Chapter 1 Systems of Measurement

Transcription

1 Chapter Systes of Measureent Conceptual Probles * Deterine the Concept he fundaental physical quantities in the SI syste include ass, length, and tie. Force, being the product of ass and acceleration, is not a fundaental quantity. (c) is correct. Picture the Proble We can express and siplify the ratio of /s to /s to deterine the final units. Express and siplify the ratio of /s to /s : s s s s and (d) is correct. s Deterine the Concept Consulting able - we note that the prefix giga eans 0 9. (c) is correct. Deterine the Concept Consulting able - we note that the prefix ega eans 0 6. (d) is correct. *5 Deterine the Concept Consulting able - we note that the prefix pico eans 0. (a) is correct. 6 Deterine the Concept Counting fro left to right and ignoring zeros to the left of the first nonzero digit, the last significant figure is the first digit that is in doubt. Applying this criterion, the three zeros after the decial point are not significant figures, but the last zero is significant. Hence, there are four significant figures in this nuber. (c) is correct.

2 Chapter 7 Deterine the Concept Counting fro left to right, the last significant figure is the first digit that is in doubt. Applying this criterion, there are six significant figures in this nuber. (e) is correct. 8 Deterine the Concept he advantage is that the length easure is always with you. he disadvantage is that ar lengths are not unifor; if you wish to purchase a board of two ar lengths it ay be longer or shorter than you wish, or else you ay have to physically go to the luberyard to use your own ar as a easure of length. 9 (a) rue. You cannot add apples to oranges or a length (distance traveled) to a volue (liters of ilk). (b) False. he distance traveled is the product of speed (length/tie) ultiplied by the tie of travel (tie). (c) rue. Multiplying by any conversion factor is equivalent to ultiplying by. Doing so does not change the value of a quantity; it changes its units. Estiation and Approxiation *0 Picture the Proble Because θ is sall, we can approxiate it by θ D/r provided that it is in radian easure. We can solve this relationship for the diaeter of the oon. Express the oon s diaeter D in ters of the angle it subtends at the earth θ and the earth-oon distance r : Find θ in radians: D θ r π rad θ rad 60 Substitute and evaluate D: D ( rad)( 8M).5 0 6

3 Systes of Measureent * Picture the Proble We ll assue that the sun is ade up entirely of hydrogen. hen we can relate the ass of the sun to the nuber of hydrogen atos and the ass of each. Express the ass of the sun M S as the product of the nuber of hydrogen atos N H and the ass of each ato M H : M N S H M H Solve for N H : N H M M S H Substitute nuerical values and evaluate N H : kg N H kg 57 Picture the Proble et P represent the population of the United States, r the rate of consuption and N the nuber of aluinu cans used annually. he population of the United States is roughly 0 8 people. et s assue that, on average, each person drinks one can of soft drink every day. he ass of a soft-drink can is approxiately.8 0 kg. (a) Express the nuber of cans N used annually in ters of the daily rate of consuption of soft drinks r and the population P: N rp t Substitute nuerical values and can 8 N ( 0 people) approxiate N: person d 0 cans ( y) 65. d y (b) Express the total ass of aluinu used per year for soft drink cans M as a function of the nuber of cans consued and the ass per can: M N

4 Chapter Substitute nuerical values and evaluate M: M ( 0 cans/y)(.8 0 kg/can) 0 9 kg/y (c) Express the value of the aluinu as the product of M and the value at recycling centers: ( $/ kg) Value M 9 ( $/ kg)( 0 kg/y) $ 0 9 / y billion dollars/y Picture the Proble We can estiate the nuber of words in Encyclopedia Britannica by counting the nuber of volues, estiating the average nuber of pages per volue, estiating the nuber of words per page, and finding the product of these easureents and estiates. Doing so in Encyclopedia Britannica leads to an estiate of approxiately 00 illion for the nuber of words. If we assue an average word length of five letters, then our estiate of the nuber of letters in Encyclopedia Britannica becoes 0 9. (a) Relate the area available for one letter s and the nuber of letters N to be written on the pinhead to the area of the pinhead: Ns π d pinhead. where d is the diaeter of the Solve for s to obtain: s πd N Substitute nuerical values and evaluate s: π ( in) s c.5 in ( ) 0 (b) Express the nuber of atos per letter n in ters of s and the atoic spacing in a etal d atoic : n s d atoic Substitute nuerical values and evaluate n: n atos/ 0atos * Picture the Proble he population of the United States is roughly 0 8 people. Assuing that the average faily has four people, with an average of two cars per

