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L7877777777 776i 7 7867 7`77 777777`x =(`(8`5775I7`B` 76`77`>J><;VU<)?<`8'x0|<&HhHh 8 8 8 8V&` aI꽌ɪVɭ&Y:&<&Y:&;: 꽌ɪ\8`&&꽌ɪɖ'*&%&,E'зЮ꽌ɪФ`+*xS&x'8*3Ixix&& 8 9: :' 9: :& :x)*++`FG8`0($ p,&"ųųೳŪŪųųij  !"#$%&'()*+,-./0123456789:;<=>?8  1 '" *"( (9"1 ( ,.(0# 2  /#0/#0 *?'#07#000'#07#0:"4<*55**5*%5)1)1)1)'#0/#0*5*75**5*:5/#0/#0'#07#0:::*::'#07#0"):$(%"%:$(%"%$$2%4%$$2%4%$(2()!)E(!8b $!H(+ "@H !D)"E` @ $ C ` DQ &J80^݌Hh < <݌ < < h < < < <`HJ>݌h Hh݌`葠葠ȔЖȔЖȠHIHHHHhHH݌hHhHh݌H6 (G FG HZXj 80x D9$xxH` >( Z>h Z>L=.xxH >hh@(LH>9L>HH/Hh/ H?-З( 8(& 8$8 H` *8HNx`* >x$50x*$50xL9JJJJ`HHjf5 >h $50x`HA>VD;;P >(ED Z> ?$0x8x D9- 8DD# H8`?E Vˆʎ55L $ 5 55555 &a*5`*5` "L&5_*b*JL%u**Lz%`** $  Q(lXJ̥KlV $  Q(lV eօ3L e3L &%R*L &%Q*L d' "L4% N'e*)n*5 5 &`@-e*f*f* 5 "L# "5f*`L %.* *t*Q*L$ %L&*L` "O**P**u** d' "L% N'e*)n*o*o*n* &8ɍ` ^&f\*555L& ^&NR* & "R*ΩL&)\*Z* ʽ LH*v 3h`0h8` " ['L & N'С55 &5`*A*@` +5L^6L65`  \* ?*0 '\* '  Q( ^&\*lZl^?*c*q)H c*h`f*5h*5j*555@O*AP*`u*@`@5&`Q*R*`E 'Ls' ' ' @DAE@u*`8` %@ @A@`@`**@*A`M5 ) "L&`8@AW*c*@8@-@HAȑ@hHȑ@ȑ@hHȑ@Ȋ@c*h8&ȑ@Hȑ@Ah@L'Hȑ@ȑ@* htphso`hMhL`9V*8U*897T*6S*67`INILOASAVRUCHAIDELETLOCUNLOCCLOSREAEXEWRITPOSITIOOPEAPPENRENAMCATALOMONOMOPRINMAXFILEFINBSAVBLOABRUVERIF!pppp p p p p`" t""#x"p0p@p@@@p@!y q q p@  LANGUAGE NOT AVAILABLRANGE ERROWRITE PROTECTEEND OF DATFILE NOT FOUNVOLUME MISMATCI/O ERRODISK FULFILE LOCKESYNTAX ERRONO BUFFERS AVAILABLFILE TYPE MISMATCPROGRAM TOO LARGNOT DIRECT COMMANč$3>L[dmx- (   Ϡ734@~3!+,W,o,*--,,9,-.-~3~3,,,,~3~3,,,,~3_*3 j.5 *H*H`Lc3 (+L3 +55555 1^3_* )3J Q*L_3Ls3443 D23455545554 70 / 7 :/354545455454555535L^/5-5I5555 55` / 4/ 2-5! / 08555454 70L3 (+50+5B5C3 2 70L35 *H*H`Lg3L{3505 *H*H` 3 ,5L3 3 1 ,H 1hBL, 0 BH [1 1h`Lo3 35 ,L3 3 1B , 1L,H 0hB@ 55 [1L133 (+34) 34 70L3 3L3 (+ 055L- (+34L{33454445 70 ^/* / 3B0 HȱBh -355 -8 /L38 25` +5 /3 /. /. 3 E7D B. /. /. 0]34S0J4 4) 3 4D4E B. 4  /. 02L3 Ν3 3`HD٤33DEEhiHLG.h ` /5B5-` /5B-` + X05I33383 33DH /4 E03744Ȍ7 X040 7 J7L35B5C`,5p` / R0-55`50` K/ R0-55`575755` 4/ K/ /( 55L/BȱBL/58` D2BH5BH :/ 75Bȭ5Bhh55 R0B5m55ȱB5m55` /LR0557755`*7*75LR0 E0(33 48`433 R0` E033LR0*7*7`777 555I7575757577** 7757`7 L35555f /55555555 ^/`855 i /B58` 41L 15ȱB5 /5555 /5`3 D23ȑB55B5 / 7 55`5555555555 5555`555`55BC55`5555`L3 /5B5C355 0Q340"B4 3` 02۰Ϭ33B438`3i#`3ЗLw35!50>5555`53 /3m3 3 3733i35 3583 /35ЉLw35`H /5h 25L/~43 533`55555555J5m55m5jn5n5n5ۭ55m5555m55m55`"L? 585H ~.(3` # d ֠z# u*`@ '3<JRXdq~ ,,$$36..$$%%..">'$$$..&%%666$$$--:?6.-= $,,-5>?<I9?'$$,$$$-557?$$$M1$">66">"">">6>>"">"">">">6>>">UX115,6745,2151,15:#':100:8M'" ************************":q$'" STATICS - METHOD".':" ************************"8'255:D11000:::(7):B'"";A$:L'31:"";A$:6">5:X5,Y:X5,Y5:X6,Y1:X6,Y6:y2#3:X1,YX3,Y:X4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:<#3:X,Y2X,Y4:X1,Y1X5,Y5:X1,Y5X5,Y1:F#3:X5,Y1X5,YX,YX3,Y3X,Y6X5,Y6X5,Y5:#3:51,15121,61OU WANT TO REVIEW,"::"TYPE 1. IF YOU'RE READY TO END THIS":b>"PROGRAM, TYPE 2: ";A:A1250H13376:D13:(7)::19:"END"RD$;"PR#0":\D11500:::D$;"RUN MENU"'#%(#3:X,Y5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,YNOWNS.";:10060:p13376:::"THAT'S IT FOR NOW. YOU SHOULD BE"::"ABLE TO RESOLVE FORCES, FIND RADIUS": "ARMS, AND CALCULATE TORQUES. OTHER"::"PROGRAMS SHOW HOW TO SOLVE COMPLETE":*"PROBLEMS.";:10060>434,0:13376:10:Z$;", IF Y)(Ty)";/X31:Y166:9000:Y182:9000:1006013376::"NOTICE THAT BOTH TORQUES ARE POSITIVE"::"(COUNTER-CLOCKWISE).";:10060::"#11 AND #12 HERE YOU WOULD SET THE SUM"::" OF ALL TORQUES EQUAL TO ZERO AND": " SOLVE FOR ANY UNK80:Y60:9010R13376:"#9 CALCULATE THE RADIUS ARM FOR EACH"::" FORCE:":" Ra = L * sin( )"::" Rb = L * cos( )";X127:Y118:9010:Y134:9010:10060:"#10 CALCULATE THE TORQUES:"::" a = (Ra)(Tx)"::" b = (Rb ROTATION POINT. WE'LL"::" CHOOSE THE HINGE POINT.";:10060{:"#8 DRAW LINES OF ACTION FOR THE FORCES."::100606:118,44118,10:105,6435,64:24:"*** NOW AGAIN: ";A$ 5:62,18116,18:47,2647,63:2:14:"Rb":7:5:"Ra":X sin( )":9:25:"Tx = T * cos( )"HX218:Y38:9020:X260:Y70:902023:"*** NOW AGAIN: ";A$:13376:"#5 AND #6 HERE'S WHERE YOU WOULD SUM":" FORCES IN THE x-DIRECTION AND THE"::" y-DIRECTION.";:10060?:"#7 SELECT THE" THE TENSION VECTOR).";:10060v::"#3 ASSUME CONVENTIONS ARE ESTABLISHED."::10060::"#4 RESOLVE TENSION INTO COMPONENTS."::100602:118,47118,64162,64:3:116,48120,48:160,63160,65"7:25:" ":X143:Y62:9020:5:19:"Ty = T *WE WON'T DO THE WHOLE PROBLEM NOW."::"LET'S JUST LOOK AT HOW TO HANDLE THE":X"LOWER END WHERE THE CABLE IS CONNECTED."::10050b62450:9100:1:118,64163,47:3:160,47162,49:7:25:"T""l13376:"#2 HERE'S THE FREE-BODY (SHOWING ONLY"::AT A"::"SIMPLE EXAMPLE.";:10060E&13376:5:20,12270,12:9100x02:X15:45X,1345X,19::1:118,64251,13:X72:Y22:9010:X210:9020D34,10:13376:"#1 HERE'S OUR PICTURE--A BEAM OF LENGTH"::" L, SUPPORTED AS SHOWN."::ON"EPS 5,6, & 11 TO SOLVE FOR ANY":I" UNKNOWN QUANTITIES.";:10060::"THIS IS A LONG ONE! BEFORE WE GO ON TO"::"AN EXAMPLE, YOU MIGHT WANT TO REVIEW."::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1250"::"OK, ";Z$;". LET'S LOOK ::" DICULAR DISTANCE FROM ORIGIN) TO"::" EACH LINE OF ACTION.":x">#10 CALCULATE EACH TORQUE = (r)(F).":">#11 SUM ALL TORQUES AND SET EQUAL TO"::" ZERO."::10050!13376:3:">#12 USE THE EQUATIONS DEVELOPED IN"::" ST IN THE y DIRECTION"::" AND SET EQUAL TO ZERO."::1005013376:3:">#7 SELECT A POINT OF ROTATION FOR"::" CALCULATION OF TORQUES.":">#8 DRAW A 'LINE OF ACTION' FOR ALL"::" FORCES.":J">#9 CALCULATE THE RADIUS ARM (PERPEN-"" COUNTER-CLOCKWISE TORQUES ARE ALL"::" POSITIVE.":">#4 RESOLVE ALL FORCES INTO x AND y"::" COMPONENTS (p AND n ON INCLINES).":">#5 SUM ALL FORCES IN THE x DIRECTION"::" AND SET EQUAL TO ZERO.":=">#6 SUM ALL FORCESE-BODY DIAGRAM. SHOW ONLY"::" THE FORCES ACTING ON THE OBJECT.":" (TENSION, GRAVITY, HINGE, FRICTION"::" SUPPORT REACTION, ETC.)";:1006013376:3:">#3 ESTABLISH CONVENTIONS: FORCES"::" POINTING UP OR TO THE RIGHT AND":@SIS. THERE"::"ARE QUITE A FEW STEPS AND IT'S EASY TO": h"GET LOST. BE SURE THAT YOU WORK"::" CAREFULLY AND ORDERLY! ";:10060:: r"METHOD - COPY IT DOWN!"::">#1 START WITH A DRAWING. SHOW ALL THE"::" FORCES.":H|">#2 DRAW A FRE FORCE OF A HINGE"::" W - WEIGHT = (M)(g)":w J" Fn - NORMAL FORCE"::" Ff - FRICTION FORCE = (u)(Fn)":: T"YOU SHOULD ALREADY BE FAMILIAR WITH"::"THESE. WRITE DOWN THE NOTATIONS."::100509 ^13376:3:"NOW ON TO THE METHOD OF ANALY0"::= "X30:Y92:9030:Y110:9030:Y126:9030:X44:9000 ,"GET THEM RIGHT, ";Z$;"? OF COURSE YOU"::"DID! LET'S LOOK AT SOME OF THE": 1"FORCES YOU'LL WORK WITH.";:10060 613376:3:"TYPES OF FORCES:"::" T - TENSION":1 @" H -::"CALCULATOR READY, LET'S GO."::10050 13376:3:"YOU SHOULD ALREADY KNOW THE CONDITIONS"::"FOR EQUILIBRIUM. WRITE THEM ON YOUR": "PAPER, THEN ";:13:10050: :"HERE THEY ARE:": " Fx = 0"::" Fy = 0"::" = HANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL DESCRIBE A METHOD FOR ANALYZING": "THE FORCES AND TORQUES WHICH ACT ON A"::"BODY IN STATIC EQUILIBRIUM. IT'S A": "LITTLE INVOLVED, SO PAY CLOSE ATTENTION!":* "IF YOU HAVE YOUR PAPER, PENCIL, AND"d- STATICS - METHOD&n:100006x:16302,0N- TURN ON ROMPLUS+ZD$(4)iD$;"PR#5"vR$(13)~R$M$"":- THERE IS A HEREM$;"1A"13376:3:"HELLO, FRIEND!"::"WHAT'S YOUR NAME? ";Z$::L "T                  ***********":21$'" PROJECTILE MOTION"\1.':" ************************"{18'255:A11000:::(7):1B'"";A$:AROBLEMS FOR PRACTICE. LOOK"::"IN YOUR TEXT OR SEE YOUR INSTRUCTOR.":q0v"GOOD LUCK. BYE FOR NOW!":24:100500:D13:(7)::20:"END"0D11500:::(4);"RUN MENU"0?0(#12:2,3737:2,3738:0':100:81'" *************3 - 1/2*9.8*(1.3)^2"::I/N"SOLVING--"::7::"Y = 3.67 METERS":::/X"THAT ALL THERE IS TO IT! TYPE 1 TO"::"REVIEW, 2 TO CONTINUE: ";A:A11570/b:6:"THAT'S THE END OF THIS PROGRAM."::Z$;", YOU SHOULD PROBABLY WORK":G0l"SOME EXTRA PME: ";A::c.