chapter01 - CHAPTER 1 Problem Solutions 1.1 v has units of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 1 Problem Solutions 1.1 v has units of ( L T ) . a has units of L/ T 2 and x has units of L. Thus, the left side of the equation has units of v 2 = ( L T) = L2 T2 , and the right side has units of 2 ( ) ax = L T2 L = L2 T2 . Therefore, from the standpoint of units alone, the equation m igh t be v a lid . 1.2 r, a, b, c and s all have units of L. So, ( s − a) ( s − b) ( s − c) s 1.3 = L× L× L = L2 = L L Substituting in dimensions, we have L = L T2 T= T2 = T Thus, the d im e nsio ns a r e co nsist e nt . 1.4 (a) Since the units of ax are m 2 s 2 , while the units of v are m s , the e q u a t io n is in co r r e ct . (b) Since the units of y are m, and cos ( kx ) is dimensionless if k has units of m -1 , t h e e q u a t io n is d im e n sio n a lly co r r e ct . 1.5 m3 ⎛ kg ⋅ m ⎞ G ( kg ) = , so cross-multiplying gives the units of G as . ⎜ 2⎟ ⎝ s ⎠ ( m )2 kg ⋅ s 2 1.6 (a) Given that a ∝ F m , we have F ∝ ma . Therefore, the units of force are those of ma, 2 ( ) F = M L T2 = M L T-2 . 3 CHAPTER (b) newton = 1.7 (a) 1 kg ⋅ m . s2 78.9 ± 0.2 has 3 sig n ifica n t fig u r e s . (b) 3.788 × 109 has 4 sig n ifica n t fig u r es . (c) 2.46 × 10−6 has 3 sig n ifica n t fig u r e s . (d) 0.0032 = 3.2 × 10 − 3 has 2 sig n ifica n t fig u r e s . 1.8 t o t a l le n g t h = 3.456 m + 4.3 m = 7.756 m which must be rounded to 7.8 m since the term 4.3 m has only one position beyond the decimal. 1.9 (a) The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal. ( ) (b) 0.0032 × 356.3 = 3.2 × 10- 3 × 356.3=1.14016 must be rounded to 1.1 because 3.2 × 10-3 has only two significant figures. (c) 5.620 × π must be rounded to 17.66 because 5.620 has only four significant figures. 1.10 c = 2 .997 924 574 × 108 m s (a) Rounded to 3 significant figures: c = 3.00 × 108 m s (b) Rounded to 5 significant figures: c = 2 .997 9 × 108 m s (c) Rounded to 7 significant figures: c = 2 .997 926 × 108 m s 1.11 The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m , but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 1.12 (a) c = 2π r = 2π ( 3.5 cm ) must be rounded to 22 cm . The number 2 and π are considered as known to many significant figures, while 3.5 is known only to two. 4 CHAPTER 1 (b) A = π r 2 = π ( 4.65 cm ) must be rounded to the number of significant figures in 4.65, 2 which gives an answer of 67.9 cm 2 . 1.13 Adding the two lengths together, we get 228.76 cm. However, 135.3 cm has only one decimal place. Therefore, only one decimal place accuracy is possible in the sum, changing 228.76 cm to 228.8 cm . 