5 Systes of Measureent 5 faily, there are about cars in the United States. If we double that nuber to include trucks, cabs, etc., we have 0 8 vehicles. et s assue that each vehicle uses, on average, about gallons of gasoline per week. (a) Find the daily consuption of gasoline G: Assuing a price per gallon P $.50, find the daily cost C of gasoline: G 8 ( 0 vehicles)( gal/d) 6 0 C GP 8 $9 0 gal/d 8 ( 6 0 gal/d)( $.50 / gal) 8 / d $ billion dollars/d (b) Relate the nuber of barrels N of crude oil required annually to the yearly consuption of gasoline Y and the nuber of gallons of gasoline n that can be ade fro one barrel of crude oil: Substitute nuerical values and estiate N: N N Y n G t n 8 ( 6 0 gal/d) ( 65.d/y) gal/barrel barrels/y 5 Picture the Proble We ll assue a population of 00 illion (fairly accurate as of Septeber, 00) and a life expectancy of 76 y. We ll also assue that a diaper has a volue of about half a liter. In (c) we ll assue the disposal site is a rectangular hole in the ground and use the forula for the volue of such an opening to estiate the surface area required. (a) Express the total nuber N of disposable diapers used in the United States per year in ters of the nuber of children n in diapers and the nuber of diapers D used by each child in.5 y: Use the daily consuption, the nuber of days in a year, and the estiated length of tie a child is in diapers to estiate the nuber of diapers D required per child: N nd diapers 65.d D.5 y d y 0 diapers/child

6 6 Chapter Use the assued life expectancy to estiate the nuber of children n in diapers:.5y n 76 y ( 00 0 children) children 7 Substitute to obtain: N ( 0 children) ( 0 diapers/child) 0 0 diapers (b) Express the required landfill volue V in ters of the volue of diapers to be buried: V NV one diaper Substitute nuerical values and evaluate V: V 0 ( 0 diapers)( 0.5/diaper) (c) Express the required volue in ters of the volue of a rectangular parallelepiped: V Ah 7 Solve and evaluate h: V A.5 0 h 0 Use a conversion factor to express this area in square iles: A i 6 i.590 k 6 Picture the Proble he nuber of bits that can be stored on the disk can be found fro the product of the capacity of the disk and the nuber of bits per byte. In part (b) we ll need to estiate (i) the nuber of bits required for the alphabet, (ii) the average nuber of letters per word, (iii) an average nuber of words per line, (iv) an average nuber of lines per page, and (v) a book length in pages. (a) Express the nuber of bits N bits as a function of the nuber of bits per byte and the capacity of the hard disk N bytes : N bits bytes( 8bits/byte) 9 ( 0 bytes)( 8bits/byte) N bits

7 Systes of Measureent 7 (b) Assue an average of 8 letters/word and 8 bits/character to estiate the nuber of bytes required per word: bits 8 character 8 characters bits 6 word word bytes 8 word Assue 0 words/line and 60 lines/page: Assue a book length of 00 pages and approxiate the nuber bytes required: words bytes page word bytes 800 page bytes 6 00pages bytes page Divide the nuber of bytes per disk by our estiated nuber of bytes required per book to obtain an estiate of the nuber of books the -gigabyte hard disk can hold: N books 9 0 bytes 6. 0 bytes/book 00 books *7 Picture the Proble Assue that, on average, four cars go through each toll station per inute. et R represent the yearly revenue fro the tolls. We can estiate the yearly revenue fro the nuber of lanes N, the nuber of cars per inute n, and the $6 toll per car C. cars in h d $6 R NnC lanes $77M in h d y car Units 8 Picture the Proble We can use the etric prefixes listed in able - and the abbreviations on page EP- to express each of these quantities. (a),000,000 watts 0 watts MW (b) 0.00gra 0 6 g g (c) 0 6 eter µ (d) 0,000seconds 0 0 s 0ks