&"REMEMBER THAT X & Y MOTIONS DO NOT"::"AFFECT EACH OTHER, BUT THE FLIGHT TIME":.0"IS THE SAME. SO, T= 1.3 SECONDS."::10050.::3:"THEREFORE--"::" Y = V(0)Y*T - 1/2*G*T^2":/D:"SUBSTITUTING--"::" Y = 9.192*1. 10/7.714":3-5::"T = 1.3 SECONDS":::10050-:4:"HOW DO WE FIND THE ELEVATION WHERE THE"::"ARROW HITS THE CLIFF? ";:10050::-"HINT: HOW MUCH TIME DOES THE ARROW HAVE"::"TO GO UP AND DOWN?": ."INPUT THE ARROW'S FLIGHT TI0)Y--":a," V(0)X = 12*COS(50) = 7.714 M/S"::" V(0)Y = 12*SIN(50) = 9.192 M/S"::10050,::"SINCE WE KNOW THE RANGE (10 METERS) AND"::"AND THE HORIZONTAL VELOCITY (7.714 M/S)," -"WE CAN FIND THE FLIGHT TIME--"::" T = RANGE/V(X) =9,20;+T11400::22:" TIME MARKERS SHOW THE MOTION."]+Y501:7:203Y,11Y2:|+Y03:7:203Y,11Y2:+T11000::24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A11620,::2:"HERE'S THE SOLUTION:":::" FIND V(0)X & V(24:100507*T::9000:1:X3035:14,36X::7:5,36*^22:" PRESS TO SHOOT THE ARROW.";A$:T11200:::(7)*hY50.5*r7:203Y,11Y2:0:203Y,11Y2:*|YO3.5*7:203Y,11Y2:0:203Y,11Y2:+(7):7:2REES ABOVE THE"::"HORIZONTAL AND FIRED WITH AN INITIAL":)6"VELOCITY OF 12 M/S TOWARD A VERTICAL"::"CLIFF 10 METERS AWAY. HOW HIGH ABOVE":)@"THE GROUND DOES THE ARROW STRIKE THE"::"CLIFF?":: *J"COPY THE PROBLEM AND THEN WE'LL WATCH.": THE RANGE--"::" RANGE = V(X)*T":(" RANGE = 12 * 4"::8::"RANGE = 36 METERS"::::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A11240(":3:"LET'S DO ONE MORE PROBLEM, THIS TIME"::"IN REVERSE:"::;),"AN ARROW IS AIMED 50 DEG;'"OR,"::" -78.4 = 0*T - 1/2*9.8*T^2"::10050:b'"THEN,"::" T^2 = 78.4/4.9":'"AND,"::5::"T = 4 SECONDS":::10050:1550':10:"GREAT WORK, ";Z$;"!"::"KEEP GOING...":22:10050*(:5:"SINCE T = 4 SECONDS, WE FINDSE: Y = V(0)Y*T - 1/2*G*T^2":::10050&::"REMEMBER THAT Y = -78.4 METERS, SINCE"::"THE FINAL POSITION IS BELOW THE START."::&" INPUT THE FLIGHT TIME: ";A::A41540':3:(7):"SORRY, ";Z$;".":::" Y = V(0)Y*T - 1/2*G*T^2"::"BECAUSE THE BALL IS KICKED HORIZONTALLY.":T%" ENTER V(0)Y: ";A::A01460b%"SURE!";%25::"V(0)Y = 0 M/S":::"SINCE THE BALL HAS NO INITIAL VERTICAL"::"VELOCITY. ";:10050*&:4:"NOW FIND THE TIME OF FLIGHT--":::"U$n::2:"NOW FOR THE SOLUTION. YOU SHOULD BE ABLE"$x"TO GET THIS ONE ON YOUR OWN.":::"START BY FINDING THE X & Y COMPONENTS":$"OF THE INITIAL VELOCITY."::" ENTER V(0)X: ";A::A121430$"OF COURSE!";.%25::"V(0)X = 12 M/S"::13X,11:#2Y05.5A#<9:162Y,11Y2:0:162Y,11Y2:a#F(7):9:26,36:T11200:#P22:" TIME MARKERS SHOW THE MOTION."#Z9:14,11:Y05:162Y,11Y2::T11000:#d24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A112805OCITY IS":]""12 METERS/SECOND?":::"COPY THE PROBLEM. THEN WE'LL LOOK AT IT."::10050"::9000:3:X515:12,36X::9:13,11" 22:" PRESS TO KICK THE BALL.";A$":T1800::(7)"X02:9:13X,11 #(T150:T:0:NNON TO SHOOT THAT"::"HIGH.":22:10050T!:3:"HOW ABOUT ANOTHER PROBLEM?"::!"A MAN KICKS A BALL HORIZONTALLY OFF THE"::"ROOF OF A BUILDING 78.4 METERS TALL.": ""HOW FAR FROM THE BASE OF THE BUILDING"::"DOES IT LAND IF THE INITIAL VEL (173.2)^2 - 2*9.8*Y"::" Y = (173.2)^2/(2)(9.8)":j "AND,":9::"Y = 1530.5 METERS"::: "IT'S A BIG CANNON!":::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A1910 1240)!:10:"EXCELLENT WORK, ";Z$;"!"::"IT'S A PRETTY BIG CAON HAS NO EFFECT ON HOW":i"HIGH IT GOES.":::"INPUT YOUR MAX. HEIGHT: ";A::A1530A15311230:4:(7):"SORRY. HERE'S THE CORRECT SOLUTION:":"AT MAXIMUM HEIGHT, V(Y) = 0. SO,"::" V(Y)^2 = V(0)Y^2 - 2*G*Y":< " 0^2 =EVIEW THIS":Yt"PROBLEM SOLUTION AGAIN, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A1910~::"AS A CONTINUATION OF THIS PROBLEM, FIND"::"THE MAXIMUM HEIGHT OF THE SHELL."::10050:3:"HINT: THE FACT THAT THE SHELL MOVES IN"::"THE X DIRECTI= 173.2/4.9 = 35.35 SEC":::kB"* NOW YOU CALCULATE THE RANGE."::" INPUT YOUR ANSWER (NO UNITS):";AL:5:A35351120V"GOOD!";`10::"RANGE = V(X)*T = 3535 METERS"::: j"THAT'S NOT SO BAD, IS IT?"::::"IF YOU WOULD LIKE TO RND,":"SO: Y = Y(0) = 0 METERS ";:10050y:3:"USE THIS EQUATION:"::" Y = V(0)Y*T - 1/2*G*T^2"::10050$:"THEN,":" 0 = 173.2*T - 1/2*9.8*T^2":."AND":" -173.2*T = -1/2*9.8*T^2"::100508:"FINALLY,"::6::"T 73.21010"YES!";I9::"V(0)Y = V(0)*SIN(60) = 173.2 M/S":::"NOW USE THE VALUE OF V(0)Y AND THE"::"EQUATIONS OF MOTION FOR GRAVITATIONAL":"ACCELERATION TO FIND THE TIME OF FLIGHT.":/:"REMEMBER THAT THE SHELL HITS THE GROU:"INPUT YOUR ANSWER FOR V(0)X: ";A::A100970A"GOOD!";x10::"V(0)X = V(0)*COS(60) = 100 M/S":::10050:3:"REMEMBER THAT THIS IS THE VALUE OF THE"::"CONSTANT HORIZONTAL VELOCITY.":: "INPUT YOUR VALUE FOR V(0)Y: ";A::A1CANNON SHELL IS FIRED WITH A MUZZLE":}"VELOCITY OF 200 M/S AT AN ANGLE OF"::"60 DEGREES ABOVE HORIZONTAL. FIND THE":"RANGE OF THE SHELL. ";:10050::"COPY THE PROBLEM. THEN TRY TO FIND THE"::"X & Y COMPONENTS OF V(0).":3 THE GROUND IS X(0) = 0, Y(0) = 0.":|z"> THE STARTING TIME IS T(0) = 0."::"> THE ANGLE (THETA) IS MEASURED FROM ":" THE X-AXIS.":::"COPY THESE CONVENTIONS IF YOU NEED TO."::"THEN, ";:10050&:3:"NOW HERE'S A PROBLEM:":::"A METHOD FOR REFERENCE: ";:10050\:2:"THE CONVENTIONS WE USE ARE THE SAME AS"::"THOSE ESTABLISHED IN PREVIOUS PROGRAMS:"::f"> UP AND RIGHT ARE POSITIVE."::"> DOWN AND LEFT ARE NEGATIVE.":'p"> THE STARTING POSITION (EVEN IF ABOVE"::" E: V(X) = CONSTANT V(0)X!"::" V(0)Y = V(0)*SIN(THETA)">" NOTE: V(Y) WILL CHANGE!"::" 4. APPLY EQUATIONS OF MOTION IN THE"H" Y DIRECTION TO FIND THE TIME OF":" FLIGHT (T)":R" 5. FIND RANGE = V(X) * T"::"COPY THE ONE! ";:10050T:2:"METHOD:"::" 1. ESTABLISH CONVENTIONS": " 2. DETERMINE KNOWN & UNKNOWN VALUES"::" 3. FIND THE X & Y COMPONENTS OF THE"*" INITIAL VELOCITY--"::" V(0)X = V(0)*COS(THETA)">4" NOTE (THETA) OF":h"THE PROJECTILE. YOU ARE TO SOLVE FOR THE":"RANGE (MAX. HORIZONTAL DISTANCE)."::"OF COURSE, THE PROBLEM CAN BE CHANGED"::"AROUND (GIVEN RANGE AND ANGLE, FIND THE": "INITIAL VELOCITY). BUT, LET'S START WITH":"THE EASY NT TO WATCH AGAIN, ";Z$;"?"::"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370:3:"NOW LET'S LOOK AT A METHOD FOR SOLVING"::"PROJECTILE PROBLEMS. ";:10050::"IN THE BASIC PROBLEM, YOU ARE GIVEN THE"::"INITIAL VELOCITY AND ANGL IN THE X DIRECTION AND BY THE"::"ACCELERATED MOTION IN THE Y DIRECTION."::"THEREFORE, THE TIME IT TAKES THE PRO-"::"JECTILE TO GO UP AND DOWN IS EXACTLY":"THE AMOUNT OF TIME THE PROJECTILE HAS"::"TO MOVE SIDEWAYS."::P"DO YOU WA11Y15:T11000:TU11:10,36203Y:7:21,3611Y2:1:203Y,11Y2:24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1620::3:"YOU SHOULD SEE THAT THE PROJECTILE"::"POSITION IS DETERMINED BY THE CONSTANT":O"VELOCITY13:203Y,11Y2:Pb24:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:A1370l::9000:1:5,36:22:" HERE IT IS IN SLOW MOTION.":T11500:vY501:T11000:T11:10,36203Y:7:4,2011Y2:1:203Y,11Y2:7:21,3613:5,36:T11000:&Y50.5Q13:203Y,11Y2:0:203Y,11Y2:_&YO5.5013:203Y,11Y2:0:203Y,11Y2::13:35,36DT11000::22:" TIME MARKERS SHOW THE MOTION."NY501:13:203Y,11Y2:XY05:1000:Y50.567:4,11Y2:0:4,11Y2:DY05.5m7:4,11Y2:0:4,11Y2::7:4,36T11000::22:" TIME MARKERS SHOW THE ACCELERATION."7:Y05:4,11Y2::24:10050:22:" NOW COMBINE THE MOTIONS." |11:5,10:T11000:( Y010.1T 11:53Y,10:0:53Y,10::11:35,10 T11000::22:" TIME MARKERS SHOW CONSTANT VELOCITY. " 11:Y010:53Y,10::24:10050 :22:" HERE'S A VERTICAL ACCELERATION."7:4,36:T1NSIDER THE PROJECTILE MOTION AS TWO":m ^"SEPERATE MOTIONS WHICH LAST FOR THE"::"SAME AMOUNT OF TIME.": h:"LET'S SEE HOW THE MOTIONS COMBINE."::"WATCH CAREFULLY, ";Z$;"!":::10050 r::9000:22:" HERE'S A CONSTANT HORIZONTAL VELOCITY."ICAL - ACCELERATION DUE TO"= ," GRAVITY"::H 610050 @:3:"THESE TWO MOTIONS OCCUR AT THE SAME"::"TIME, AND THE RESULT IS PARABOLIC": J"PROJECTILE MOTION. ";:10050::& T"THE TRICK IN SOLVING PROBLEMS IS TO"::"CO SOLVE