1.14 (a) ( 2.437 × 10 ) ( 6.5211 × 10 ) ( 5.37 × 10 ) = 2.9594 × 10 9 (b) ( 3.14159 × 10 ) ( 27.01 × 10 ) (1234 × 10 ) = 6.8764 × 10 −2 1.15 4 9 2 4 4 6 9 = 2.96 × 10 = 6.876 × 10−2 ⎛ 5 280 ft ⎞ ⎛ 1 fa t h om ⎞ d = ( 250 000 m i ) ⎜ = 2 × 10 8 fa t h o m s ⎟⎜ ⎝ 6 ft ⎟ ⎠ ⎝ 1.000 m i ⎠ The answer is limited to one significant figure because of the accuracy to which the conversion from fathoms to feet is given. 1.16 (a) l = ( 348 m i ) ⎛ ⎜ 1.609 k m ⎞ 2 5 7 = 5.60 × 10 km = 5.60 × 10 m = 5.60 × 10 cm ⎝ 1.000 m i ⎟ ⎠ ⎛ 1.609 k m ⎞ 0.4912 k m = 491.2 m = 4.912 × 104 cm (b) h = (1612 ft ) ⎜ = ⎝ 5280 ft ⎟ ⎠ (c) ⎛ 1.609 k m ⎞ 6.192 k m = 6.192 × 103 m = 6.192 × 105 cm h = ( 20 320 ft ) ⎜ = ⎝ 5280 ft ⎟ ⎠ ⎛ 1.609 k m ⎞ 3 5 (d) d = ( 8 200 ft ) ⎜ = 2 .499 k m = 2 .499 × 10 m = 2.499 × 10 cm ⎝ 5280 ft ⎟ ⎠ In (a), the answer is limited to three significant figures because of the accuracy of the original data value, 348 miles. In (b), (c), and (d), the answers are limited to four significant figures because of the accuracy to which the kilometers-to-feet conversion factor is given. 1.17 m = 325 m g = 325 × 10 − 3 g = 0.325 g 5 CHAPTER 1.18 1 ⎛ 1 in ⎞ ⎛ 1 d a y ⎞ ⎛ 1 h ⎞ ⎛ 2.54 cm ⎞ ⎛ 10 9 n m ⎞ rat e = ⎜ = 9.2 n m s ⎟⎜ ⎟ ⎜ ⎟⎜ 2 ⎝ ⎠ ⎝ 32 d a y ⎟ ⎝ 24 h ⎠ ⎝ 3600 s ⎠ ⎝ 1.00 in ⎠ ⎜ 10 cm ⎟ ⎠ This means that the proteins are assembled at a rate of many layers of atoms each second! ( ) 1.19 ⎛ 3.281 ft ⎞ 1 × 10 17 ft Dist a n ce = 4 × 10 16 m ⎜ = ⎝ 1m ⎟ ⎠ 1.20 ⎛ ⎞ ⎛ 1 d ay ⎞ ⎛ 1 h ⎞ 1 yr 9 Ag e o f ea r t h = 1 × 10 17 s ⎜ ⎟ ⎜ 24 h ⎟ ⎜ 3600 s ⎟ = 3 × 10 y r . ⎠ ⎝ ⎠⎝ ⎝ 365.242 d a ys ⎠ 1.21 m ⎞ ⎛ 3600 s ⎞ ⎛ 1 k m ⎞ ⎛ 1 m i ⎞ ⎛ 8 c = ⎜ 3.00 × 10 8 = 6.71 × 10 m i h ⎟⎜ ⎝ ⎠ ⎝ 1 h ⎟ ⎜ 103 m ⎟ ⎜ 1.609 k m ⎟ ⎠⎝ ⎠⎝ ⎠ s 1.22 ⎛ 2.832 × 10 − 2 m 3 ⎞ Vo lu m e o f h o u se = ( 50.0 ft ) ( 26 ft ) ( 8.0 ft ) ⎜ ⎟ 1 ft 3 ⎝ ⎠ ( ) ( 3 ) ⎛ 100 cm ⎞ 2 8 = 2.9 × 10 m 3 = 2.9 × 10 2 m 3 ⎜ = 2.9 × 10 cm 3 ⎝ 1m ⎟ ⎠ 1.23 ⎛ 43 560 ft 2 ⎞ ⎛ 1 m ⎞ 3 4 3 Vo lu m e = ( 25.0 a cr e ft ) ⎜ ⎟ ⎜ 3.281 ft ⎟ = 3.08 × 10 m ⎝ ⎠ ⎝ 1 a cr e ⎠ 1.24 Volu me of p yr a mid = 1 ( a r e a o f b a s e ) ( h e ig h t ) 3 ( ) 1 = ⎡(13.0 a cr es ) 43 560 ft 2 a cr e ⎤ ( 481 ft ) =9.08 × 10 7 ft 3 ⎦ 3⎣ ⎛ 2.832 × 10 − 2 m 3 ⎞ 6 3 = 9.08 × 10 7 ft 3 ⎜ ⎟ = 2 .57 × 10 m 1 ft 3 ⎝ ⎠ ( ) 6 CHAPTER 1.25 1 Vo lu m e o f cu be = L3 = 1 q u ar t (Where L = length of one side of the cube.) ⎛ 1 g a llon ⎞ ⎛ 3.786 lit er ⎞ ⎛ 1000 cm 3 ⎞ 3 Thus, L3 = (1 q u a r t ) ⎜ ⎟ ⎜ 1 g a llon ⎟ ⎜ 1 lit er ⎟ = 946 cm , ⎠ ⎝ 4 q u a r ts ⎠ ⎝ ⎠⎝ and 1.26 1.27 L = 