8 8 Chapter 9 Picture the Proble We can use the definitions of the etric prefixes listed in able - to express each of these quantities without prefixes. (a) 0 µ W 0 0 (b) ns s W W s (c) MW 0 6 W (d) 5k 5 0,000,000 W 5,000 *0 Picture the Proble We can use the definitions of the etric prefixes listed in able - to express each of these quantities without abbreviations. (a) 0 boo picoboo (e) 0 6 phone egaphone (b) 0 9 low gigalow (f) 0 9 goat nanogoat (c) 0 6 phone icrophone (g) 0 bull terabull (d) 0 8 boy attoboy Picture the Proble We can deterine the SI units of each ter on the right-hand side of the equations fro the units of the physical quantity on the left-hand side. (a) Because x is in eters, C and C t ust be in eters: C is in ; C is in /s (b) Because x is in eters, ½C t ust be in eters: C is in /s (c) Because v is in /s, C x ust be in /s : C is in /s (d) he arguent of trigonoetric function ust be diensionless; i.e. without units. herefore, because x C is in ; C is in s

9 Systes of Measureent 9 is in eters: (e) he arguent of an exponential function ust be diensionless; i.e. without units. herefore, because v is in /s: C is in /s; C is in s Picture the Proble We can deterine the US custoary units of each ter on the right-hand side of the equations fro the units of the physical quantity on the left-hand side. (a) Because x is in feet, C and C t ust be in feet: C is in ft; C is in ft/s (b) Because x is in feet, ½C t be in feet: ust C is in ft/s (c) Because v is in ft /s, C x ust be in ft /s : C is in ft/s (d) he arguent of trigonoetric function ust be diensionless; i.e. without units. herefore, because x is in feet: C is in ft; C is in s (e) he arguent of an exponential function ust be diensionless; i.e. without units. herefore, because v is in ft/s: C is in ft/s; C is in s Conversion of Units Picture the Proble We can use the forula for the circuference of a circle to find the radius of the earth and the conversion factor i.6 k to convert distances in eters into distances in iles. (a) he Pole-Equator distance is one-fourth of the circuference: c 0 7

10 0 Chapter (b) Use the forula for the circuference of a circle to obtain: R c π 7 0 π (c) Use the conversion factors k 000 and i.6 k: c k i i i.6k and 6 k i R k Picture the Proble We can use the conversion factor i.6 k to convert speeds in k/h into i/h. Find the speed of the plane in k/s: v ( 0 /s) k s s 0 h 50 k/h 680 /s Convert v into i/h: k v 50 h 50 i/h i.6k *5 Picture the Proble We ll first express his height in inches and then use the conversion factor in.5 c. Express the player s height into inches: in h 6 ft + 0.5in 8.5in ft Convert h into c:.5c h 8.5in in 0c 6 Picture the Proble We can use the conversion factors i.6 k, in.5 c, and.09 yd to coplete these conversions.