PROBLEMS."::m "PROJECTILE MOTION CAN BE CONSIDERED AS"::"TWO SIMULTANEOUS MOTIONS WHICH ARE": :"INDEPENDENT": 12:13:"EXECPT FOR THE TIME FACTOR.":::"THE MOTIONS ARE:": "" 1. HORIZONTAL - CONSTANT VELOCITY"::" 2. VERTFFICULTY WITH":_ "THEM--SO PAY CLOSE ATTENTION.":::"YOU SHOULD HAVE ALREADY WATCHED:": :"KINEMATICS & GRAVITATIONAL ACCELERATION":::"WHEN YOU'RE READY, ";:10050 :3:"THE BASIC IDEA IS SIMPLE, BUT YOU MUST"::"UNDERSTAND IT WELL TO)d- PROJECTILE MOTION%n10000[:3:"HELLO!"::"PLEASE TELL ME YOUR NAME: ";Z$8:"THANK YOU, ";Z$;". THIS PROGRAM"::"IS DEVOTED TO THE SOLUTION OF PROJECTILE" "PROBLEMS. ALTHOUGH THEY'RE REALLY NOT "::"TOO HARD, STUDENTS HAVE DI                   MOMENTUM"42.':" ************************"S28'255:A11000:::(7):h2B'"";A$:A19X,13.75X1 O1!23:"TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A18400U1 !1(#" P1 + P2 = P1' + P2'"::" M1*V1 + M2*V2 = M1*V1' + M2*V2'"::1':100:81'" ************************": 2$'" CONSERVATION OF LINEAR:10,11154X:10,1116X:10,1115X:x0D 15,1612:15,169:23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A18200~0N 0 ::21:10:"2-DIMENSIONAL RECOIL":15:19,130 T1800::(7);(7)0 X115.51 1:19X,13:2:19,131.60X:12:110:10,111:12:10,11306/ T1800::(7):(7)/& X014:1:10,11X:10,11X1:12:10,1130X:T140::0:10,11X:10,11X1:10,1130X:/0 (7):21:15:"COLLISION"(0: X06.25:12:10,11154X:1:10,1116X:10,1115X:T135::0(7)P.^X010.25:1:10,11(10.9X):10,11(11.9X):13:10,11(122.7X).h0:10,11(10.9X):10,11(11.9X):10,11(122.7X):X.r23:" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE?";A:A18000.|.. ::21:15:"COLLISION"/ 1:10,"::"SHOULD HOLD THE RIFLE FIRMLY!":::^-"THAT'S IT FOR NOW. SEE YOU LATER!":23:10050~-:D13:(7)::20:"END"-"D11500:::(4);"RUN MENU"-?-@::21:17:"RECOIL"-J1:10,1110:10,1111:13:10,1112.TT1800::(7): THE RECOIL WHEN A 95 KG MAN HOLDS":,"THE RIFLE FIRMLY AGAINST HIS SHOULDER"::"(SO THE RECOIL MASS IS 100 KG).":::10050,:4:"ANSWERS:"::" 1. V2' = -.90 M/S":," 2. V2' = -.045 M/S"::*-"THE DIFFERENCE EXPLAINS WHY A SHOOTERTINUE: ";A:A11160p+:4:"WELL, ";Z$;", THAT'S ABOUT IT.":::"HERE'S ONE LAST PROBLEM FOR PRACTICE:"::+"A 5 KG RIFLE FIRES A 15 GM BULLET WITH"::"A MUZZLE VELOCITY OF 300 M/S. FIND THE":%,"RECOIL VELOCITY OF THE RIFLE. ALSO,"::"FIND THETA = ARCTAN (4.267/-6.8)"::8::"THETA = 32.1 DEGREES"::*"WHERE THETA IS THE ANGLE ABOVE THE"::"NEGATIVE X-AXIS. ";:10050*8400:*:6:"IF YOU WOULD LIKE TO SEE THE LAST "::"PROBLEM AGAIN, TYPE 1. TYPE 2 TO":+"CON:24:10050d)x:3:"NOW WE MUST FIND THE TOTAL V3. WE USE"::"TRIG & THE PYTHAGOREAN THEOREM:":)10050::" V3 = SQRT";(91);" V3(X)^2 + V3(Y)^2 ]"::" V3 = SQRT";(91);" -6.8^2 + 4.267^2 ]":)3::"V3 = 8.03 M/S":::"AND,":D*" ::10050Y(P:4:"SUBSTITUTING:"::" 8*0 + 15*0 = 8*(-8) + 15*V3(Y)'"::10050(Z:"FINALLY,"::" 0 + 0 = -64 + 15*V3(Y)'":(d"AND,"::" V3(Y)' = 64/15 = 4.267 M/S":::10050 )n:"SO M3'S Y COMPONENT IS POSITIVE." 15*V3(X)' = -102":U'("AND,":" V3(X)' = -6.8 M/S"::10050::'2"SO THE X COMPONENT OF M3'S VELOCITY IS"::"IN THE NEGATIVE DIRECTION."::'<"NOW FOR THE Y DIRECTION:"::10050 (F:" M2*V2 + M3*V3(Y) = M2*V2' + M3*V3(Y)'"NOW LET'S LOOK AT THE X DIRECTION:"::10050g&:" M1*V1 + M3*V3(X) = M1*V1' + M3*V3(X)'"::10050:& "SUBSTITUTING:"::" 17*0 + 15*0 = 17*6 + 15*V3(X)'":&"OR,":" 0 + 0 = 102 + 15*V3(X)'"::10050':4:"THEN,":" "BOTH PARTS--IT WILL HAVE MOMENTUM COM-"::"PONENTS IN BOTH DIRECTIONS & WILL FLY":q%"OFF AT AN ANGLE."::%"FIND THE MASS OF PART #3."::"ENTER YOUR ANSWER WITHOUT UNITS: ";A%:3:"MASS OF #3:"::" M3 = 40 - 17 - 8 = 15 KG"::,&":"COPY THE PROBLEM, MAKE A SKETCH, AND"::10050=$8400:$:3:"YOU SHOULD NOTICE THAT PART #1 HAS ONLY"::"AN X COMPONENT WHILE PART #2 HAS ONLY A":$"Y COMPONENT. SINCE THE INITIAL TOTAL"::"MOMENTUM IS ZERO, PART #3 MUST BALANCE":U%IN THE POSITIVE X"::"DIRECTION AT 6 M/S. THE SECOND PART":#"(M = 8 KG) MOVES IN THE NEGATIVE Y"::"DIRECTION AT 8 M/S. WHAT IS THE VELOCITY"#"(MAGNITUDE & DIRECTION) OF THE THIRD"::"PART IF THE SHELL HAD A TOTAL MASS OF":1$"40 KG?":22SMALL. WE USUALLY"7"t"NEGLECT THIS EFFECT."::10050"~20:"GETTING TIRED, ";Z$;"? WELL, WE'RE"::"ALMOST DONE!":24:10050":3:"HERE'S A GOOD 2-DIMENSIONAL PROBLEM:"::"A SHELL EXPLODES INTO 3 PARTS. ONE PART":=#"(M = 17 KG) FLIES OFF :3:"SINCE THE SHELL HAS AN UPWARD MOMENTUM,"::"THERE MUST BE A BALANCING MOMENTUM DOWN."!`"BUT IT IS THE GUN AND THE EARTH WHICH"::"RECOIL IN THE Y DIRECTION. AND BECAUSE":"j"THE EARTH IS SO MASSIVE, IT'S RECOIL"::"VELOCITY IS VERY, VERY ." 0 + 0 = 1040 + 600*V2(X)'"::"OR,"::" 600*V2(X)' = -1040":} 8"AND,":8::"V2(X)' = -1.73 M/S"::: B"SO THE GUN RECOILS BACKWARDS WITH A"::"VELOCITY OF -1.73 M/S. WHAT ABOUT THE": L"Y DIRECTION? ";:10050]!VYOU SHOULD VERIFY THESE VALUES."::10050::|"NOW LET'S CONSIDER ONLY THE X-AXIS:"::" M1*V1(X) + M2*V2(X) =":" M1*V1(X)' + M2*V2(X)'"::10050$:3:"SUBSTITUTING:"::" 4*0 + 600*0 = 4*260 + 600*V2(X)'"::10050:OCITY OF THE GUN?":24:"COPY THE PROBLEM AND ";:10050:3:"SINCE THE SHELL'S VELOCITY IS AT AN"::"ANGLE, THE FIRST STEP IS TO FIND THE":"X & Y COMPONENTS:"::" V1(X) = V1*COS(30) = 260 M/S":-" V1(Y) = V1*SIN(30) = 150 M/S"::" M/S":::10050::n"NOW LET'S DO A 2-DIMENSIONAL PROBLEM:"::"A 600 KG GUN IS MOUNTED ON WHEELS. IT":"FIRES A 4 KG SHELL WITH A MUZZLE"::"VELOCITY OF 300 M/S AT AN ANGLE OF 30":"DEGREES ABOVE HORIZONTAL. WHAT IS THE"::"RECOIL VELOCOIL VELOCITY OF THE GUN?"::"ENTER YOUR ANSWER WITHOUT UNITS: ";A:A1.5960]::(7)"SORRY!"::" .015*0 + 6*0 = .015*600 + 6*V2'"::" 0 + 0 = 9 + 6*V2'"::10050:3:"THE CORRECT ANSWER IS:"::10::"V2' = -1.5400*V2'":" 0 + 0 = 100 + 400*V2'"::" 400*V2' = -100"::13::"V2' = -.25 M/S"::::10050:4:"LET'S CONTINUE:"::"A 6 KG RIFLE FIRES A 15 GM BULLET AT A":O"MUZZLE VELOCITY OF 600 M/S. WHAT IS "::"THE REULDN'T"::"THE SIGN BE NEGATIVE?"6f::10050:900p"EXCELLENT, ";Z$;"! THE RECOIL VELOCITY"::"IS -.25 M/S, ASSUMING YOU SAID THE MAN":z"HAD A POSITIVE VELOCITY.":::10050:920 :4:"SOLUTION:"::9000::" 50*0 + 400*0 = 50*2 + SO, YOUR NUMERICAL VALUE IS NOT"::"CORRECT."74850>"YOU SEEM TO UNDERSTAND THAT THE"::"VELOCITY OF THE BOAT SHOULD BE":H"NEGATIVE, BUT YOU HAVE THE WRONG VALUE."R::"TRY AGAIN.":::10050:720"\"YOU GOT THE RIGHT VALUE, BUT SHO4A.25880KNN1:(7);"SORRY, ";Z$;", YOUR ANSWER IS WRONG.":XN3900g A.25860tA0830 "SINCE THE INITIAL SYSTEM MOMENTUM IS"::"ZERO, AND THE MAN HAS A (+) VELOCITY,"::"THE BOAT SHOULD HAVE A (-) VELOCITY.":.*"AL":`"400 KG BOAT WITH A VELOCITY OF 2 M/S."::"WHAT IS THE RECOIL VELOCITY OF THE BOAT?":"RECOGNIZE THAT THE TOTAL MOMENTUM IS"::"INITIALLY ZERO, AND THAT YOU MAY WORK":"IN ONE DIMENSION.":20:"ENTER YOUR ANSWER WITHOUT UNITS: ";A::AL":^"MOMENTUM OF THE SYSTEM WAS ZERO, THE"::"'POSITIVE' MOMENTUM OF THE 15 KG MASS":"MUST BE BALANCED BY THE 'NEGATIVE'"::"MOMENTUM OF THE 5 KG MASS.) ";:10050:N0:4:"NOW YOU TRY ONE:"::"A 50 KG MAN DIVES HORIZONTALLY FROM Az#??          Š+ӯͧӠנ Š"ҠϠҠŠ! Š àӠ"Ϡ  ɭӠǠȠӠ7͠,+ 5*V2'":D"OR,":" 0 + 0 = 45 + 5*V2'":21:10050x:3:"FINALLY,"::" 5*V2' = -45":"AND,":15::"V2' = -9 M/S":::10050::"YOU SHOULD RECOGNIZE THAT THE 5 KG MASS"::"MOVES TO THE LEFT. (SINCE THE INITICE THAT WE HAVE ASSUMED THE INITIAL"::"VELOCITY IS ZERO (SINCE IT ISN'T GIVEN),"v"AND THE 15 KG MASS HAS A (+) VELOCITY.":23:10050:3:"NOW THE CONSERVATION LAW:"::9000::10050:: "SUBSTITUTE THE DATA:"::" 15*0 + 5*0 = 15*3 THE RECOIL PROBLEM YOU HAVE":PD"SEEN?