Sears - ex. resolvidos

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Física I – Sears, Zemansky,
Young & Freedman.

PHYSICS ACT.
http//physicsact.wordpress.com

Capítulo 1

(

)

1.1: 1 mi × (5280 ft mi ) × (12 in. ft ) × (2.54 cm in .) × 1 km 105 cm = 1.61 km
Although rounded to three figures, this conversion is exact because the given conversion
from inches to centimeters defines the inch.
3

 1000 cm 3   1in 
3
0.473 L × 


 1 L ×  2.54 cm  = 28.9 in .




1.2:

1.3: The time required for light to travel any distance in a vacuum is the distance
divided by the speed of light;
103 m
= 3.33 × 10− 6 s = 3.33 × 103 ns.
8
3.00 × 10 m s
3

g  1 kg   100 cm 
4 kg
11.3
×
3
 1000 g  ×  1 m  =1.13 × 10 m 3 .


cm 



1.4:

1.5:

1.6:

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

3

3

 1000 L   1 gal   128 oz.   1 bottle 
1 m3 × 
 1 m 3  ×  3.788 L  ×  1 gal  ×  16 oz. .









= 2111.9 bottles ≈ 2112 bottles

The daily consumption must then be
bottles  1 yr 
bottles
2.11 × 10 3
×
 365.24 da  = 5.78 da .

yr



1.7:

(1450 mi hr ) × (1.61 km mi) = 2330 km hr .
2330 km hr × (103 m km)× (1 hr 3600 s ) = 648 m s.

1.8:

180,000

1.9:

15.0

furlongs  1 mile   1 fortnight   1 day 
mi
×
 8 furlongs  ×  14 day  ×  24 h  = 67 h .



fortnight 




km  1 mi   3.788 L 
mi
×
×
 1 gal  = 35.3 gal .

L  1.609 km  


 mi 
1.10: a)  60 
 hr 

 1h 

 3600 s 




 5280 ft 
ft

 1 mi  = 88s




m
 ft   30.48 cm   1 m 
b)  32 2  
 1ft   100 cm  = 9.8 s 2


 s 


g   100 cm 

c) 1.0


cm 3   1 m 


3

 1 kg 
3 kg

 1000 g  = 10 m 3




1.11: The density is mass per unit volume, so the volume is mass divided by density.
V = 60 ×103 g 19.5 g cm 3 = 3077 cm 3
4
Use the formula for the volume of a sphere, V =πr 3 ,
3
1/ 3
to calculate r : r = (3V 4π ) = 9.0 cm

(

)(

)

1.12:

(3.16 × 10 7 s − π × 10 7 s) (3.16 × 10 7 s) × 100 = 0.58%

1.13: a)

10 m
= 1.1× 10 −3 %.
3
890 × 10 m

b) Since the distance was given as 890 km, the total distance should be 890,000
meters.
To report the total distance as 890,010 meters, the distance should be given as
890.01 km.
1.14: a) (12 mm ) ×(5.98 mm ) = 72 mm 2 (two significant figures).
.98 mm
b) 512 mm = 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
0.13%. b) If the chemical balance can measure to the nearest milligram, the error will be
about 8.3 × 10−3%. c) If a handheld stopwatch (asopposed to electric timing devices) can
measure to the nearest tenth of a second, the error will be about 2.8 × 10 −2 %.

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
0 07 cm 2
area is 9..69 cm 2 = 0.72%, and the fractional uncertainties in the length and width are0.01 cm
5.10 cm

= 0.20% and

0.01 cm
1.9 cm

= 0.53%.

1.17: a) The average volume is
(8.50 cm )2 (0.050 cm ) = 2.8 cm 3
π
4
(two significant figures) and the uncertainty in the volume, found from the extreme
values of the diameter and thickness, is about 0.3 cm3 , and so the volume of a
cookie is 2.8 ± 0.3 cm3 . (This method does not use the usual form for progation of errors,which is not addressed in the text. The fractional uncertainty in the thickness is so much
greater than the fractional uncertainty in the diameter that the fractional uncertainty in the
volume is 10% , reflected in the above answer.)

b)

8.50
.05

= 170 ± 20.

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