11 Systes of Measureent (a) k k i h h.6k i 6. h (b) in 60 c 60c.5c.6in (c) 00 yd 00 yd.09 yd 9. 7 Picture the Proble We can use the conversion factor.609 k 580 ft to convert the length of the ain span of the Golden Gate Bridge into kiloeters. Convert 00 ft into k:.609 k 00 ft 00ft 580ft.8k *8 Picture the Proble et v be the speed of an object in i/h. We can use the conversion factor i.6 k to convert this speed to k/h. Multiply v i/h by.6 k/i to convert v to k/h: i i.6k v v h h i.6v k/h 9 Picture the Proble Use the conversion factors h 600 s,.609 k i, and i 580 ft to ake these conversions. 5 k 5 k (a) h h h 600s k 6.0 h s 5 k 5 k h 0 (b) h h 600s k s (c) i i 580ft h h h i 600s ft 88.0 s (d) i 60 h i.609 k 0 60 h i k h 600s 6.8 s

12 Chapter 0 Picture the Proble We can use the conversion factor.057 qt to convert gallons into liters and then use this gallons-to-liters conversion factor to convert barrels into cubic eters. qt gal.057 qt (a) gal ( gal).78 gal.78 0 barrel barrel barrel gal (b) ( ) Picture the Proble We can use the conversion factor given in the proble stateent and the fact that i.609 k to express the nuber of square eters in one acre. Multiply by twice, properly chosen, to convert one acre into square iles, and then into square eters: acre ( acre) 050 i acres i Picture the Proble he volue of a right circular cylinder is the area of its base ultiplied by its height. et d represent the diaeter and h the height of the right circular cylinder; use conversion factors to express the volue V in the given units. (a) Express the volue of the cylinder: Substitute nuerical values and evaluate V: (b) Use the fact that.8 ft to convert the volue in cubic feet into cubic eters: V V V π d π π h ( 6.8in) ( ft) ( 6.8in) ( ft) 0.50ft ( 0.50 ft ) 0.0 ft in.8ft (c) Because 0 : V ( )

13 Systes of Measureent * Picture the Proble We can treat the SI units as though they are algebraic quantities to siplify each of these cobinations of physical quantities and constants. (a) Express and siplify the units of v /x: ( s) s s (b) Express and siplify the units of x a : s /s s (c) Noting that the constant factor has no units, express and siplify the units of at : s s () s ( s ) Diensions of Physical Quantities Picture the Proble We can use the facts that each ter in an equation ust have the sae diensions and that the arguents of a trigonoetric or exponential function ust be diensionless to deterine the diensions of the constants. (a) x C + C (b) x C t (c) C x v t (d) x C cos C t (e) v C exp( C t) 5 Picture the Proble Because the exponent of the exponential function ust be diensionl the diension of λ ust be.

14 Chapter *6 Picture the Proble We can solve Newton s law of gravitation for G and substitute the diensions of the variables. reating the as algebraic quantities will allow us to express the diensions in their siplest for. Finally, we can substitute the SI units for the diensions to find the units of G. Solve Newton s law of gravitation for G to obtain: Substitute the diensions of the variables: Fr G M G M M Use the SI units for, M, and : Units of G are kg s 7 Picture the Proble et represent the ass of the object, v its speed, and r the radius of the circle in which it oves. We can express the force as the product of, v, and r (each raised to a power) and then use the diensions of force F, ass, speed v, and radius r to obtain three equations in the assued powers. Solving these equations siultaneously will give us the dependence of F on, v, and r. Express the force in ters of powers of the variables: F a v b r c Substitute the diensions of the physical quantities: Siplify to obtain: M M M M a a b+ c b c b Equate the exponents to obtain: a, b + c, and b Solve this syste of equations to obtain: a, b, and c Substitute in equation (): F v r v r

15 Systes of Measureent 5 8 Picture the Proble We note fro able - that the diensions of power are M /. he diensions of ass, acceleration, and speed are M, /, and / respectively. Express the diensions of av: Fro able -: M [ av ] M M [ P ] Coparing these results, we see that the product of ass, acceleration, and speed has the diensions of power. 9 Picture the Proble he diensions of ass and velocity are M and /, respectively. We note fro able - that the diensions of force are M/. Express the diensions of oentu: [ ] v M M Fro able -: [ F ] M Express the diensions of force ultiplied by tie: M Ft [ ] M Coparing these results, we force ultiplied by tie. see that oentu has the diensions of 0 Picture the Proble et X represent the physical quantity of interest. hen we can express the diensional relationship between F, X, and P and solve this relationship for the diensions of X. Express the relationship of X to force and power diensionally: Solve for [ X ]: [ F ][ X ] [ P] [ ] X [ P] [ F]