--OF COURSE YOU DO!":22:10050:8000N::5:"HERE'S HOW TO SOLVE THE PROBLEM:":X"DATA:"::" M1 = 15 KG V1 = 0 M/S V1' = +3 M/S":b" M2 = 5 KG V2 = 0 M/S V2' = ?"::10050::Rl"NOTI5 KG MASS ARE PLACED":s&"ON A FRICTIONLESS SURFACE. AN EXPLOSION"::"DRIVES THEM APART, AND THE 15 KG MASS":0"MOVES TO THE RIGHT AT 3 M/S. WHAT IS"::"THE VELOCITY OF THE 5 KG MASS?":::"COPY THE PROBLEM. DO YOU REALIZE THAT"::"THIS ISWRITE DOWN THE STEPS!":Q"ENTER 1 TO REVIEW, 2 TO CONTINUE? ";A:::A1370"NOW WE'RE READY TO SOLVE A SIMPLE"::"PROBLEM. WE'LL BEGIN WITH 1 DIMENSIONAL":"MOTION (NO Y COMPONENT). ";:10050:3:"PROBLEM:"::"A 5 KG MASS & A 1":3" P(X)=M*V(X) P(Y)=M*V(Y)":24:10050:3:" C. APPLY THE CONSERVATION LAW"::" P1(X)+P2(X)=P1(X)'+P2(X)'":" P1(Y)+P2(Y)=P1(Y)'+P2(Y)'":::" D. SOLVE FOR UNKNOWNS (USUALLY A":" VELOCITY).":::"BE SURE TO GN CONVENTION"::" IS NEEDED (UP & RIGHT ARE POSITIVE,":o11:" WHILE DOWN & LEFT ARE NEGATIVE.)"::"3. STEPS TO SOLUTION:"::" A. RESOLVE VELOCITIES INTO X & Y":" COMPONENTS - V(X) & V(Y)."::" B. FIND MOMENTUM COMPONENTS -M AFTER"::" THE CHANGE.":T9000::" WHERE (') MEANS AFTER THE CHANGE.":23:"WRITE DOWN THESE IDEAS AND ";:10050:3:"2. MOMENTUM IS A VECTOR. THEREFORE, WE"::" MUST WORK WITH X & Y COMPONENTS":<" SEPERATELY. ALSO, A SIET'S GET DOWN TO BUSINESS!":23:10050 r:2:"HERE'S WHAT YOU NEED TO KNOW:":::"1. IN THE ABSENCE OF EXTERNAL FORCES,": |" MOMENTUM IS CONSERVED. THEREFORE, THE":" TOTAL MOMENTUM BEFORE SOME CHANGE IS":" EQUAL TO THE TOTAL MOMENTUN BOTH SITUATIONS, MOMENTUM WILL BE"::"CONSERVED--THAT'S THE BASIS FOR SOLVING":m T"PROBLEMS."::10050 ^12:"IN THIS PROGRAM, WE'LL ONLY LOOK AT "::"RECOIL TYPE PROBLEMS. YOU SHOULD ALSO":& h"WATCH THE PROGRAMS ON COLLISIONS.":::"NOW, LBLEMS INVOLVE 2 OR "::"EVEN 3 DIMENSIONS. ";:10050 ,:6:"DON'T GET NERVOUS, ";Z$;". WE"::"WILL WORK ONLY IN 1 & 2 DIMENSIONS!" 613:"NOW LET'S LOOK AT AN EXAMPLE OF EACH"::"TYPE OF SITUATION.":20:10050 @8000:8200S J::3:"I TWO OR MORE BODIES"::" ARE BLOWN APART.": " 2. COLLISIONS, WHERE TWO OR MORE "::" BODIES RUN INTO EACH OTHER."::10050 ::"SIMPLER PROBLEMS INVOLVE ONLY ONE"::"DIMENSION--MOTION ALONG THE X-AXIS.":; ""MORE COMPLEX PROMENTUM LAW'"::d "IF YOU KNOW WHAT THE LAW SAYS, AND IF"::"YOU HAVE YOUR PAPER, PENCIL AND": "CALCULATOR HANDY, LET'S GO!":24:10050 :3:"MOST MOMENTUM PROBLEMS IN BASIC PHYSICS"::"FALL INTO TWO CATEGORIES:":4 " 1. RECOIL, WHEREi*(d- CONSERVATION OF LINEAR MOMENTUM3n10000]:3:"HI! WHAT'S YOUR NAME? ";Z$::"WELCOME, ";Z$;". IN THE NEXT FEW"::"PROGRAMS WE'LL BE LOOKING AT PROBLEMS": "WHICH CAN BE SOLVED MOST EASILY USING:":::" 'THE CONSERVATION OF MO               H THE":" SUPPORT OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">A]2::':100:8>'" ************************":j$'" CONSERVATION OF ENERGY - METHOD".':" ************************"8'255:A11000:::(7):B'"";A$:L'32:"";A$:6">NSERVATIVE & NON-CONSERVATIVE FORCES.":::"SEE YOU LATER!":22:10050mH13376:D13:(7)::20:"END"~RD$;"PR#0":\D11500:::D$;"RUN MENU"'#(#5:X,YX10,YX10,Y8X,Y8X,Y:Z#X,84X,94X8,94X8,84:Y84922:X,YX8,Y"::10050:810a "CORRECT! WORK IS EQUIVALENT TO THE"::"AMOUNT OF ENERGY GAINED."::10060:*34,0:13376:10:"THAT'S IT FOR THIS PROGRAM, ";Z$;".":4"YOU SHOULD NOW WATCH THE CONSERVATION"::"OF ENERGY PROGRAMS WHICH DEAL WITH":G>"COTING THE"::"MASS FROM h = 0 TO h = 1.53 METERS?":Y" ANSWER? ";A:13376:A6800 "SINCE WORK & ENERGY ARE EQUIVALENT, THE"::"AMOUNT OF WORK DONE IS THE AMOUNT OF":"ENERGY GAINED BY THE MASS:"::" W = Ug = 0.4 * 9.8 * 1.53 = 6 JOULESh":D12500:s" OR,":" 6 = 3.92 * h":" AND,":" h = 1.53 METERS":1005013376:"THAT IS EASY!"::"LET'S DO ONE FINAL PROBLEM FOR THIS"::"PROGRAM.";:10060413376:"HOW MUCH WORK IS DONE IN LIF10060o13376:25:"(K + Ug + Us + Ef) = (K + Ug + Us + Ef)"::"(0 + 0 + Us + 0 ) = (0 + Ug + 0 + 0 )":"OR,":" 1/2 * K * x^2 = M * g * h"::255:1005013376:"#5 NOW WE SUBSTITUTE VALUES:"::" 1/2 * 1200 * (.1)^2 = .4 * 9.8 * X. HEIGHT),"::" SO K = 0."::"AND...";:10060g13376:"> THE MASS HAS GRAVITATIONAL ENERGY."::"> THERE IS NO STORED ELASTIC ENERGY."::"> THERE ARE NO FRICTIONAL LOSSES."::"#4 SO EQUATING THE INITIAL & FINAL"::" ENERGIES...";:.";:10060Zv13376:"> THE SPRING IS COMPRESSED, SO THERE"::" IS ELASTIC ENERGY.":"> THERE ARE NO FRICTIONAL LOSSES."::1006013376:"#3 IN THE FINAL CONDITION:"::"> NOTHING IS MOVING (THE MASS IS":1" MOMENTARILY AT REST AT MAT YOU WISH."::10050xX13376:11:33:"h = 0":15:1:"WE'VE CHOSEN THE GROUND AS OUR REFERENCE":"HEIGHT.";:10060b13376:"#2 NOW WE NOTE THAT INITIALLY,"::"> NOTHING IS MOVING, SO K = 0.": l"> THE MASS IS AT h = 0, SO Ug = 0."::"AND..X72:9050:X162:9050:162,82170,84x:::" HERE IT IS.";:10060:::" THE PROBLEM IS TO FIND (h)."D3:178,22190,22:1:185,23185,82:7:29:"h = ?":22:10060N13376:"#1 SELECT A REFERENCE HEIGHT. CHOOSE"::" ANY HEIGHRING WHICH IS THEN RELEASED.":m:"HOW HIGH DOES THE BLOCK GO?":20:"COPY THE PROBLEM AND ";:22:1005013376:3:35,83215,83:0:72,8380,83:162,83170,83:X71:Y82:9000:X161:Y22:9000&2:2:"INITIAL";:31:"FINAL":14:34,14%02: THE ENERGIES"::"ASSOCIATED WITH EACH BODY IF MORE THAN":f"ONE OBJECT IS INVOLVED!"::1005013376:3:"HOW ABOUT A SAMPLE PROBLEM?":::"A SPRING (K = 1200 NT/M) IS COMPRESSED":"10 CM. A 0.4 KG BLOCK IS PLACED ON TOP"::"OF THE SP76:3:"METHOD:"::"1. ESTABLISH A REFERENCE HEIGHT (h = 0).":"2. DETERMINE THE INITIAL K, Ug, Us, Ef.":"3. DETERMINE THE FINAL K, Ug, Us, Ef."::"4. EQUATE #2 WITH #3.":"5. SOLVE FOR THE UNKNOWNS."::10050::>"BE SURE THAT YOU FIND OF ANY ACTION. ADD UP"::"ALL THE ENERGIES AT THE START, AND SET":"THEM EQUAL TO ALL THE ENERGIES AT THE"::"END. IT'S REALLY EASY, ";Z$;"!"::10050::"NOW WE'LL LIST THE METHOD IN DETAIL."::"BE SURE TO COPY IT DOWN."::10050k133ERNAL WORK IS"::"DONE ON OR BY THE SYSTEM, IT MUST BE":X"ADDED IN.)";:10060::"COPY THE BASIC EQUATION AND ";:29:1005013376:4:"THE BASIC EQUATION SUGGESTS A METHOD"::"FOR SOLVING PROBLEMS. SIMPLY ANALYZE":E"THE END-POINTSAT FRICTIONAL ENERGY IS ENERGY"::"LOST TO HEAT DUE TO A NON-CONSERVATIVE ":e ^"FORCE.";:10060 h13376:4:"SO WE HAVE:"::" K + Ug + Us + Ef (Before) =": r" K + Ug + Us + Ef (After)"::10050:::|"(DON'T FORGET THAT IF EXTRE OTHER FORMS, BUT THESE":@ 6"WILL DO FOR NOW.";:10060:: @"KINETIC - (K) = 1/2 * M * V^2"::"GRAVITATIONAL - (Ug) = M * g * h": J"ELASTIC - (Us) = 1/2 * K * x^2"::"FRICTIONAL - (Ef) = u * Fn * d"::10060:O T"NOTE TH THE AMOUNT OF ENERGY IN"::"A SYSTEM BEFORE AN EVENT MUST BE THE":q "SAME AS THE AMOUNT AFTER AN EVENT.": ":"SO,"::"ALL ENERGY (Before) = ALL ENERGY (After)"::10060 ,13376:3:"HERE ARE THE ENERGIES YOU'LL NEED. OF"::"COURSE THERE AYOU HAVE YOUR PAPER, PENCIL AND"::"CALCULATOR READY, LET'S GO!";:10060Y :16302,0 13376:3:"YOU KNOW THAT THE LAW SAYS:":::" ENERGY CANNOT BE CREATED OR DESTROYED,": " BUT IT CAN BE CHANGED IN FORM."::10060:E "THIS MEANS THAT";Z$::X "THANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL LOOK AT A SIMPLE METHOD FOR": "SOLVING PROBLEMS USING THE LAW OF"::"CONSERVATION OF ENERGY. YOU'LL SEE THAT": "IT'S MUCH SIMPLER THAN USING DYNAMICS"::"AND KINEMATICS."I ::"IF )d - CONSERVATION OF ENERGY - METHOD6n:10000Fx:16302,0^- TURN ON ROMPLUS+jD$(4)yD$;"PR#5"R$(13)R$M$"":- THERE IS A HEREM$;"1A" 13376:3:"HELLO, FRIEND!"::"WHAT'S YOUR NAME?      