9.82 cm . V = A t . Hence, t = (a) V 3.786 × 10 − 3 m 3 = = 1.51 × 10 − 4 m = 0.115 m m 2 A 25.0 m ⎛ 1.0 × 10 − 3 k g ⎞ 3 ma ss = ( d en sit y ) ( v olum e ) = ⎜ 3 ⎟ 1.0 m ⎝ 1.0 cm ⎠ ( ) 3 ⎛ 10 2 cm ⎞ kg ⎞ ⎛ 3 = ⎜ 1.0 × 10 − 3 1.0 m 3 ⎜ 3⎟ ⎟ = 1.0 × 10 k g ⎝ cm ⎠ ⎝ 1m ⎠ ( ) (b) As rough calculation, treat as if 100% water. kg ⎞ 4 ⎛ −6 cell: m a ss = d en sit y × v olu m e= ⎜ 103 ⎟ π 0.50 × 10 m ⎝ m3 ⎠ 3 ( ) 3 = 5.2 × 10 −16 kg kg ⎞ 4 ⎛ π 4.0 × 10−2 m kidney: ma ss = d en sit y × v olume = ⎜ 103 3⎟ ⎝ ⎠3 m ( ( fly: ma ss = d en sity × v olume = ( d en sit y ) π r 2 h kg ⎞ ⎛ = ⎜ 103 π 1.0 × 10−3 m 3⎟ ⎝ ⎠ m ( 1.28 3 = 0.27 k g ) ) ( 4.0 × 10 2 ) −3 ) m = 1.3 × 10 − 5 kg Since you have only 16 hours (57,600 s) per day, you can count only $57,600 per day. So, the time it would take would be: t= 1.00 × 109 d olla r s ⎛ 1 y r ⎞ ⎜ ⎟≈ 5.76 × 104 d o lla r s/ d a y ⎝ 365 d a ys ⎠ 47.6 y r . Right now, you are at least 18 yr old, so you would be at least age 65 when you finished counting the money. It would provide a nice retirement, but a very boring life until then. W e w o u ld a d vise a g a in st it . 7 CHAPTER 1.29 1 n u m b er o f b a lls n eed ed = ( n u m b er lo st p er h it t er ) ( n u m b er h it t er s ) ( g a m es ) ga m es ⎞ h it t er s ⎞ ⎛ innings ⎞ ⎛ ⎛1 ⎞⎛ = ⎜ ba ll p er h it t er ⎟ ⎜ 10 ⎟ ⎜ 9 g a m e ⎟ ⎜ 81 yea r ⎟ ⎝4 ⎠⎝ inning ⎠ ⎝ ⎠⎝ ⎠ = 1800 ba lls 3 b alls , or ~10 . y ea r y ea r Assumptions are 1 ball lost for every four hitters, 10 hitters per inning, 9 innings per game, and 81 games per season. 1.30 n u m b er o f p o u n d s = ( n u m b er o f b u r g er s ) ( w eigh t b u r ger ) ( ) = 5 × 10 10 bu r ger s ( 0.25 lb bu r ger ) = 1.25 × 10 10 lb , or ~10 10 lb n u m ber of h ea d of ca t t le = ( w eigh t n eed ed ) ( w eig h t p er h ea d ) ( = 1.255 × 10 10 lb ) ( 300 lb h ead ) = 4.17 × 10 7 h ea d , or ~10 7 h ea d Assumptions are 0.25 lb of meat per burger and a net of 300 lb of meat per head of cattle 1.31 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make ( 50 000 m i ) ( 5280 ft 1.32 m i ) (1 r e v 8 ft ) = 3 × 10 7 r e v , o r ~10 7 r ev A blade of grass is ~ 1 4 inch wide, so we might expect each blade of grass to require at least 1 16 in 2 = 4.3 × 10 − 4 ft 2 . Since 1 a cr e = 43 560 ft 2 , the number of blades of grass to be expected on a quarter-acre plot of land is about n= ( ( 0.25 a cr e ) 43 560 ft 2 a cr e total a r ea = ar ea p er b lad e 4.3 × 10 - 4 ft 2 b la d e = 2 .5 × 10 7 b la d es, or ~10 7 bla d es . 