Hw2 - Chapter 3 Problem 20(a What is the sum of the...

Unformatted text preview: Chapter 3, Problem 20 (a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the angle in degrees, and (d) the angle in radians? Positive angles are counterclockwise from the positive direction of the x axis; negative angles are clockwise. E16.00mat+ 0.900rad F:5.00mat- 750° (324.00mat-I-120rad H:6.00mat— 210° (a) Number i j Units m (b) Number Lila“? Units m (c) Number Units ° (degrees) (d) Number gmgkégmi Units rad Wylie 6&0“ Ma‘l'of in I .2 ’~ A f A V A E: G as (5,4043? 4 ésfw(o.~~w ; —. 3.43A+ (/90) if ‘ / fl 1 A A ‘ {LDC (-jgjyfc .i. 86"» ('7‘537): .. . V ‘4 /\ 7: ~ fill? r2944 4' Hi 5,“ (llzrk’fi‘ T: }' /\ ¥ [’5 i " v «‘- \ /‘ f a \ \= -- /\‘ A J i ,x ~~ /§.Zo,« + “a 2‘ (a \ L2“ ”" “ "7‘ \ ' J . \i : %< J ___',_,....—.. I“ 2 _) __> A Chapter 3, Problem 32 Page 1 of 1 KHWZ~ZK | Chapter 3, Problem 32 1 In the product F = qax E, take q = -6, -‘ E’ 2 203+ 403+ 6.012 and F = 4.01 — 20j+121c. What then is E in unit-vector notation if Bx = By ? Number 1i1j1.333333333 k LINK 1D » Answer i1: significant digits are disabled; the tolerance is +/-2% Answer i2: significant digits are disabled; the tolerance is +/-2% Answer i3: significant digits are disabled; the tolerance is +/-2% taw‘wwfim. \vaw$ylmWNaw,vW”¢lytvaflWV‘Aanyygxmlnmuu»«\mlc’x\»»> WWW. no.” “Makes wwwwflywww.» "WW, 7 "MW , (w ,5 , M»»¢,,,,M,,;Wwwfiwxm WM“ MMWMWMW,Mwwwmm m'wxxwxawtv Qh,a,,ng.e,,l_/al.ues Cooyrégl‘ié @v by John :3. Sons: inc. or cempanées. All réghts seservecl, A A ,A A F W E *— A j ’4 x J I; 4- E “E ’% 'iv;\/y\/+ ’1 Vi V‘jvf x ‘3' (BY B) ‘32 B)‘ BX b% «+1-20’Stl2’l; = —(a ’M Z Ll Bxgx : -tgqet-wfi 42312—12863 + (fix-4303] Slarl mm Q (mpoml' A , .ZQ , -43 (Lax/Ami: ,> 12 =~ 1le =7 Bx=l :3; _/l New Solve %r 97: M9“) A CW‘y7cWU-‘VH A - an: -(o Cage—égaa‘t —> «i = —z% is; n36 _> -3437} :_5z_v _> 53 c 3;: ; M? p? b 3% :> g: 11r13+l.23l~:— http://cdugcn.wiIcy.com/edugen/shared/assignment/test/qpreview.uni?id=5624&operation=... 2/4/2005 [szr37 Chapter 3, Problem 66 What are (a) the x component, (b) the y component, and (c) the 2 component of F = E - b + E if 5 = 3.1i + 8.5j - 7.11;}: = -7.oi + 8.7j + 7.91:, and E = 3.oi + 2.4j + 2.31:. (dLCalculate the angle between F and the positive 2 axis. (e) What is the component of 5 along the direction of 39? (f) \LVhat is the magnitude of the component of E perpendicular to the direction of b but in the plane of a? and b? (a) Number iMin—3‘1 Units This answer has no units (b) Number Units This answer has no units (c) Number f""‘"-"i“2“§7”§ Units This answer has no units (d) Number Units ° (degrees) (e) Number Units This answer has no units (f) Number "i372‘9‘75h‘2701277"? Units This answer has no units A i:3 1:;ng : CoxJoy-FCY):» (a5 Jail-NEAR +(Q; —\o}+C2)l¢ :(si—{ikoii at)? + (2.3 %.1+Z.4\3+(*1~|—1‘q4 233i; : iaii + 2.21 411’: (M isl (M 12 (a) _\7. 1 /\ FA 4 A " A :5 wSoé: r:E/ii: eizi+ MW” ____ 2 _, - 60 K“ (flew—i~~ \ > ‘7,'— CMJ < “‘3 ’ Q7273 2.7, + i2 ail : ‘Sglq'o Ce) : 005$: Ought T ajL7L1+a}\72 ........._.‘.,W.v--—w-.e-.____‘ —-—---—-—-—~-—-—»—‘ [air Vain glsh —7.i : iiiqi lei-Alva {(23 1&1 43,? $1569" Rabe * 035;) 4' “5%) /lcrlib\] z mtg/3i (4H 3541+ (~4.\3¥.Ci]/i'€i}ti-iigz r CW" " i015 [Ci is \Qiuseft 'iiLg‘us‘ii.