16 6 Chapter Substitute the diensions of force and power and siplify to obtain: [ ] M X M Because the diensions of velocity are /, we can conclude that: [ P ] [ F][ v] Rearks: While it is true that P Fv, diensional analysis does not reveal the presence of diensionless constants. For exaple, if P πfv, the analysis shown above would fail to establish the factor of π. * Picture the Proble We can find the diensions of C by solving the drag force equation for C and substituting the diensions of force, area, and velocity. Solve the drag force equation for the constant C: C F Av air Express this equation diensionally: Substitute the diensions of force, area, and velocity and siplify to obtain: [ ] C [ Fair ] [ A][ v] M M [ C ] Picture the Proble We can express the period of a planet as the product of these factors (each raised to a power) and then perfor diensional analysis to deterine the values of the exponents. Express the period of a planet as a b c the product of r, G,and M : Solve the law of gravitation for the constant G: Express this equation diensionally: S a b c Cr G M S () where C is a diensionless constant. Fr G [ ] G [ F][ r] [ ][ ]

17 Systes of Measureent 7 Substitute the diensions of F, r, and : [ ] M ( ) G M M M Noting that the diension of tie is represented by the sae letter as is the period of a planet, substitute the diensions in equation () to obtain: M a ( ) ( M ) c b Introduce the product of M 0 and 0 in the left hand side of the equation and siplify to obtain: M M 0 0 c b a+ b b Equate the exponents on the two sides of the equation to obtain: Solve these equations siultaneously to obtain: 0 c b, 0 a + b, and b a, b, and c Substitute in equation (): C Cr G M S r GM Scientific Notation and Significant Figures * Picture the Proble We can use the rules governing scientific notation to express each of these nubers as a decial nuber. S (a) 0 0, 000 (c) (b) (d) , 000 Picture the Proble We can use the rules governing scientific notation to express each of these easureents in scientific notation. (a) 9 5.GW. 0 W (c).fs. 0 s

18 8 Chapter 6 (b) 0p (d) µ s 0 s 5 Picture the Proble Apply the general rules concerning the ultiplication, division, addition, and subtraction of easureents to evaluate each of the given expressions. (a) he nuber of significant figures in each factor is three; therefore the result has three significant figures: (b) Express both ters with the sae power of 0. Because the first easureent has only two digits after the decial point, the result can have only two digits after the decial point: 5 (.)( ) (.78 0 ) ( 5. 0 ) ( ) (c) We ll assue that is exact. Hence, the answer will have three significant figures: π (d) Proceed as in (b): ( ) Picture the Proble Apply the general rules concerning the ultiplication, division, addition, and subtraction of easureents to evaluate each of the given expressions. (a) Note that both factors have four significant figures. (b) Express the first factor in scientific notation and note that both factors have three significant figures. 5 ( 00.9)( 569.). 0 7 ( )( 6. 0 ) 7 7 ( 5. 0 )( 6. 0 ).0 0

19 (c) Express both ters in scientific notation and note that the second has only three significant figures. Hence the result will have only three significant figures. (d) Because the divisor has three significant figures, the result will have three significant figures. 80+ Systes of Measureent 9 ( ) (.8 0 ) + ( ) ( ) *7 Picture the Proble et N represent the required nuber of ebranes and express N in ters of the thickness of each cell ebrane. Express N in ters of the thickness of a single ebrane: in N 7 n Convert the units into SI units and siplify to obtain: N in 7 n 9 0.5c in 6 00c n 0 8 Picture the Proble Apply the general rules concerning the ultiplication, division, addition, and subtraction of easureents to evaluate each of the given expressions. (a) Both factors and the result have three significant figures: (b) Because the second factor has three significant figures, the result will have three significant figures: (c) Both factors and the result have three significant figures: (d) Write both ters using the sae power of 0. Note that the result will have only three significant figures: (.00 0 )( ) (.59)(.00 0 ) ( 5. 0 ) + (.78 0 ) ( 5. 0 ) + ( ) ( )