8 ) CHAPTER 1.33 1 Consider a room that is 12 ft square with an 8.0 ft high ceiling. The volume of this room is 3 Vroom ⎛ 1m ⎞ = (12 ft ) (12 ft ) ( 8.0 ft ) ⎜ = 33 m 3 . ⎝ 3.281 ft ⎟ ⎠ A ping pong ball has a radius of about 2.0 cm, so its volume is ( 4 4 Vball = π r 3 = π 2 .0 × 10−2 m 3 3 ) 3 = 3.4 × 10−5 m 3 . The number of balls that would easily fit into the room is therefore n= 1.34 Vroom 33 m 3 = = 9.7 × 105 or ~106 Vball 3.4 × 10-5 m 3 Assume an average of 1 can per person each week and a population of 250 million. ⎛ nu m ber ca ns p er son ⎞ n u m ber ca n s yea r = ⎜ ⎟ ( p op u la t ion ) ( w eeks yea r ) ⎝ ⎠ w eek ⎛ ca n p er son ⎞ 8 = ⎜1 ⎟ 2 .5 × 10 p eop le ( 52 w eeks yr ) w eek ⎠ ⎝ ( ) = 1.3 × 1010 ca n s y r , or ~ 10 10 ca n s y r nu m b e r o f t o ns = ( w e ig ht ca n ) ( n u m b e r ca ns y e a r ) o z ⎞ ⎛ 1 lb ⎞ ⎛ 1 t o n ⎞ ⎤ ⎛ ⎡⎛ 10 ca n ⎞ = ⎢⎜ 0.5 ⎟ ⎟⎜ ⎟⎜ ⎟ ⎥ ⎜ 1.3× 10 ca n ⎠ ⎝ 16 o z ⎠ ⎝ 2000 lb ⎠ ⎦ ⎝ yr ⎠ ⎣⎝ = 2 × 10 5 t o n y r , o r ~10 5 t o n yr Assumes an average weight of 0.5 oz of aluminum per can. 1.35 The x coordinate is found as x = r co s θ = ( 2 .5 m ) co s 35o = 2.0 m and the y coordinate y = r sin θ = ( 2 .5 m ) sin 35o = 1.4 m . 9 CHAPTER 1.36 The x distance out to the fly is 2.0 m and the y distance up to the fly is 1.0 m. Thus, we can use the Pythagorean theorem to find the distance from the origin to the fly as, d = x2 + y2 = 1.37 1 ( 2 .0 m ) 2 + (1.0 m ) 2 = 2 .2 m The distance from the origin to the fly is r in polar coordinates, and this was found to be 2.2 m in problem 36. The angle θ is the angle between r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as ta n θ = y 1.0 m = = 0.5 and θ = tan −1 ( 0.50) = 27° x 2.0 m The polar coordinates are r = 2 .2 m a n d θ = 27 ° . 1.38 The x distance between the two points is 8.0 cm and the y distance between them is 1.0 cm. The distance between them is found from the Pythagorean theorem: d = x2 + y2 = 1.39 8.1 cm . (a) From the Pythagorean theorem, the unknown side is b = c2 − a2 = b) 1.40 ( 8.0 cm ) 2 + (1.0 cm ) 2 = 65 cm 2 = tan θ = ( 9.00 m ) 2 − ( 6.00 m ) 2 = 6.71 m . 6.00 m = 0.894 6.71 m From the diagram, sin ( 20.0°) = (c) sin φ = 6.71 m = 0.746 9.00 m ∆y . 30.0 m 0m 30.0 Thus, ∆y = ( 30.0 m ) sin ( 20.0°) = 10.3 m 1.41 ∆y 20.0° From the diagram at the left, θ = 60.0° θ Thus, 6.00 m i = tan ( 60.0°) , and h 30.0° h 6.00 m i h= = 3.46 m i , or about 18,300 ft. ta n ( 60.0°) 10 A B 6.00 mi CHAPTER 1 1.43 sin θ = sid e op p osit e so, sid e op p osit e = ( 3.00 m ) ( sin 30.0°) = 1.50 m . hyp ot enu se (b) cos θ = 1.42 a d ja cen t sid e so, ad jacen t sid e = ( 3.00 m ) ( cos 30.0°) = 2.60 m . hyp ot enu se (a) (a) The side opposite θ = 3.00 . (c) 4.00 = 0.800 5.00 (e) 1.44 cos θ = ta n φ = (b) The side adjacent to φ = 3.00 . 