THE":" SUPPORT OF NATIONAL SC>">66">"">">>>"6>>>>">>">>"A8X,Y:':100:8?'" ************************":d$'" DYNAMICS - METHOD".':" ************************"8'255:A11000:::(7):B'"";A$:L'31:"";A$:6">H 4,Y1X4,Y5:X1,Y6X3,Y6:X,Y1X,Y5:X,Y3X4,Y3:u<#3:X,Y2X,Y4:X1,Y1X5,Y5:X1,Y5X5,Y1:F#3:X5,Y1X5,YX,YX3,Y3X,Y6X5,Y6X5,Y5:#3:A3.143.14.1:X11.6(A),Y10(A)::#X,YX10,YX10,Y8X,YE YOU IN THE NEXT PROGRAM.":20:10050L 13376:D13:(7)::20:"END"]D$;"PR#0":~ D11500:::D$;"RUN MENU"'#(#3:X,Y5:X1,Y6:X2,Y6:X3,Y1X3,Y6:X4,Y:X4,Y5:X5,Y:X5,Y5:X6,Y1:X6,Y6:=2#3:X1,YX3,Y:XEQUATIONS FOR EACH OBJECT AND SOLVE.":k"WE'LL SEE HOW TO DO THIS IN FOLLOWING"::"PROGRAMS.";:1006034,0:13376:6:"IF YOU WANT TO REVIEW THIS METHOD,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1250&12:"THAT'S IT FOR NOW, ";Z$;"."::"SE13376:"#3 CONVENTIONS:"::" WE CHOOSE DOWN FOR M1, LEFT FOR M2,"::" AND COUNTER-CLOCKWISE FOR THE WHEEL":" AS POSITIVE SINCE THAT IS THE WAY"::" THEY'LL MOVE."::10060'13376:"THE REST IS PRETTY EASY. SIMPLY SET UP"::"144,24144,26:114,51116,51:214,59216,59:214,12216,121:31:"Fr":8:"T1":22:"T2":8:16:"T1":5:26:"T2";:35:"Ff":9:7:"Fg1";:28:"Fn = Fg2"13376:"SINCE THE WHEEL DOESN'T TRANSLATE,"::"WE'VE LEFT OFF Fg AND Fr.";:10060z10060>62450:X128:Y35:9100:X48:Y39:9200:X210:92001:53,4053,60:53,3153,16:209,35195,35:221,35233,35:131,25145,25:115,35115,525:215,11215,31:215,40215,60;3:52,5954,59:52,1754,17:196,34196,36:232,34232,36: UNKNOWN.";:10060vv13376:" THEREFORE, WITH FOUR UNKNOWNS, WE'LL"::" NEED FOUR EQUATIONS. WE'LL GET ONE":" EQUATION FOR EACH BODY & a = (r)( ).":X253:Y127:9020::10060:"#2 DRAW A FREE-BODY DIAGRAM FOR EACH"::" OBJECT.";:2:"I":4:14:"r":4:30:"u":6:11:"T1":9:11:"M1"X34,11:13376:"#1 HERE'S THE PICTURE. THE PROBLEM IS"::" TO FIND THE ACCELERATION OF THE":b" MASSES. ALSO NOTE THAT THE ANGULAR"::" ACCELERATION AND TWO TENSIONS ARE":l" 50F&::"OK, ";Z$;". LET'S LOOK AT A"::"SIMPLE EXAMPLE.";:10060013376:X100:Y30:9100:X161:Y24:9200:X82:Y60:9200:2:200,25120,25120,45100,45:100,30100,75200,75D1:87,5287,30:101,20161,208N2:18:"T2 M2":3:1RTIA":t" ROTATES, THE TENSION IN THE CABLE"::" IS NOT THE SAME ON EACH SIDE OF THE"::" PULLEY.";:1006013376:8:Z$;", THIS IS A LONG ONE. YOU"::"MIGHT WANT TO LOOK AT IT AGAIN.":::"TYPE 1 TO REVIEW, 2 TO CONTINUE: ";A:A12 THE EQUATIONS DEVELOPED IN"::" STEPS #5, 6, & 8 FOR ANY UNKNOWNS."::10060::"SPECIAL NOTES:"::"> IF A PROBLEM INVOLVES TRANSLATION AND":" ROTATION, YOU'LL NEED a = (r)( ).":X225:Y110:9020:::"> IF A PULLEY WITH A MOMENT OF INEUAL TO (M)(ay)"::10060::t">#7 CALCULATE THE TORQUES ACTING ON ANY"::" BODY WHICH ROTATES.";:10060:::">#8 FOR EACH ROTATING BODY, SUM THE"::" TORQUES AND SET EQUAL TO (I)( )":X232:Y166:9020::10060U13376:3:">#9 SOLVE":{" x-DIRECTION AND SET EQUAL TO (M)(ax)"::"(SINCE IT'S ACCELERATING, F IS NOT 0.)":X182:Y134:9030::1006013376:3:">#6 FOR EACH BODY WHICH MOVES (i.e."::" TRANSLATES), SUM THE FORCES IN THE":" y-DIRECTION AND SET EQ IF THE BODIES GO IN DIFFERENT"::" DIRECTIONS.";:1006013376:3:">#4 RESOLVE ALL FORCES INTO x AND y"::" COMPONENTS (p AND n ON INCLINES)."::10060::">#5 FOR EACH BODY WHICH MOVES (i.e."::" TRANSLATES), SUM THE FORCES IN THE."::10060m13376:3:">#3 ESTABLISH CONVENTIONS: CHOOSE A"::" CONVENTION SO THAT THE DIRECTION":" OF MOTION IS POSITIVE. WHEN MASSES"::" ARE CONNECTED BY A CABLE, CHOOSE":<" THE DIRECTION OF CABLE MOVEMENT,"::" EVENORK":" h"ORDERLY.";:10060:: r"COPY DOWN THIS METHOD:":::">#1 START WITH A DRAWING. SHOW ALL THE"::" FORCES.";:10060: |">#2 DRAW A FREE-BODY DIAGRAM FOR EACH"::" OBJECT. SHOW ONLY THE FORCES": " ACTING ON THE OBJECTS110:9030:Y126:9030:X44:9000:X106:9020[ O:"GOOD! NOW REMEMBER THAT a AND ARE": T"RELATED: a = (r)( ).";:X211:Y159:9020:X120:Y175:9020:10060 ^13376:4:"NOW FOR THE METHOD. BE CAREFUL, THERE"::"ARE SEVERAL STEPS. YOU NEED TO W:3:"YOU SHOULD ALREADY KNOW THE BASIC"::"RELATIONS FOR BODIES WHICH ARE NOT IN": ;"EQUILIBRIUM. REMEMBER THEM?";:10060: @:"HERE THEY ARE:": E" Fx = (M)(ax)"::" Fy = (M)(ay)"::" = (I)( )"::, JX30:Y94:9030:Y Fx - FORCE IN THE x-DIRECTION"::" ax - ACCELERATION IN THE x-DIRECTION": " I - MOMENT OF INERTIA"::" - ANGULAR ACCELERATION": " - TORQUE":: "X28:Y94:9020:Y111:9000 ,18:"COPY THIS NOTATION.";:10060U 613376 ";Z$;"! IN THIS PROGRAM"::"WE'LL DESCRIBE A METHOD FOR ANALYZING": "THE FORCES AND TORQUES WHICH ACT ON "::"ACCELERATING BODIES. PAY CLOSE"::"ATTENTION.":: "IF YOU'RE READY, LET'S GO..."::10050 13376:3:"NOTATION:"::N " d- DYNAMICS - METHOD (n:100008x:16302,0P- TURN ON ROMPLUS+\D$(4)kD$;"PR#5"xR$(13)R$M$"":- THERE IS A HEREM$;"1A"13376:4:"HELLO!"::"WHAT'S YOUR NAME? ";Z$::F "THANKS,          7,13:<%780,75:888,75:996,75:10104,75:2112,75:J':100:8t'" ************************":$'" CIRCULAR MOTION - CONSTANT SPEED".':" ************************"8'255:D1900:::(7):B'"";A$:3.143.14.1:13069.6(T),8060(T)::>#1:0:T3.14z#712875.5(T),8365(T):712875.5(T),8365(T)$TT.2:T14.859240$9210$7186,41:T$1175,21:5185,21:3191,21:6198,21:$182,40151,13151,18:151,1315NSTANT SPEED.":`"YOU SHOULD NOW WATCH:":::3::"CIRCULAR MOTION - CHANGING SPEED":::"SEE YOU LATER, ";Z$;"!":24:10050:D13:(7)::20:"END"D11500:::(4);"RUN MENU"'#(#:3:128,80132,80:130,78130,82,2#T60162,60Nx21:"COPY THIS DIAGRAM FOR ROTATIONS":"WITH CONSTANT SPEED.":a10050:::10"IF YOU WOULD LIKE TO REVIEW, TYPE 1."::"TO CONTINUE, TYPE 2: ";A:A1290:7"THIS CONCLUDES OUR DISCUSSION OF"::"CIRCULAR MOTION WITH COIS NOT CONSTANT"::" - ANGULAR ACCELERATION IS ZERO"::" - TANGENTIAL ACCELERATION IS ZERO"::" - RADIAL ACCELERATION = V^2/R":::10050<9000:9200:9300:9500:2160,71:5169,71:4176,71:6184,71P5:9400 d2:182,41156,60156,55:156,"::" ACCELERATION. A(R) = V^2/R ."::" IT ALWAYS POINTS TOWARDS THE CENTER"::" OF THE CURVE. **"::10050:3:"TO SUMMARIZE;"::"WHEN THE ROTATION SPEED IS CONSTANT:"::(" - ANGULAR VELOCITY IS CONSTANT"::" - TANGENTIAL VELOCITY THE TAN-"::" GENTIAL VELOCITY, V(T), IS CHANGING,"::" THERE MUST BE AN ACCELERATION."::" ** THIS ACCELERATION, WHICH IS":" ASSOCIATED WITH A CHANGE IN"::" DIRECTION BUT NOT MAGNITUDE, IS":u" CALLED A RADIAL OR CENTRIPETALTION (ALPHA) IS"::" ZERO."::10050:"2. SINCE THE MAGNITUDE OF THE"::" TANGENTIAL VELOCITY IS CONSTANT,"::" THERE CAN BE NO ACCELERATION IN"::" THE DIRECTION OF THE TANGENT: A(T)=0"::10050::3"3. SINCE THE DIRECTION OF TANGENTIAL VELOCITY IS NOT CONSTANT"::::10050t:"NOW LET'S CONSIDER ACCELERATIONS WHEN"::"THE SPEED OF ROTATION IS CONSTANT."::)"1. SINCE THE ANGULAR VELOCITY (OMEGA)"::" IS CONSTANT, IT MUST BE TRUE THAT"::" THE ANGULAR ACCELERA IS CHANGING, THE VELOCITY IS":X"NOT CONSTANT. LET'S WATCH AGAIN."::::100509000:9200:9300:9500:5:9400:22:10050L::7:"TO SUMMARIZE;"::"WHEN THE ROTATION SPEED IS CONSTANT:"::2`" - ANGULAR VELOCITY IS CONSTANT"::" -BER THAT VELOCITY IS A"::"VECTOR. ALTHOUGH THE MAGNITUDE OF THE":"VELOCITY (SPEED) IS CONSTANT, THE"::"DIRECTION (TANGENT TO THE CIRCLE) IS":"CONSTANTLY CHANGING."::::10050:880 "VERY GOOD! SINCE THE DIRECTION OF THE"::"TANGENTE":CH"ANGULAR VELOCITY (OMEGA) IS CONSTANT."::::10050:780\"GOOD. THE ANGULAR VELOCITY IS CONSTANT."::::10050p:5z"IS THE TANGENTIAL VELOCITY CONSTANT?"::"ANSWER YES OR NO.";B$::(B$,1)"N"980D"SORRY. REMEM."::::10050:660D"YES! THE SPEED IS CONSTANT."::::10050O :7:z"IS THE ANGULAR VELOCITY CONSTANT?": :"ANSWER YES OR NO.";B$:::*(B$,1)"Y"8604"WRONG. SINCE THE OBJECT SWEEPS OUT"::"EQUAL ANGLES IN EQUAL TIMES, THRE ARE SOME QUESTIONS FOR YOU:"::S"DOES THE OBJECT HAVE A CONSTANT SPEED?":u:"ANSWER YES OR NO.";B$::(B$,1)"Y"760"WRONG. SINCE THE OBJECT TRAVELS AROUND"::"THE CIRCLE IN EVEN TIME INTERVALS, IT":"HAS A CONSTANT SPEEDCONSTANT SPEED."::10050# 09000x D21:"HERE WE HAVE AN OBJECT TRAVELLING":"WITH CONSTANT SPEED AROUND A CIRCLE." X9200:::"REMEMBER THAT THE VELOCITY VECTOR IS":"TANGENT TO THE CIRCLE." l9300:5:9400 :10050 ::7#"NOW HE IS THE RATE OF CHANGE":h " OF VELOCITY WITH RESPECT TO TIME.":::"OK SO FAR, ";Z$;"?"::10050 :6:"NOW WE WANT TO LOOK AT AN OBJECT": "TRAVELLING IN A CIRCLAR PATH."::"LET'S TAKE THE SIMPLEST CASE: THE": "OBJECT TRAVELS WITH RST, LET'S REVIEW SOME BASIC":h "CONCEPTS YOU SHOULD KNOW."::"1. DISPLACEMENT MAY BE EXPRESSED": " TRANSLATIONALLY OR ROTATIONALLY."::"2. VELOCITY IS THE RATE OF CHANGE OF": " DISPLACEMENT WITH RESPECT TO TIME."::"3. ACCELERATION OMEGA":Z T" - TANGENTIAL VELOCITY V(T)"::" - ANGULAR ACCELERATION ALPHA": h" - TANGENTIAL ACCELERATION A(T)"::" - RADIAL ACCELERATION A(R)":" (CENTRIPETAL)":: |"WRITE THESE NOTATIONS DOWN AND THEN"::10050 :2:"FI:"IN PARTICULAR, WE'LL LOOK AT THE":j "ACCELERATIONS IN MOTION ALONG A"::"CIRCULAR PATH."::10050 ":2:"IN THE END, YOU SHOULD BE ABLE TO WORK"::"WITH 2 KINDS OF VELOCITY AND 3 KINDS": @"OF ACCELERATION:":::" - ANGULAR VELOCITY )d- CIRCULAR MOTION - CONSTANT SPEEDAg(4);"BLOAD SHAPE"Ti232,0:233,64_n10000:3:"HI! WHAT'S YOUR NAME? ";Z$::Z$;", THIS PROGRAM IS"::"DESIGNED TO HELP YOU REVIEW THE KINDS":& "OF ACCELERATIONS WE FIND IN PHYSICS.":      ************************":B#8'255:D11500:::(7):W#B'"";A$:10:%"T$1,2932:V132:V230:9020:C"$8,395:V13:V21:9040:n"%6,2732:26,31:25,30:26,33:25,34:"%10,315:6,11:7,12:4,11:3,12:"'"':100:8"'" ************************":"$'" VECTOR RESOLUTION"##.':"12,V24:V13,V21:V14,V2:b!d#V2,V24V1:V2,V24V12:V11,V13V22:V11,V2:V11,V24:n!