SOLUTIONS TO PROBLEMS Intoduction and Vectos Section 1.1 Standads of Length, Mass, and Time *P1.4 Fo eithe sphee the volume is V = 4! and the mass is m =!V =! 4. We divide this equation fo the lage sphee by the same equation fo the smalle: Then l = s 5 m l = 4.50 cm( 1.71) = 7.69 cm. =!4 l m s!4 s = l = 5. s Section 1.2 Dimensional nalysis P1.7 (a) This is incoect since the units of [ ax] ae m 2 s 2, while the units of [ v] ae m s. This is coect since the units of [ y] ae m, and cos( kx) is dimensionless if [ k] is in m!1. Section 1. Convesion of Units *P1.14 N atoms = m Sun m atom = 1.99! 100 kg 1.67! 10 27 kg = 1.19! 1057 atoms 1
2 Intoduction and Vectos Section 1.4 Ode-of-Magnitude Calculations P1.19 Model the oom as a ectangula solid with dimensions 4 m by 4 m by m, and each ping-pong ball as a sphee of diamete 0.08 m. The volume of the oom is 4! 4! = 48 m, while the volume of one ball is 4! 0.08 m 2 & ' = 2.87 ( 10 )5 m. 48 Theefoe, one can fit about 2.87! 10 ~ 5 106 ping-pong balls in the oom. s an aside, the actual numbe is smalle than this because thee will be a lot of space in the oom that cannot be coveed by balls. In fact, even in the best aangement, the so-called best packing faction is 1 6! above estimate educes to 1.67! 10 6! 0.740 ~ 10 6. 2 = 0.74 so that at least 26 of the space will be empty. Theefoe, the *P1.21 ssume: Total population = 10 7 ; one out of evey 100 people has a piano; one tune can seve about 1 000 pianos (about 4 pe day fo 250 weekdays, assuming each piano is tuned once pe yea). Theefoe,! 1 tune! 1 piano tunes ~ 1 000 pianos & 100 people & (107 people) = 100. Section 1.5 Significant Figues P1.2 (a) 4 (c) (d) 2 Section 1.6 Coodinate Systems P1.2 x = cos! = ( 5.50 m)cos 240 = ( 5.50 m) (0.5) = 2.75 m y = sin! = ( 5.50 m)sin 240 = ( 5.50 m) (0.866) = 4.76 m P1. The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m. (a) We can use the Pythagoean theoem to find the distance fom the oigin to the fly. distance = x 2 + y 2 = ( 2.00 m) 2 + ( 1.00 m) 2 = 5.00 m 2 = 2.24 m! = tan 1 1 & 2' ( = 26.6 ; = 2.24 m, 26.6
Chapte 1 P1.5 We have = x 2 + y 2 and! = tan 1 y & x ' (. (a) The adius fo this new point is (!x) 2 + y 2 = x 2 + y 2 = and its angle is tan!1 y!x & ' = 180!(. (!2x) 2 + (!2y) 2 = 2. This point is in the thid quadant if ( x, y) is in the fist quadant o in the fouth quadant if ( x, y) is in the second quadant. It is at an angle of 180 +!. (c) (x) 2 + (!y) 2 =. This point is in the fouth quadant if ( x, y) is in the fist quadant o in the thid quadant if ( x, y) is in the second quadant. It is at an angle of!. P1.8 To find these vecto expessions gaphically, we daw each set of vectos. Measuements of the esults ae taken using a ule and potacto. (Scale: 1 unit = 0.5 m ) (a) + B = 5.2 m at 60 (c) (d)! B =.0 m at 0 B! =.0 m at 150! 2 B = 5.2 m at 00. FIG. P1.8
4 Intoduction and Vectos Section 1.9 Components of a Vecto and Unit Vectos P1.41 x =!25.0 = 40.0 = 2 x + 2 y = (!25.0) 2 + ( 40.0) 2 = 47.2 units We obseve that tan! = x. So! = tan 1 y x & ( ' = tan 40.0 25.0 = tan1 1.60 ( ) = 58.0. FIG. P1.41 The diagam shows that the angle fom the +x axis can be found by subtacting fom 180 :! = 180 58 = 122. P1.4 We have B = R! : P1.45 (a) Theefoe, (c) x = 150 cos120 =!75.0 cm = 150sin120 = 10 cm R x = 140cos 5.0 = 115 cm R y = 140sin 5.0 = 80. cm B = [ 115! (!75)]î + 80.! 10 B = 190 2 + 49.7 2 = 196 cm = tan!1! 49.7 & 190 ' ( =!14.7. + B ( ) = î! 2ĵ ( ) +!î! 4ĵ ( ) = 2î! 6ĵ! B ( ) = î! 2ĵ ( )!!î! 4ĵ ( ) = 4î + 2ĵ + B = 2 2 + 6 2 = 6.2 ( ) cm [ ] ĵ = 190î! 49.7ĵ FIG. P1.4 (d)! B = 4 2 + 2 2 = 4.47 (e)! + B = tan1 6 & 2' ( = 71.6 = 288! B = 2 & tan1 4 ' ( = 26.6
P1.46 (a) D = + B + C = 2î + 4ĵ D = 2 2 + 4 2 = 4.47 m at! = 6.4 E =!! B + C =!6î + 6ĵ E = 6 2 + 6 2 = 8.49 m at! = 15 Chapte 1 5 Section 1.10 Modeling, ltenative Repesentations, and Poblem-Solving Stategy P1.65 The volume of the galaxy is! 2 t =! ( 10 21 m) 2 ( 10 19 m) ~ 10 61 m. If the distance between stas is 4! 10 16 m, then thee is one sta in a volume on the ode of The numbe of stas is about ( 4! 10 16 m) ~ 10 50 m. 10 61 m 10 50 m sta ~ 1011 stas. P1.66 Let θ epesent the angle between the diections of and B. Since and B have the same magnitudes,, B, and R = + B fom an isosceles tiangle in which the angles ae 180!,! 2, and! 2. The magnitude of R is then R = 2cos! 2& '. [Hint: apply the law of cosines to the isosceles tiangle and use the fact that B =. ] gain,, B, and D =! B fom an isosceles tiangle with apex angle θ. pplying the law of cosines and the identity ( 1! cos ) = 2sin 2 & 2' ( D R! B FIG. P1.66 B gives the magnitude of D as D = 2sin! 2& '. The poblem equies that R = nd o cos! 2& ' = n sin! 2 & ' giving! = 2 tan 1 1 & n ' (. F =! F = (!9.î! 181ĵ ) N
6 Intoduction and Vectos P1.72 Since we have giving and + B = 6.00ĵ, ( x + B x ) î + ( + B y ) ĵ = 0î + 6.00ĵ Since both vectos have a magnitude of 5.00, we also have Fom x =!B x, it is seen that Theefoe, x 2 + 2 = B x 2 + B y 2 gives! + B = 2!! B FIG. P1.72 x + B x = 0 o x =!B x [1] + B y = 6.00. [2] x 2 + 2 = B x 2 + B y 2 = 5.00 2. x 2 = B x 2. 2 = B y 2. Then, = B y and Eq. [2] gives = B y =.00. Defining θ as the angle between eithe o B and the y axis, it is seen that cos! = = B y B =.00 = 0.600 and! = 5.1. 5.00 The angle between and B is then! = 2 = 106.