Capitulo 1 fisica univer vol 12

UNITS, PHYSICAL QUANTITIES AND VECTORS
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 .
EXECUTE: (a) 2 3
1.00 mi (1.00 mi) 5280 ft 12 in. 2.54 cm 1 m 1 km 1.61 km
1 mi 1 ft 1 in. 10 cm 10 m
= ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
(b)
3 2
1.00 km (1.00 km) 10 m 10 cm 1 in. 1 ft 3.28 103ft
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.
1.2. IDENTIFY: Convert volume units from L to in.3 .
SET UP: 1 L =1000 cm3 . 1 in. = 2.54 cm
EXECUTE:
3 3
0.473 L 1000 cm 1 in. 28.9 in.3.
1 L 2.54 cm
⎛ ⎞ ⎛ ⎞ ×⎜ ⎟×⎜ ⎟ =
⎝ ⎠ ⎝ ⎠
EVALUATE: 1 in.3 isgreater than 1 cm3 , so the volume in in.3 is a smaller number than the volume in cm3 ,
which is 473 cm3 .
1.3. 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 .
EXECUTE: 9
8
0.3048 m 1.02 10 s 1.02 ns
3.00 10 m/s
t = = × − =
×
EVALUATE: In 1.00 s light travels3.00×108 m = 3.00×105 km =1.86×105 mi .
1.4. IDENTIFY: Convert the units from g to kg and from cm3 to m3 .
SET UP: 1 kg =1000 g . 1 m =1000 cm .
EXECUTE:
3
4
3 3
11.3 g 1 kg 100 cm 1.13 10 kg
cm 1000 g 1m m
⎛ ⎞⎛ ⎞
×⎜ ⎟×⎜ ⎟ = ×
⎝ ⎠⎝ ⎠
EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3 .
1.5. IDENTIFY: Convert volume units from in.3 to L.
SET UP: 1 L=1000 cm3 . 1 in. = 2.54 cm .
EXECUTE: (327 in.3 )×(2.54 cm in.) 3 ×(1 L 1000 cm3 ) = 5.36 L
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 .
1.6. IDENTIFY: Convert ft2 to m2 and then to hectares.
SET UP: 1.00 hectare =1.00×104 m2 . 1 ft = 0.3048 m .
EXECUTE: The area is
2 2
4 2
(12.0 acres) 43,600 ft 0.3048m 1.00 hectare 4.86 hectares
1 acre 1.00 ft 1.00 10 m
⎛ ⎞⎛ ⎞ ⎛ ⎞ = ⎜ ⎟⎜ ⎟ ⎜ × ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
.
EVALUATE: Since 1 ft = 0.3048 m , 1 ft2 = (0.3048)2 m2 .
1.7. IDENTIFY: Convert seconds to years.
SET UP: 1 billion seconds =1×109 s . 1 day = 24 h . 1 h = 3600 s .
EXECUTE: 1.00 billion seconds (1.00 109 s) 1 h 1 day 1 y 31.7 y
3600 s 24 h 365 days
⎛ ⎞⎛ ⎞⎛ ⎞ = × ⎜ ⎟⎜ ⎟⎜ ⎟ =
⎝ ⎠⎝ ⎠⎝ ⎠
.
11-2 Chapter 1
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.
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) 0.125 mi 1 fortnight 1day 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.
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) 1.609 km 1 gallon 23.4 km/L
1 mi 3.788 L
= ⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
.
(b) The volume of gas required is 1500 km 64.1 L
23.4 km/L
= . 64.1 L 1.4 tanks
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 24
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 =1kg .
EXECUTE: (a) 60mi 1h 5280 ft 88 ft
h 3600s 1mi s
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟=
⎝ ⎠⎝ ⎠⎝ ⎠
(b) 2 2
32 ft 30.48cm 1 m 9.8m
s 1ft 100 cm s
⎛ ⎞ ⎛ ⎞ ⎛ ⎞=
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(c)
3
3
3 3
1.0 g 100 cm 1 kg 10 kg
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 relation 32 ft/s2 = 9.8 m/s2 is...