#T021|!#T6,31T!#!#24,2710!#10,1427!#V132:V25:9000:!#27,17:27,21:27,25:27,29"#8,10:11,10:14,10:17,10:20,10:23,NU" ? '#p (#V1,V2:V1,V21:V11,V22:V11,V23:V12,V24:V13,V23:V13,V22:V14,V21:V14,V2: <#V1,V2:V11,V21:V12,V22:V13,V23:V14,V24:V1,V24:V11,V23:V13,V21:V14,V2:!P#V1,V2:V11,V21:V12,V22:V12,V23:V ONE"::"MORE TIME. IF YOU'RE STILL HAVING ":T"TROUBLE, SEE YOUR INSTRUCTOR."20:"TO REVIEW THIS PROGRAM, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A1250:10:"BYE FOR NOW, ";Z$;"!":X13:(7)::20:"END" X12500:::(4);"RUN MEUNDS":::ij" V(N) = V * COS(THETA)":" V(N) = 400 * COS(40)"::" V(N) = 306 POUNDS"::xt23:10050~:6:"HOW DID YOU DO? IF YOU GOT THE RIGHT"::"ANSWERS, YOU'RE DONE WITH THIS LESSON.":/"IF NOT, YOU SHOULD RUN THE PROGRAMRB"IF THE VECTOR IS 400 POUNDS AND"::"THE ANGLE THETA IS 40 DEGREES."::L" FIND V(P) AND V(N). THEN,":16:4:10050V:6:"HERE ARE THE ANSWERS:":: `" V(P) = V * SIN(THETA)":" V(P) = 400 * SIN(40)"::" V(P) = 257 POE COSINE?":::"ANSWER P OR N:";::B$8B$"N"3640$:(7):21:"SORRY. REMEMBER THAT THE COMPONENT":"NEXT TO THE ANGLE USES THE COSINE.".24:10050::36008::5:"GREAT, ";Z$;". YOUR'RE GETTING"::"THE IDEA! NOW SOLVE THE PROBLEM":CULAR TO THE AXES."" 24:100504 :20,2:20,3@ T013R .65T21,T3e .65T10,16Tk u 18,2 13,1629:27,2916 13,169:9,1116 :21:"FINALLY, THE COMPONENTS ARE FOUND."24:10050':21:"WHICH ONE USES TH N":"AXES WHICH ARE AT RIGHT-ANGLES."5 24:10050= 15M V122:V214W 9060 :21:"AND, HERE'S OUR ANGLE THETA."> 24:10050H 1R 15,25:13,22:11,19:23,25:25,22:27,19\ :21:"NATURALLY, WE NEED TO SKETCH LINES":"PERPENDI18,0& T124!0 .65T19,T': 5D 33,3425CN 22,2535iX 27,3235:27,3238:36,28:37,29p 3| T025 .65T4,25T  4,525 22,253 27,321:27,293:2,27:2,29& :21:"HERE ARE THE AXES WE'LL USE--THE P &E--THE STEPS ARE THE SAME."::q "*** AS YOU DO THIS EXAMPLE, TRY TO"::" PREDICT EACH STEP!!":24:10050z :2 2,3019:20,29:21,28:18,29:17,28 V118:V233:9000 :21:"HERE'S THE VECTOR WE'LL CONSIDER." 24:10050 12::" V(Y) = 5.74 METERS"::9 "IT'S EASY!":24:10050 :3:"LET'S LOOK AT ANOTHER COMMON SITUATION."::"WE'LL CONSIDER THE COMPONENTS OF": "A VECTOR WHEN THE AXES ARE NOT THE X-"::"AND Y-AXES. IT REALLY DOESN'T MAKE ANY": "DIFFERENC YOU HAVE THE ANSWER,"::10050B :3:"HERE'S THE ANSWER:":: " V(X) = V * COS(THETA)":" V(X) = 10 * COS(35)":" V(X) = 10 * .819"::" V(X) = 8.19 METERS":: " V(Y) = V * SIN(THETA)":" V(Y) = 10 * SIN(35)":" V(Y) = 10 * .574":NENT NEXT TO THE MARKED ANGLE "::"WILL ALWAYS USE THE COSINE.":12 "NOW LET'S CONSIDER A REAL PROBLEM."::"FOR THE DRAWING WE'VE BEEN USING, FIND": "THE X- & Y-COMPONENTS IF V = 10 METERS"::"AND THETA = 35 DEGREES." 22:"WHEN YOU THINK9300:3:9400:0:5,32Fx 15:V117:V225:9060:1:9200:9600:9700 :21:" V(X) = V * COS(THETA)":" V(Y) = V * SIN(THETA)"::10050 ::3:"NOTE THAT THE X-COMPONENT WILL NOT"::"ALWAYS USE THE COSINE TERM! THE":D "COMPOEN"::"LOOKING AT, THE X-COMPONENT WOULD BE:"::}P " V(X) = V * COS(THETA)":::"BECAUSE THE ANGLE THETA IS SHOWN ":Z "BETWEEN THE VECTOR AND THE X-AXIS."::"NATURLLY, THEN:"::d " V(Y) = V * SIN(THETA)":24:10050n :2:9100:12: VECTOR AND":c( "AXIS #1, THEN THE COMPONENT IN THE"::"DIRECTION OF #1 WILL USE THE COSINE":2 "OF THE ANGLE.**":::" V(1) = V * COSINE(THETA), AND":< " V(2) = V * SIN(THETA)":24:100502F :3:"FOR EXAMPLE, IN THE DRAWING WE'VE BEAD' OF THE"::" VECTOR PERPENDICULAR TO THE AXES."D 24:10050Q Q2:999\ ::3 "4. FIND THE COMPONENTS GRAPHICALLY"::" (MEASURE THEM), OR USE TRIGONOMETRY.": :"** HERE'S ANY EASY TRICK: IF THE ANGLE"::"(THETA) IS BETWEEN THEHE SPECIFIC INSTRUCTIONS:":::"1. SKETCH THE RIGHT-ANGLE AXES WITH THE": " ORIGIN (0,0) AT THE 'TAIL' OF THE"::" VECTOR.": "2. MEASURE THE ANGLE BETWEEN THE VECTOR"::" AND ONE AXIS (EITHER ONE).":5 "3. SKETCH LINES FROM THE 'HENG."24:10050:b$ 21:"THEY CAN BE EXCHANGED FOR EACH OTHER":"WHENEVER IT'S CONVENIENT!"m. X17}8 9600:9700B 1:9100L 9600:9700V 2:9100` :24:10050 :::"ISN'T THAT NEAT? OF COURSE IT IS!"K 5:"NOW HERE ARE TNY MORE!"24:10050:*49300:9400x>21:"IN FACT, YOU DON'T NEED THE AXES, ONCE":"YOU HAVE THE COMPONENTS."f1:5,2732:10,31524:10050:2:910021:"THAT'S BECAUSE THE VECTOR AND ITS":"COMPONENTS REPRESENT THE SAME THI Q22570 1:9600:9700z 21:"STEP 4 - USE GRAPHICAL METHODS OR TRIG-":"ONOMETRY TO FIND THE TWO COMPONENTS." 24:10050: 0:9100:9200:V117:V225:9060 21:"ONCE YOU HAVE THE COMPONENTS, YOU DON'T":"NEED THE ORIGINAL VECTOR A4:10050:$ V117:V225:15:9060v 21:"STEP 2 - MEASURE THE ANGLE BETWEEN THE":"VECTOR AND ONE AXIS (THETA)." F24:10050: x1:9200 21:"STEP 3 - DRAW PERPENDICULAR LINES FROM":"THE HEAD OF THE VECTOR TO THE AXES." 24:10050: 0E :21:"CONSIDER ANY VECTOR , POINTING":"IN ANY DIRECTION."V 24:10050:d L12:9300q ~3:9400~ 0:5,32 21:"STEP 1 IS TO ADD A SET OF AXES. YOU DO":"NOT HAVE TO USE X & Y, BUT THE AXES" "SHOULD BE AT RIGHT-ANGLES." 2ST"::"ALSO BE POSSIBLE TO REPLACE ONE VECTOR": "BY TWO EQUIVALENT VECTORS. THIS TECH-"::"NIQUE, CALLED RESOLVING A VECTOR INTO": "COMPONENTS, IS VERY USEFUL IN PHYSICS."::"LET'S LOOK AT THE PROCESS:" "24:10050  2 :910Z "TRIG BEFORE YOU START THIS LESSON. BE"::"SURE YOU HAVE YOUR CALCULATOR HANDY!": :"WHEN YOU'RE READY TO BEGIN,"::10050 :4:Z$;", THE BASIC IDEA IS"::"VERY SIMPLE. SINCE TWO VECTORS CAN BE":1 "ADDED TOGETHER TO MAKE ONE, IT MUXd - VECTOR RESOLUTION&n10000N:3:"HELLO! WHAT'S YOUR NAME?";Z$6:"THIS PROGRAM IS FOR YOU, ";Z$;"."::"IT SHOULD HELP YOU UNDERSTAND HOW TO": "RESOLVE A VECTOR INTO TWO COMPONENTS."::"YOU SHOULD BE ABLE TO DO RIGHT-TRIANGLE":          '" ONE DIMENSIONAL KINEMATICS"::" ************************"j=.'255:D11000:::(7):=B'"";A$:=L'" TYPE 1 TO SEE AGAIN, 2 TO CONTINUE: ";A:X014<$9:10,129X:T1100:T:0:10,119X:F< $12:18,1419T<$X301<$9:10,1235(2XX2):T1100:T:0:10,1135(2XX2):<$9:10,1237<"$22:10060:A19200<,$<':100:8<'" ************************":K=$18,1412;#X04O;#7:10,12122XX2:T160:T:0:10,11122XX2:h;#22:10060:A19100n;#;#::9:10,120:21:"A NEGATIVE ACCELERATION ADDED TO A"::"CONSTANT VELOCITY. PRESS .";A$;#12:14,180<#T1800:::(7):0:10,11X22X:+:F#22:10060:A190001:P#:#::7:10,120:21:"A POSITIVE ACCELERATION ADDED TO A"::"CONSTANT VELOCITY. PRESS .";A$:#13:14,180:#T1800:::(7):X05:#7:10,122X:T1100:T:0:10,112X: ;#13: V - FINAL":::" ACCELERATION A - ACCELERATION":::10050::9(#::12:10,120:21:"AN OBJECT AT REST IS GIVEN A POSITIVE"::"ACCELERATION. PRESS TO START.";A$9-#7:14,18092#T1800:::(7):X05:<#12:10,12X22X:2 + 2*A*X"::10050::^8l :"HERE ARE THE VARIABLES:":::" POSITION X(0) - INITIAL":8v " X - FINAL":::" TIME T(0) - INITIAL":8 " T - FINAL":::" VELOCITY V(0) - INITIAL":H9 " " 2. DETERMINE KNOWN VARIABLES"::" 3. ORGANIZE DATA":7" 4. APPLY EQUATIONS--YOU MUST KNOW THEM!":" 5. SOLVE FOR UNKNOWNS"::10050::7 :"EQUATIONS:"::" 1. X = V(0)*T + 1/2*A*T^2"::8 " 2. V = V(0) + A*T":::" 3. V^2 = V(0)^ T * 0 * *":" ------*-------*-------*"6^" V * * *":" ------*-------*-------*"6h" A * *":" ------*---------------*"::6:" 1. ESTABLISH CONVENTIONS":87LATER, ";Z$;". BYE!"