20 0 Chapter (e) Follow the sae procedure used in (d): 5 (.99 0 ) + ( ) (.99 0 ) + ( ).99 0 *9 Picture the Proble Apply the general rules concerning the ultiplication, division, addition, and subtraction of easureents to evaluate each of the given expressions. (a) he second factor and the result have three significant figures: (.) (b) We ll assue that is exact. herefore, the result will have two significant figures: (c) We ll assue that / is exact. herefore the result will have two significant figures: π (.) 5. 6 (d) Because.0 has two significant figures, the result has two significant figures: General Probles (.0) Picture the Proble We can use the conversion factor i.6 k to convert 00 k/h into i/h. Multiply 00 k/h by i/.6 k to obtain: k 00 h k 00 h 6.i/h i.6k *5 Picture the Proble We can use a series of conversion factors to convert billion seconds into years. Multiply billion seconds by the appropriate conversion factors to convert into years:

21 Systes of Measureent s 0 s h 600s day h y 65.days.7 y 5 Picture the Proble In both the exaples cited we can equate expressions for the physical quantities, expressed in different units, and then divide both sides of the equation by one of the expressions to obtain the desired conversion factor. (a) Divide both sides of the equation expressing the speed of light in the two systes of easureent by 86,000 i/s to obtain: 8 0 /s i/h k.6 0 i 0.6k/i /i (b) Find the volue of.00 kg of water: Volue of.00 kg 0 g is 0 c Express 0 c in ft : ( 0c) in.5 c ft in 0.05ft Relate the weight of ft of water to the volue occupied by kg of water:.00 kg 0.05ft lb 6. ft Divide both sides of the equation by the left-hand side to obtain: lb 6. ft.00 kg 0.05ft.0lb/kg 5 Picture the Proble We can use the given inforation to equate the ratios of the nuber of uraniu atos in 8 g of pure uraniu and of ato to its ass. Express the proportion relating the nuber of uraniu atos N U in 8 g of pure uraniu to the ass of ato: NU 8g ato kg

22 Chapter Solve for and evaluate N U : N U ( 8g) ato kg 5 Picture the Proble We can relate the weight of the water to its weight per unit volue and the volue it occupies. Express the weight w of water falling on the acre in ters of the weight of one cubic foot of water, the depth d of the water, and the area A over which the rain falls: lb w 6. Ad ft Find the area A in ft : A ( acre).56 0 i 580ft 60acre i ft Substitute nuerical values and evaluate w: lb w 6. ft ft 5 (.56 0 ft )(.in).7 0 lb in 55 Picture the Proble We can use the definition of density and the forula for the volue of a sphere to find the density of iron. Once we know the density of iron, we can use these sae relationships to find what the radius of the earth would be if it had the sae ass per unit volue as iron. (a) Using its definition, express the density of iron: ρ V Assuing it to be spherical, express the volue of an iron nucleus as a function of its radius: V π r Substitute to obtain: ρ () π r

23 Substitute nuerical values and evaluate ρ: ρ π Systes of Measureent 6 ( 0 kg) 5 ( 5. 0 ) kg/ (b) Because equation () relates the density of any spherical object to its ass and radius, we can solve for r to obtain: r πρ Substitute nuerical values and evaluate r: r ( 0 kg) 7 (. 0 kg/ ) 5.98 π 6 56 Picture the Proble Apply the general rules concerning the ultiplication, division, addition, and subtraction of easureents to evaluate each of the given expressions. (a) Because all of the factors have two significant figures, the result will have two significant figures: (b) Because the factor with the fewest significant figures in the first ter has two significant figures, the result will have two significant figures. Because its last significant figure is in the tenth s position, the difference between the first and second ter will have its last significant figure in the tenth s position: 5 ( ) ( ) ( )( ) (.)( 6. 0 )( 8. 0 ) (c) Because all of the factors have two significant figures, the result will have two significant figures: 6 ( 6. 0 ) (.6 0 ) (.6 0 ).9 0 8