4.00 = 1.33 3.00 (d) sin φ = 4.00 = 0.800 5.00 Using the diagram at the right, the Pythagorean Theorem yields c= ( 5.00 m ) 2 + ( 7.00 m ) 2 = c 5.00 m 8.60 m θ 7.00 m 1.45 From the diagram given in Problem 1.44 above, it is seen that ta n θ = 1.46 5.00 and θ = 35.5° . 7.00 Using the sketch at the right: Tree w = t a n 35.0° , or 100 m River w w = (100 m ) t a n 35.0° = 70.0 m 35.0° 100 m 1.47 Consider tissue in the shape of a cube 1 cm on a side. The number of cells along each edge of this cubical volume is then n= 1 cm ⎛ 1 µm ⎞ ⎛ 1 m ⎞ = 1 × 10 4 , ⎜ -6 ⎟ ⎜ 2 ⎝ 10 m ⎠ ⎝ 10 cm ⎟ ⎠ 1 µm and the number of cells within the 1 cm 3 volume is ( N = n 3 = 1 × 10 4 ) 3 12 = 1 × 10 12 , or ~10 . 11 CHAPTER 1.48 1 ( ) 3 4 4 (a) The volume of Saturn is V = π r 3 = π 5.85 × 10 7 m =8.39 × 10 2 3 m 3 and the 3 3 density is 3 26 3 m ⎛ 5.68 × 10 kg ⎞ ⎛ 1 m ⎞ ⎛ 10 g ⎞ 3 =⎜ ⎜2 ⎟⎜ 23 3⎟ ⎟ = 0.677 g cm . ⎝ 10 cm ⎠ ⎝ 1 kg ⎠ V ⎝ 8.39 × 10 m ⎠ (b) The surface area of Saturn is ( A = 4π r 2 = 4π 5.85 × 10 7 m 1.49 ) 2 2 ⎛ 3.281 ft ⎞ 17 2 = 4.63 × 10 ft ⎜ ⎝ 1m ⎟ ⎠ The term s has dimensions of L, a has dimensions of LT − 2 , and t has dimensions of T. Therefore, the equation, s = k am t n has dimensions of ( L= LT - 2 ) m ( T) n L1T 0 = Lm T n − 2m . or The powers of L and T must be the same on each side of the equation. Therefore, L1 =Lm and m = 1 . Likewise, equating terms in T, we see that n − 2m = 0 . Thus, n = 2m = 2 . Dim en sion a l an a lysis can n ot d eter m in e th e v alu e of k , a dimensionless constant. 1.50 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. (a) It then has a volume V = 0.2 m 3 and contains a mass of water ( )( ) m = ρV = 10 3 kg m 3 0.2 m 3 = 200 kg, or ~10 2 kg . (b) Pennies are now mostly zinc, but consider copper pennies filling 80% of the volume of the tub. Their mass is ( )( ) m = 0.80 8.93 × 10 3 kg/ m 3 0.2 m 3 = 1400 kg, or 12 ~10 3 kg CHAPTER 1.51 The volume of oil equals V = 1 9.00 × 10-7 k g = 9.80 × 10-10 m 3 . 3 918 k g m If the diameter of a molecule is d, then this volume must also equal () d π r 2 = ( th ickn ess of slick ) ( a r ea of oil slick ) , where r = 0.418 m . Thus, d = 1.52 9.80 × 10−10 m 3 = 1.78 × 10 − 9 m , o r 2 π ( 0.418 m ) ~10 −9 m . (a) The amount paid per year would be d olla r s ⎞ ⎛ 8.64 × 10 4 s ⎞ ⎛ 365.25 d a ys ⎞ ⎛ 10 d o lla r s a m ou n t = ⎜ 1000 . ⎟⎜ ⎟ = 3.16 × 10 ⎟⎜ s ⎠ ⎝ 1.00 d a y ⎠ ⎝ yr yr ⎝ ⎠ Therefore, it would take 6.00 × 10 12 d o lla r s = 190 yr , or 