given conversion factors. SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.

1.9.

⎛ 0.125 mi ⎞⎛ 1fortnight ⎞⎛ 1 day ⎞ furlongs fortnight ) ⎜ ⎟ = 67 mi/h ⎟⎜ ⎟⎜ ⎝ 1 furlong ⎠⎝ 14 days ⎠ ⎝ 24 h ⎠ 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. IDENTIFY: Convert miles/gallon to km/L. SET UP: 1 mi = 1.609 km . 1 gallon = 3.788 L.
EXECUTE:

(180,000

⎛ 1.609 km ⎞⎛ 1 gallon ⎞ (a) 55.0 miles/gallon = (55.0 miles/gallon)⎜ ⎟⎜ ⎟ = 23.4 km/L . ⎝ 1 mi ⎠⎝ 3.788 L ⎠ 1500 km 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,
EXECUTE: 1.10.

so 1 mi/gal ∼ 2 km/L , which is roughly our result. 4 IDENTIFY: Convert units. SET UP: Use the unit conversions given in the problem.Also, 100 cm = 1 m and 1000 g = 1 kg .
EXECUTE:

ft ⎛ mi ⎞ ⎛ 1h ⎞ ⎛ 5280 ft ⎞ (a) ⎜ 60 ⎟ ⎜ ⎟ ⎜ ⎟ = 88 h ⎠ ⎝ 3600s ⎠ ⎝ 1mi ⎠ s ⎝ m ⎛ 1m ⎞ ⎜ ⎟ = 9.8 2 100 cm ⎠ s ⎝
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⎛ ft ⎞ ⎛ 30.48cm ⎞ (b) ⎜ 32 2 ⎟ ⎜ ⎟ ⎝ s ⎠ ⎝ 1ft ⎠

g ⎞ ⎛ 100 cm ⎞ ⎛ 1 kg ⎞ ⎛ 3 kg (c) ⎜1.0 3 ⎟ ⎜ ⎟ = 10 3 ⎟ ⎜ cm ⎠ ⎝ 1 m ⎠ ⎝ 1000 g ⎠ m ⎝
EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The...