#5v20:10050C5:T13:(7)::20:"END"W5(4);"RUNMENU"]5?5@:" *INITIAL* FINAL *":" ------*-------*-------*"5J" X * 0 * *":" ------*-------*-------*"=6T" TE BEFORE IT":;4N"STOPS?":16:"WHAT'S YOUR ANSWER? ";Au4X20:"THE CORRECT ANSWER IS 2400 RADIANS.":24:100504b:6:"THAT'S THE END OF THIS PROGRAM. YOU"::"SHOULD ALSO WATCH:"::5l" KINEMATICS & GRAVITATIONAL ACCELERATION":::"SEE YOU REVERSED.":22:"TYPE 1 TO REVIEW THIS PROBLEM. TYPE 2"::"TO CONTINUE: ";A:A115803::3:"TRY ONE LAST PROBLEM ON YOUR OWN:":::"A WHEEL ROTATING AT 120 RAD/SEC IS GIVEN"4D"AN ACCELERATION OF -3 RAD/S/S. THRU HOW ":"MANY RADIANS DOES IT ROTA:3:"FINDING THE FINAL VELOCITY IS EASY."::"WHAT ANSWER DO YOU GET?";A:::A141820s2"GOOD, ";Z$;".":2"SINCE V = V(0) + A*T"::" V = 10 + (-3)*8":2&7::"V = -14 M/S":::"WHICH IS WHAT WE EXPECT IF THE OBJECT":V30"HAS 2":G1"SUBSTITUTE THE DATA--"::" X = 10*8 + 1/2*(-3)*8^2":s1"THEN--"::" X = 80 - 96":1"FINALLY--"::6::"X = -16 METERS"::1"WHICH MEANS THE OBJECT REVERSES AND"::"ENDS UP TO THE LEFT OF START.":110050\2ITION IS -16 METERS,"::"WHICH MEANS TO THE LEFT OF THE START.":0"SO, THE OBJECT HAS REVERSED DIRECTION."::10050:18000::"THAT'S NOT RIGHT. LET'S SEE WHERE YOU"::"WENT WRONG."::100501:3:"SOLUTION--"::" X = V(0)*T + 1/2*A*T^"EXCELLENT!";[/14::"USE EQUATION #1":::"BECAUSE #2 AND #3 REQUIRE YOU TO KNOW":|/"THE FINAL VELOCITY.":::/"NOW YOU SOLVE FOR THE FINAL POSITION."::"INPUT YOUR ANSWER WITHOUT UNITS:";A/A161730B0::"VERY GOOD! THE POS THE PROBLEM AND FILL IN"::"THE CHART.":::10050C.T:2:8000o.^7:27:"8":9:18:"+10":11:22:"-3".h15:"BY NOW THIS SHOULD BE EASY!"::10050::.r"WHICH EQUATION ARE YOU GOING TO USE TO"::"FIND THE POSITION? ";A::3:A11670/|C-,:3:"HERE'S A PROBLEM WITH A NEGATIVE"::"ACCELERATION:"::-6"AN OBJECT WITH A VELOCITY OF 10 M/S IS ":"GIVEN A NEGATIVE ACCELERATION OF":-@"-3 M/S^2. WHERE IS IT AND HOW FAST IS"::"IT GOING AFTER 8 SECONDS?":::4.J"AS USUAL, COPY";A::A141560X,:"SOLUTION--"::" V = V(0) + A*T"::" V = 6 + 2*4":,6::"V = 14 M/S":::10050:1570,12:"OF COURSE! V = 14 M/S","20:"IF YOU WANT TO REVIEW THIS PROBLEM,"::"TYPE 1. TYPE 2 TO CONTINUE: ";A:A113604 OR -10 SECONDS"::10050::x+"NATURALLY, WE CHOOSE THE +4 SECONDS"::"ANSWER. YOU SHOULD VERIFY THE ANSWER BY":+"PUTTING IT BACK INTO EQUATION #1"::10050,:3:"WHAT IS THE FINAL VELOCITY? SOLVE"::"EQUATION #2, AND INPUT YOUR ANSWER: *2*T^2"::10050:\*:"NOTICE THAT THIS IS A QUADRATIC--"::" T^2 + 6*T - 40 = 0":*"SOLVE IT BY FACTORING AND INPUT YOUR"::"ANSWER (WITHOUT UNITS):";A*:3:"FACTORING--"::" (T + 10)(T - 4) = 0":+"FINALLY--"::" T = + EQUATIONS. WHICH ":"ONE IS BEST FOR FINDING THE TIME?";A:::A11440X)"SURE!";)11::"#1 CAN BE USED DIRECTLY.":::10050):3:"LET'S SOLVE IT--"::" Y = V(0)*T + 1/2*A*T^2":*"SUBSTITUTE THE DATA--"::" 40 = 6*T + 1/2"AN ACCELERATION OF +2 M/S^2. HOW LONG":Y(d"DOES IT TAKE TO TRAVEL 40 METERS?":::(n"TRY TO FILL OUT THE CHART. PRESS"::" TO CHECK YOUR WORK.";A$(x:2:8000(5:26:"40":9:19:"6":11:22:"+2"J)15:"O.K? NOW LOOK AT THE:6::"REVOLUTIONS = 1.59":::31:10050'F:10:"TO REVIEW THIS LAST PROBLEM, TYPE 1."::"TYPE 2 TO CONTINUE: ";A:A11040'P:3:"HERE'S ANOTHER PROBLEM. IT'S A LITTLE"::"MORE COMPLICATED:"::*(Z"AN OBJECT TRAVELLING AT 6 M/S IS GIVEN ":"FOR THE NUMBER OF REVOLUTIONS AND NOT":S&"NUMBER OF RADIANS?";:31:10050::&("REMEMBER HOW TO FIND REVOLUTIONS FROM"::"RADIANS?";:31:10050::&2" REVOLUTIONS = RADIANS/2*PI":::"THEREFORE--":*'<" REVOLUTIONS = 10/2*3.14":::" OMEGA^2 = OMEGA(0)^2 + 2*A*THETA":%" (2)^2 = (0)^2 + 2*0.2*THETA"::" THETA = 4/0.4 = 10 RADIANS"::10050:1300% 10:"VERY GOOD!";:12:"THETA = 10 RADIANS":::10050)&:3:"DID YOU NOTICE THAT THE PROBLEM ASKED"::"NOW FOR THE DISTANCE. WHAT EQUATION DO ":"YOU WANT TO USE? ";A::A21240Z$"O.K!";$10::"#1 OR #3 ARE GOOD.":::10050::$"YOU USE #3 AND SOLVE FOR THE DISTANCE."::10050$:3:"INPUT YOUR ANSWER: ";A::A101290+%::"NOPE!":"ALPHA. INPUT YOUR ANSWER (NO UNITS): ";A::A.21210o#3:"WRONG, ";Z$;"."::"HERE'S THE SOLUTION--":#" 2 = 0 + ALPHA*10"::6::"ALPHA = 2/10 = 0.2 M/S^2":::10050:1220#6:"GREAT! ALPHA = 0.2 M/S^2"::10050M$13:;!"~11::"#2 IS THE BEST"::}""WE CHOOSE #2, SINCE #1 AND #3 NEED THE"::"POSITION. REMEMBER THAT WE'RE WORKING IN"""THE ROTATIONAL SYSTEM, SO--"::" OMEGA = OMEGA(0) + ALPHA*T"::10050::9#"YOU SUBSTITUTE THE DATA AND SOLVE FOR": GET IT RIGHT, ";Z$;"?"::"IF NOT, THINK ABOUT WHAT YOU DID WRONG.":!`"ANALYZING YOUR ERRORS IS A GOOD WAY TO ":"LEARN. O.K?"::31:10050!j:3:"NOW SELECT THE BEST EQUATION TO USE FOR ":"FINDING THE ACCELERATION: ";A::A21150"t"YES." REVOLUTIONS ":"DID IT MAKE IN THE FIRST 10 SECONDS?":: .:"COPY THE PROBLEM AND TRY TO FILL IN THE ":"CHART YOURSELF. WHEN YOU'RE READY TO": 8"CHECK YOUR WORK, ";:10050 B:2:8000 L7:26:"10":9:19:"0";:27:"2"F!V15:"DID YOU"TYPE 1. TYPE 2 TO CONTINUE: ";A:A1790:3:"HERE'S A ROTATIONAL PROBLEM:":::"A WHEEL, INITIALLY AT REST, IS GIVEN AN ""ANGULAR ACCELERATION. AFTER 10 SECONDS ":"IT IS ROTATING AT 2 RAD/SEC. WHAT WAS "; $"THE ACCELERATION? HOW MANY510::"#3 IS FINE, BUT #2 IS EASIER.":::10050q:" V = V(0) + A*T"::"SUBSTITUTING THE DATA--":" V = 0 + 3*5"::"THEN--":5::"V = 15 M/S (MOVING TO THE RIGHT)"::::"EASY! IF YOU WANT TO REVIEW THIS PROBLEM")7.5960V"WRONG, ";Z$;"."::" Y = 0 + 1/2*3*25 = 37.5 METERS"::10050:970"GOOD!":12::"Y = 37.5 METERS":::10050:3:"NOW FOR THE SECOND PART--TO FIND THE ":"VELOCITY. WHICH EQUATION DO YOU WANT?";A::A1990"O.K!";. BUT EQUATION #3 ":"NEEDS THE FINAL VELOCITY 'V', SO WE WILL"e"USE EQUATION #1:";:31:10050:" Y = V(0)*T + 1/2*A*T^2"::"NOW WE SUBSTITUTE THE DATA--":" Y = 0 * 5 + 1/2*3*T^2"::"WHAT'S THE ANSWER (WITHOUT UNITS)?";A::A3YOU THINK WOULD ":"BE MOST HELPFUL TO FIND THE DISTANCE: ";A::A1890Xp"GOOD!";z11::"#1 IS THE BEST";::31:10050:2:"SINCE WE'RE LOOKING FOR DISTANCE (X), WE":"WOULD LOOK AT EQUATIONS #1 & #3 (SINCE "?"#2 DOES NOT INCLUDE X):J>"NOW LET'S FILL IN THE CHART WITH WHAT WE":"KNOW...";:31:10050XH::8000R7:27:"5":9:19:"0":11:22:"+3"\14:"NOTICE THAT WE HAVE ONLY TWO UNKNOWNS! "::"NOW LOOK AT THE EQUATIONS. INPUT THE":Jf"NUMBER OF THE EQUATION :3:"NOW HERE'S OUR FIRST PROBLEM:":: "AN OBJECT, INITIALLY AT REST, IS GIVEN ":"A POSITIVE ACCELERATION OF +3 M/S^2. HOW"*"FAR HAS IT GONE AFTER 5 SECONDS? HOW ":"FAST IS IT GOING AT THAT TIME?"::4"COPY THE PROBLEM.";:31:10050::8200P"COPY THE EQUATIONS. WE'LL REFER TO THEM ":"BY NUMBER.";:31:10050:5:"THAT'S THE DISCUSSION ON METHOD. IF YOU ":"WOULD LIKE TO REVIEW IT BEFORE WE BEGIN " "TO SOLVE PROBLEMS, TYPE 1. TYPE 2 TO ":"CONTINUE: ";A:A1300(BEGIN ANY"::"CALCULATIONS, CHECK TO SEE THAT YOU HAVE""OBTAINED ALL THE INFORMATIION IN THE"::"PROBLEM STATEMENT. FOR EXAMPLE, A BODY":"'INITIALLY AT REST' MEANS V(0) = 0."::10050:3:"NOW WE'RE READY FOR THE EQUATIONS."::10050:"CHART AND ";:10050~:3:"DID YOU NOTICE THAT THERE ARE FIVE"::"EMPTY PLACES IN THE CHART? GOOD!"::10050::"IN ANY PROBLEM, YOU WILL BE GIVEN THREE"::"OF THE VALUES. THEN YOU MUST SOLVE FOR":8"THE OTHER TWO. SO, BEFORE YOU HELPFUL TO ORGANIZE YOUR DATA.":m"A CHART IS A HANDY WAY TO KEEP TRACK"::"OF THE INFORMATION.":::"HERE'S ONE YOU CAN USE..."::10050:2:800017:"NOTICE THAT TWO VALUES ARE FILLED IN."::"THIS WILL ALWAYS BE TRUE. COPY THE":SIMPLIFIES THE EQUATIONS":"WE USE. JUST REMEMBER THAT IF THE FINAL":l"POSITION WORKS OUT TO HAVE A NEGATIVE"::"VALUE, IT IS LOCATED TO THE LEFT OF THE":v"INITIAL POSITION.":::10050":6:"IN ORDER TO ORGANIZE YOUR THINKING,"::"IT ISTHUS, VELOCITY TO THE RIGHT IS POSITIVE,":"WHILE ACCELERATION TO THE LEFT IS":lD"NEGATIVE.":31:10050N:3:"** INITIAL CONDITIONS:":::" THE INITIAL TIME, T(0), AND THE":X" INITIAL POSITION, X(0), ARE ZERO."::Fb"THIS CONVENTION 050:2:8300m:3:"REMEMBER THAT THE FIRST STEP IN OUR"::"METHOD IS TO ESTABLISH CONVENTIONS.":&"HERE ARE TWO USEFUL ONES:":::"** DIRECTIONS:"::" AND ARE POSITIVE. ":0" AND ARE