3.16 × 10 10 d olla r s yr ~ 10 2 y r . (b) The circumference of the Earth at the equator is ( ) C = 2π r = 2π 6.38 × 10 6 m = 4.01 × 10 7 m . The length of one dollar bill is 0.155 m so that the length of six trillion bills is 9.30 × 10 11 m . Thus, the six trillion dollars would encircle the earth 9.30 × 10 11 m n= = 2.32 × 10 4 , o r 7 4.01 × 10 m 1.53 The number of tuners is found by dividing the number of residents of the city by the number of residents serviced by one tuner. We shall assume 1 tuner per 10,000 residents and a population of 7.5 million. Thus, nu m ber of t u ner s= 1.54 ~10 4 t im es . 7.5 × 10 6 3 = 7.5 × 10 2 ~10 t u n er s 4 1.0 × 10 (a) For a sphere, A = 4π R 2 . In this case, R 2 = 2 R1 . 4π R 2 R 2 ( 2 R1 ) A 2 Hence, 2 = = 2= = 4. 2 2 2 A 1 4π R 1 R 1 R1 2 13 CHAPTER (b) For a sphere, 1 4 V = π R3 . 3 3 3 V2 ( 4 3) π R 2 R 2 ( 2R1 ) = = 3= = 8. 3 3 V1 ( 4 3) π R 1 R 1 R1 3 Thus, 1.55 ⎛ 365.2 d a ys ⎞ ⎛ 8.64 × 104 s ⎞ 7 (a) 1 y r= (1 y r ) ⎜ ⎟ ⎜ 1 d a y ⎟ = 3.16 × 10 s . 1 yr ⎝ ⎠⎝ ⎠ (b) Let us consider a segment of the surface of the moon which has an area of 1 m 2 and a depth of 1 m . When filled with meteorites, each having a diameter 10-6 m , the number of meteorites along each edge of this box is n= len gt h of a n ed ge 1m = -6 = 10 6 . m et eor it e d ia m et er 10 m The total number of meteorites in the filled box is then () N = n 3 = 10 6 3 = 10 18 . At the rate of 1 meteorite per second, the time to fill the box is 1y ⎛ ⎞ 10 t = 1018 s= ( 1018 s ) ⎜ ⎟ = 3 × 10 yr , or 7 ⎝ 3.156 × 10 s ⎠ 14 ~1010 yr . CHAPTER 1 Answers to Even Numbered Conceptual Questions 2. It was good in the sense that it was easily understood and barley grains were readily available to everyone. It was a poor way to define a standard because barley grains are not uniform in size. Thus, the length of one person’s three barley grains may differ from that of another person’s three grains. 4. (a ) ~10 − 3 lb which is ~ 10 − 3 kg or ~ 1 gr a m (b) ~ 0.5 lb ≈ 0.25 kg or ~ 10 -1 k g (c) ~ 200 lb ≈ 100 kg or ~ 10 2 kg (d ) ~ 2000 lb ≈ 1000 k g or ~ 10 3 k g 6. Let us assume the atoms are solid spheres of diameter 10−10 m . Then, the volume of each 4 1 atom is of the order of 10-30 m 3 . (More precisely, volume = π r 3 = π d 3 .) Therefore, 3 6 10−6 since 1 cm 3 = 10-6 m 3 , the number of atoms in the solid is on the order of −30 = 1024 atoms. 10 A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10. 8. Realistically, the only lengths you might be able to verify are the length of a football field and the length of a housefly. The only time intervals subject to verification would be the length of a day and the time between normal heartbeats. 