24 Chapter (d) Because the factor with the fewest significant figures has two significant figures, the result will have two significant figures. ( ) (.8 0 )( 90 0 ) 5 ( 6. 0 ) (.8 0 )( 90 0 ) 0.5 *57 Picture the Proble We can use the relationship between an angle θ, easured in radians, subtended at the center of a circle, the radius R of the circle, and the length of the arc to answer these questions concerning the astronoical units of easure. (a) Relate the angle θ subtended by an arc of length S to the distance R: S θ () R Solve for and evaluate S: S Rθ in ( parsec)( s) 60s π rad 60in parsec (b) Solve equation () for and evaluate R: (c) Relate the distance D light travels in a given interval of tie t to its speed c and evaluate D for t y: S R θ.96 0 in π rad ( s) 60s 60in D c t 0 8 s ( y) s y

25 (d) Use the definition of AU and 5 c y ( ) the result fro part (c) to obtain: Systes of Measureent AU AU.96 0 (e) Cobine the results of parts (b) and (c) to obtain: parsec 6 (.08 0 ) c y c y 5 58 Picture the Proble et N e and N p represent the nuber of electrons and the nuber of protons, respectively and ρ the critical average density of the universe. We can relate these quantities to the asses of the electron and proton using the definition of density. (a) Using its definition, relate the required density ρ to the electron density N e /V: ρ V N e V e Solve for N e /V: N V e ρ () e Substitute nuerical values and evaluate N e /V: (b) Express and evaluate the ratio of the asses of an electron and a proton: Rewrite equation () in ters of protons: N V e p kg/ kg/electron electrons/ e kg N V p 9. 0 kg ρ () p Divide equation () by equation () to obtain: N V p V N e e p e e or p N V N V p

26 6 Chapter Substitute nuerical values and use the result fro part (a) to evaluate N p /V: N V p ( ) ( protons/ ).59 protons/ *59 Picture the Proble We can use the definition of density to relate the ass of the water in the cylinder to its volue and the forula for the volue of a cylinder to express the volue of water used in the detector s cylinder. o convert our answer in kg to lb, we can use the fact that kg weighs about.05 lb. Relate the ass of water contained in the cylinder to its density and volue: Express the volue of a cylinder in ters of its diaeter d and height h: Substitute to obtain: ρv V A base π ρ d π h d h h Substitute nuerical values and evaluate : ( 0 kg/ ) ( 9.) (.) π kg Convert kg to tons: lb kg kg ton ton 000lb he 50,000 to 55,000 tons. ton clai is conservative. he actual weight is closer 60 Picture the Proble We ll solve this proble two ways. First, we ll substitute two of the ordered pairs in the given equation to obtain two equations in C and n that we can solve siultaneously. hen we ll use a spreadsheet progra to create a graph of log as a function of log and use its curve-fitting capability to find n and C. Finally, we can identify the data points that deviate the ost fro a straight-line plot by exaination of the graph.