‘i°: *‘OiZX o Cg a '5: peat—vi: L“; lq‘Slfld: 4H»? SM (“’40: (Hui? Chapter 4, Problem 14 A J second and t (> 0) in seconds. (a) What is the acceleration when t = 2.3 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) does the speed equal 10 m/s? The velocity .3 of a particle moving in the xy plane is given by ,3 = (5.5 t - 4.7 :2) + 8.9 , with ,3 in meters per (a) Number -16.12 i Units m/s/‘Z (b) Number 0585106382979 Units s (c) Number 1.730740816453 Units s Answer a1: significant digits are disabled; the tolerance is +/-.2% Answer b1: significant digits are disabled; the reference is +/-2‘.7/a Answer :1: significant digits are disabled; the tolerance is +/'-2% (q\ a : : ,5 .—— Chapter 4, Problem 22 Page 1 of 1 lez~§l l Chapter 4, Problem 22 1 In Fig. 4-33, a stone is projected at a cliff of height h with an initial speed of 44.0 m/s directed at an angle 90 = 59.0° above the horizontal. The stone strikes at A, 5.35 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground. “Wka Fig. 4-33 x, Problem 22. (a) Number 61.526932585 Units m (b) Number 27.019846804 Units m/s (c) Number 72.573901672 Units m LINK TO L] Answer a1: significant digits are disabled; the tolerance is +/-2% Answer b1: significant digits are disabled; the tolerance is +/—2% Answer c1: significant digits are disabled; the tolerance is +/—2% «4 «\\«“:(w6>o<°¢w'\-wt\\«« , - . » WWW“ » / Wm \ 4°» -\«w4 m: (MM/w. va: vowsé)’ v50 : VOEMG- j. XDHD 90 :0 (a) use g=5°+th+gat2, 1 in 2 O J—(vosuwfi ~ ‘th7- z (445,}, 905.55— gw/szs) :éisr‘w (b) v). :VYO v5 = vac-3i: u = + (W =i6i4cuss90Y+ w w w m : 2?. O m/S V910 0’“ ‘LIDP. Till/U ‘Ld 4'0? [5 PKG“ :VBb—a‘t => 1‘: = (VBO‘W/Cj ~-; viii/3 ‘- Hekfikl = ‘ 2 V 1 - V b ’ Do 9=904w¢—gg€= wfihwg —73(%§, “P 2.. (LIL/gm 56!) _ M z . 9.8 http:// edugen.wiley.com/edugen/shared/assignment/test/qpreview.uni?id=5 65 1&operation=... 2/7/2005 .-. ’ {szré I Chapter 4, Problem 26 During a tennis match, a player serves the ball at 27.1 m/s, with the center of the ball leaving the racquet horizontally 2.43 m above the court surface. The net is 12.0 m away and 0.900 m high. When the ball reaches the net, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racquet at 500° below the horizontal. When the ball reaches the net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number. (a) Number? 0.5692287_6_8673 ‘Eumts m (b) Number {Aw-04879891924423 Units m (a) m w la ruck lion w u t: L/Vb A-L “Hug. LIME, ‘ H's kuskL l’s‘. ‘ ’ ‘ I , (L>2. (1:30 ' at > “its ' ECY T9 7. \:%—'4\/, : AK ~, g’umlr f '-%(,‘i- 0.900 :1 0.54? m = 2.4% I a‘i‘é 003 Mug (Arc MN 'l'uro leomkh G V M ‘LHM ‘1-«3 \LLL “51,15 L g; L/ Va 0339 L. \ L" 7’ : l/Xo -' (\los'mg3 (VoUfie‘ ’ if) WU$Q> lHWZ-H Chapter 4, Problem 31 A rifle that shoots bullets at 488 m/s is to be aimed at a target 40.0 m away. If the center of the target is level with the rifle, how high (in cm) above the target must the rifle barrel be pointed so that the bullet hits dead center? Number :_3.g\9‘2\1_28003342 Units cm _ l 1 :8 V3“ (joivjt’igh l ~_, —;> O = O *‘Vgt "23 ’0 szifl Chapter 4, Problem 42 Two seconds after being projected from ground level, a projectile is displaced 54 m horizontally and 68 m vertically above its point of projection. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point? (a) Number 27 Units m/s (b) Number 43.8 Units m/s (c) Number 120.673469387755 Units m Answer a1: significant digits are disabled; the tolerance is +/-2% Answer b1: significant digits are disabled; the tolerance is +/'—2% Answer c1: significant digits are disabled; the tolerance is +/—20€> / (a) U: alt = 59/1 t 1? m/s (bi bi” “lo/W655 “is” tine-‘0 -lcll 1‘0 27' V03 7’ Z) s (0% 7 3.9m: “l7 1 (a) m wt lop, v: O ...
View Full Document