10. No. Dimensional analysis cannot reveal the presence of numerical constants in an equation. For example, a freely-falling object, starting from rest, obeys the equation v 2 = 2ax . Dimensional analysis would say that the equations v 2 = 3ax or v 2 = ax might be valid. 12. In general, the manufacturers are using significant figures correctly. The length and width of the aluminum foil both include three significant figures and the cited value of their product (the area) correctly includes three significant figures in both the metric and English versions. The manufacturer of the tape also gives three significant figures in the dimensions of the tape (metric version) and retains three significant figures in their product (area). However, the width of the tape is given in English units as 1 2′′ , which gives no indication of the accuracy of the measurement. To be consistent with the three significant figure accuracy cited in the metric width dimension, the width should be given as 0.500″ . 15 CHAPTER 1 Answers to Even Numbered Problems 2. Both sides have units of length, L, so the equation may be valid. 4. Only (b) is dimensionally correct. 6. (a) 8. 7.8 m MLT -2 (b) kg ⋅ m s 2 10. (a) 3.00 × 108 m s (b) 2 .9979 × 108 m s 12. (a) 22 cm (b) 67.9 cm 2 14. (a) 2 .96 × 10 9 (b) 6.876 × 10−2 16. (a) 5.60 × 10 2 k m , 5.60 × 10 5 m , 5.60 × 10 7 cm (b) 0.4912 k m , 491.2 m , 4.912 × 10 4 cm (c) 6.192 k m , 6.192 × 103 m , 6.192 × 105 cm (d) 2 .499 k m , 2.499 × 103 m , 2.499 × 105 cm 18. 9.2 n m s 20. 3 × 10 9 y r 22. 2 .9 × 10 2 m 3 , 2 .9 × 10 8 cm 3 24. 2 .57 × 10 6 m 3 26. 1.51 × 10 − 4 m 28. It would require about 47.6 yr to count the money. We advise against it. 30. ~10 10 lb of beef, ~10 7 head of cattle (assumes 0.25 lb per burger and a net of 300 lb of meat per head of cattle) 32. ~10 7 bla d es (assumes 1 16 in 2 per blade) 34. ~10 10 ca n s yr , ~10 5 t on yr (Assumes an average of 1 can per person each week, a population of 250 million, and 0.5 oz of aluminum per can) 36. 2.2 m 38. 8.1 cm 16 (c) 2.997925 × 108 m s CHAPTER 40. 10.3 m 42. (a) 44. 8.60 m 46. 70.0 m 48. (a) 0.677 g cm 3 (b) 4.63 × 10 17 ft 2 50. (a) ~10 2 kg (b) ~10 3 kg 52. (a) ~10 2 yr (b) ~10 4 t im es 54. (a) 4 (b) 8 1.50 m (b) 2.60 m 17 1 18 ...
View Full Document

This note was uploaded on 03/03/2010 for the course PHY P221 taught by Professor Dr.ha during the Spring '05 term at Indiana University South Bend.

Ask a homework question - tutors are online