27 Systes of Measureent 7 st Solution for (a) (a) o estiate C and n, we can apply the relation C n to two arbitrarily selected data points. We ll use the st and 6 th ordered pairs. his will produce siultaneous equations that can be solved for C and n. n C and n C 6 6 Divide the second equation by the first to obtain: 6 C C n 6 6 n Substitute nuerical values and solve for n to obtain: Substituting this value into the second equation gives:.75s 0.56s or kg 0.kg n.5 0 n n and so a judicial guess is that n C 5 so.75s C( kg) 0. 5 Solving for C gives: 0.5 C.75 s/kg nd Solution for (a) ake the logarith (we ll arbitrarily use base 0) of both sides of C n and siplify to obtain: n ( ) log( C ) n log logc + log n log + logc which, we note, is of the for y x + b. Hence a graph of log vs. log should be linear with a slope of n and a log - intercept log C. he graph of log vs. log shown below was created using a spreadsheet progra. he equation shown on the graph was obtained using Excel s Add rendline function. (Excel s Add rendline function uses regression analysis to generate the trendline.)

28 8 Chapter log 0.987log log log Coparing the equation on the graph generated by the Add rendline function to log() n log + log C, we observe: n 0.99 and C or s/kg (.77 s/kg ) (b) Fro the graph we see that the data points that deviate the ost fro a straight-line plot are: 0.0 kg, 0.7s, and.50 kg,. s (b) Fro the graph we see that the points generated using the data pairs (0.0 kg, 0.7s) and (0. kg,.05 s) deviate the ost fro the line representing the best fit to the points plotted on the graph. Rearks: Still another way to find n and C is to use your graphing calculator to perfor regression analysis on the given set of data for log versus log. he slope yields n and the y-intercept yields log C. 6 Picture the Proble We can plot log versus log r and find the slope of the best-fit line to deterine the exponent n. We can then use any of the ordered pairs to evaluate C. Once we know n and C, we can solve Cr n for r as a function of.

29 (a) ake the logarith (we ll arbitrarily use base 0) of both sides of Cr n and siplify to obtain: Systes of Measureent 9 n ( ) log( Cr ) n log logc + log r n log r + logc Note that this equation is of the for y x + b. Hence a graph of log vs. log r should be linear with a slope of n and a log -intercept log C. he graph of log versus log r shown below was created using a spreadsheet progra. he equation shown on the graph was obtained using Excel s Add rendline function. (Excel s Add rendline function uses regression analysis to generate the trendline.) y.506x +. log log r Fro the regression analysis we observe that: n.50 and C 0 ( ). 7.0 y/ G. 50 ( 7 ) r () or.0 y/ ( G) (b) Solve equation () for the radius r of the planet s orbit: 7.0 y /( G) Substitute nuerical values and r 6.0 y evaluate r: 7.0 y/ ( G ) 0.50 G

30 0 Chapter *6 Picture the Proble We can express the relationship between the period of the pendulu, its length, and the acceleration of gravity g as C a g b and perfor diensional analysis to find the values of a and b and, hence, the function relating these variables. Once we ve perfored the experient called for in part (b), we can deterine an experiental value for C. (a) Express as the product of and g raised to powers a and b: Write this equation in diensional for: C a g b () where C is a diensionless constant. [ ] [ ] a [ g] b Noting that the sybols for the diension of the period and length of the pendulu are the sae as those representing the physical quantities, substitute the diensions to obtain: a b Because does not appear on the left-hand side of the equation, we can write this equation as: 0 a+ b b Equate the exponents to obtain: a + b 0 and b Solve these equations siultaneously to find a and b: a and b Substitute in equation () to obtain: C g C g () (b) If you use pendulus of lengths and 0.5 ; the periods should be about: ( ) and s ( 0.5).s (c) Solve equation () for C: C g

31 Systes of Measureent Evaluate C with and s: C 9.8/s ( s) 6.6 π Substitute in equation () to obtain: π g 6 Picture the Proble he weight of the earth s atosphere per unit area is known as the atospheric pressure. We can use this definition to express the weight w of the earth s atosphere as the product of the atospheric pressure and the surface area of the earth. Using its definition, relate atospheric pressure to the weight of the earth s atosphere: Solve for w: Relate the surface area of the earth to its radius R: Substitute to obtain: w P A w PA A π R w π R P Substitute nuerical values and evaluate w: w π 0 k lb in 9 ( 670 k) lb 9.7in

32 Chapter