NEGATIVE."::P:"$;"!";:31:10050r:7:"SINCE THE METHOD WE USE WORKS FOR BOTH"::"SYSTEMS, WE'LL EXPRESS MOTION IN TERMS":"OF 'X', 'V', AND 'A'. JUST REMEMBER THAT":"YOU CAN SUBSTITUTE 'THETA', 'OMEGA',":"AND 'ALPHA' FOR ROTATIONAL PROBLEMS.":22:10"ACCELERATION. ";:10050::"IN THE ROTATIONAL SYSTEM, WE MUST USE:"::" OMEGA - ANGULAR VELOCITY (RAD/SEC)":" ALPHA - ANGULAR ACCELERATION (RAD/S/S)"::10050::"BE SURE TO WRITE DOWN ANYTHING YOU DON'T":"KNOW, ";Z"::" Z - FOR MOTION ON THE Z-AXIS":j:"** ROTATION"::" THETA - FOR MOTION AROUND AN":" AXLE (IN RADIANS)"::10050:3:"IN THE TRANSLATIONAL SYSTEM, WE USE THE"::"FAMILIAR 'V' FOR VELOCITY AND 'A' FOR":-INE PATH":A"CAN SPEED UP, SLOW DOWN, AND REVERSE."::10050:4:"SO, IF WE WANT TO REPRESENT THE POSITION":"OF AN OBJECT, WE CAN USE:"::"** TRANSLATION"::" X - FOR MOTION ON THE X-AXIS":+" Y - FOR MOTION ON THE Y-AXIS:"FOR ROTATIONAL ONE DIMENSIONAL MOTION,"::"THE OBJECT ROTATES AROUND A FIXED AXLE": r"WITH A CONSTANT RADIUS--FOR EXAMPLE,"::"A WHEEL. IT CAN SPEED UP, SLOW DOWN, ": |"AND REVERSE ROTATION DIRECTION JUST"::"LIKE AN OBJECT ON A STRAIGHT LON MEANS AN OBJECT"::"CAN MOVE IN ONE DIMENSION ONLY.": T"FOR TRANSLATIONAL MOTION, WE SAY THAT"::"THE OBJECT MAY MOVE BACK AND FORTH ALONG" ^"A STRAIGHT LINE--USUALLY THE X-AXIS."::"(WE COULD ALSO CHOOSE THE Y-AXIS, ETC.)"::10050X h:6"HERE'S THE METHOD:"::8100 6:"WRITE DOWN THESE STEPS. BE SURE TO NOTE"::"THAT YOU MUST LEARN THE EQUATIONS!"::10050:: @:5:"BEFORE WE LOOK AT THE EQUATIONS, LET'S"::"DISCUSS THE VARIABLES. ";:10050::: J"ONE DIMENSIONAL MOTIMETHOD FOR SOLVING PROBLEMS."x "BE SURE TO COPY DOWN THIS METHOD AS WE"::"GO ALONG. YOU'LL WANT TO REFER TO THE": "METHOD WHEN WE SOLVE PROBLEMS."::10050: :"NOW LET'S LOOK AT A FEW EXAMPLES."::10050 "9000:9100:9200 ,::3:L ONLY LOOK AT THE CASE":r "WHERE ACCELERATION IS CONSTANT. YOU"::"WILL NEED TO KNOW A FEW EQUATIONS AND": "SOME BASIC ALGEBRA.":::"GET YOUR CALCULATOR READY AND ";:10050 :4:"THE MOST IMPORTANT PART OF THIS PROGRAM"::"DESCRIBES A 5#d- ONE DIMENSIONAL KINEMATICS.n10000f:3:"WELCOME!"::"PLEASE GIVE ME YOUR NAME: ";Z$8:"THANK YOU, ";Z$;". IN THIS PROGRAM"::"WE'LL SPEND SOME TIME LEARNING A METHOD": "OF SOLVING ONE-DIMENSIONAL MOTION "::"PROBLEMS. WE'L               $:6">H THE":" SUPPORT OF NATIONAL SC>">66">"">">6>>"">"">">">6>>">D16000::13376?H(4);"BLOAD CHAIN,A520":520"GRAPHING 2"E'#S':100:8}'" ************************":$'" GRAPHING MOTION - PART I".':" ************************"8'255:D11200:::(7):B'"";A370l*:13376:7:"LET'S LOOK AT A POSITION GRAPH AND FIND"::"BOTH THE VELOCITY AND ACCELERATION...":4:"WE'LL NEED SOME MORE MEMORY SPACE, SO"::"PLEASE WAIT A MOMENT WHILE WE LOAD IN"::"PART II OF THIS PROGRAM.">22:15:"-LOADING-":D"ACCELERATES FROM ZERO TO THE CONSTANT"::"VALUE AT 'A'.";:31:10050\ 34,0:133768:"NOW LET'S APPLY THESE IDEAS TO"::"A COMPLETE PROBLEM.":: "IF YOU WOULD LIKE TO REVIEW BEFORE WE"::"GO ON, TYPE 1. TYPE 2 TO CONTINUE: ";A:A11:100507:4:"0")34,9:133769:"NOW HERE'S A QUESTION: "::"WHAT IS THE VELOCITY AT THE ORIGIN?":"INPUT YOUR ANSWER: ";A::"SINCE THE SLOPE AT THE ORIGIN IS ZERO,"::"THE VELOCITY MUST BE ZERO. THE OBJECT":JDUALLY FROM THE VALUE AT 'B'":x"--THE CONSTANT VELOCITY(A-B)--TO THE"::"VALUE AT 'C'--THE ZERO VELOCITY(C-D).":10050:::"SO, THE VELOCITY IS DECREASING. IF THE"::"ACCELERATION IS CONSTANT, THE VELOCITY":"WILL DECREASE EVENLY.";:3 TO DRAW SEVERAL"::"TANGENT LINES TO THE CURVE AND FIND THE":"SLOPE AT EACH POINT. SINCE THE SLOPE IS"::"CHANGING, THE VELOCITY MUST BE CHANGING.":100509:34,9:133769:"YOU SHOULD NOTICE THAT THE SLOPE ":"CHANGES GRA2,39:"SIMPLY FIND THE SLOPE OF THE STRAIGHT"::"LINE: (CHANGE IN Y) / (CHANGE IN X)."::10050::l"SO, VELOCITY(A-B) = CONSTANT";:31:10050v9:34,9:133769:"WHAT ABOUT 'B' TO 'C'? ";:31:10050::@"THE PROPER METHOD IS0)D"YES! THE SLOPE IS ZERO."::10050cN:"SO THE VELOCITY FROM 'C' TO 'D' IS ZERO.":10050;xS8:34,9:13376X75,39111,16:9:"THINK ABOUT THE SLOPE FROM 'A' TO 'B'."::"HOW WOULD YOU FIND IT?";:31:10050::db111,16111,39:75,3911GRAPH TO CONSIDER."::"> STUDY IT A MINUTE.";:31:10050::|&5:149,6191,6:"WHAT IS THE SLOPE OF THE GRAPH FROM 'C'"0:"TO 'D'? INPUT YOUR ANSWER: ";A::A0580:"NO! SINCE THE LINE IS HORIZONTAL, THE"::"SLOPE IS ZERO.";:31:10050:59%3:X3272:X,50(X32)2160:W72,40112,16:X112152:X,6(152X)2160:h152,6192,62:74,3674,42:112,13112,19:148,3148,9:192,3192,95:10:"A":2:15:"B":3:22:"C";:28:"D"9:34,9=9:"HERE'S A POSITION F":U"THE VELOCITY. PLOT THAT VALUE ON THE"::"VELOCITY GRAPH.";:31:10050::"REMEMBER:"::" THE SLOPE OF X IS THE VALUE OF V."::"HERE'S WHAT WE MEAN..."::31:1005013376::2:32,532,50200,504:3:"X":7:31:"TIME""THE DERIVATIVE OF A GRAPH MEANS THE"::"SLOPE OF THE GRAPH.";:31:1005013376:3:"SO TO FIND THE VELOCITY AT ANY TIME,"::"DRAW A LINE TANGENT TO THE POSITION":"GRAPH AT THE TIME IN QUESTION. THE"::"SLOPE OF THE LINE WILL BE THE VALUE OMENT. SO IF YOU HAVE A GRAPH":w|"OF POSITION VS. TIME, YOU SIMPLY TAKE"::"ITS DERIVATIVE TO FIND THE GRAPH OF":"VELOCITY.";:31:10050::"REMEMBER YOUR CALCULUS. WHAT IS THE"::"GRAPHICAL MEANING OF THE DERIVATIVE?"::10050::J50:370e ^:"WRONG! IT IS TRUE. IF YOU MISSED BOTH"::"QUESTIONS, YOU SHOULD LEARN THE DEFINI-": h"TIONS BEFORE YOU CONTINUE. TYPE 1 TO"::"QUIT, 2 TO CONTINUE: ";A:A1820r13376:3:"YOU KNOW THAT VELOCITY IS THE DERIVATIVE":"OF DISPLACE:A1820D 613376:3:"HERE'S ANOTHER QUESTION. IS THIS TRUE?": @:" dX":" V = --":" dt": J"INPUT T OR F: ";L$:::(L$,1)"F"350 T:"VERY GOOD. THIS IS THE RELATIONSHIP"::"WE'LL START WITH.":::100S :"OF COURSE! VELOCITY IS THE INTEGRAL"::"OF ACCELERATION.":::10050:310 ":"THAT'S INCORRECT. PERHAPS YOU SHOULD"::"STUDY BASIC DEFINITIONS BEFORE YOU RUN": ,"THIS PROGRAM. TYPE 1 IF YOU WANT TO"::"QUIT. TYPE 2 TO CONTINUE: ";A::3 "FIND THE VELOCITY AND ACCELERATION?"::10050 13376:4:"FIRST WE'LL CHECK YOUR KNOWLEDGE.":::"IS THIS STATEMENT TRUE OR FALSE?": " dA":" V = --":" dt": "INPUT T OR F: ";L$:(L$,1)"T"290Z$::V "THANKS, ";Z$;"! IN THIS PROGRAM"::"WE'LL LOOK AT A SIMPLE METHOD FOR": "DEVELOPING GRAPHS WHICH ARE THE"::"RESULT OF TAKING THE DERIVATIVE OF": "ANOTHER GRAPH. IN OTHER WORDS, GIVEN"::"A POSITION VS. TIME GRAPH, HOW DO WE":!d- GRAPHING MOTION - PART I.n:10000>x:16302,0\- TURN ON ROMPLUS+ BOARDhD$(4)wD$;"PR#5"R$(13)R$M$"":- THERE IS A HEREM$;"1A" 13376:3:"HELLO, FRIEND!"::"WHAT'S YOUR NAME? ";         SCREEN B1B1%B39B392A1A1AA39A39[ BLANK OLD CURSOR l0:A1,B1:3 PLOT NEW CURSOR  A,B630 A7A:BB19240* CHANGE SCALE TO HGR2 4::> THE END. SITION THE ORIGIN WITH THE L R U D & F KEYS Ml:3:20,20:B20:A20UvA$cA1A:B1BzA$"U"BB1:710A$"D"BB1:710A$"L"AA1:710A$"R"AA1:710A$"F"800  CHECK TO SEE IF POINT IS ON THE LO RESRT")24:14:"TYPE ANY KEY":A$:"":=(7);(7);(7)::" REMEMBER :":"60 DEF FN NUM(X)=";::" LEGAL BASIC EXPRESSION"0:"70 DEF FN DEN(X)=";::" LEGAL BASIC EXPRESSION":"TYPE 'RUN 60' TO RESTART":I12000:D8303X POWISH TO GRAPH"."THE FOLLOWING FUNCTION :"I::" (X-1)(X+2)"`" Y = ----------"u" X(X-7)"::"YOU WOULD TYPE THE FOLLOWING""60 DEF FN NUM(X)=(X-1)*(X+2) ""70 DEF FN DEN(X)=X*(X-7)""TYPE 'RUN 60' TO RESTAANT TO SHIFT THE":"ORIGIN OR CHANGE THE SCALE (Y/N) ? ";A$SA$"Y"A$"YES"90[60:"DIRECTIONS FOR RATIONAL FUNCTIONS""YOU MUST DEFINE YOUR FUNCTION ":"IN TERMS OF A NUMERATOR AND A":"DENOMINATOR FUNCTION""FOR EXAMPLE: IF YOU ILL ON THE SCREEN C,JV191V03505TH,V<^HBh:`r POKE BOTTOM OF GRAPH r|16302,0:A$z:I14:(7)::12"DO YOU WISH TO CHANGE THE FUNCTION BEING GRAPHED (Y/N)? ";AN$:AN$"Y"440:12;"DO YOU W279 :G  TRANSFER THE HGR COORD. TO THE REAL VALUE _ X(HA)S:DDEN(X)e : " DRAW THE ASYMPTOTES ,D0Ē3:H,0H,191:7:350 6YNUM(X)D:VBYS ;: @ TRANSFORM THE REAL Y ATO HGR & SEE IF IT IS BSTSTANCE" :7. REAL AXIS AT (A,B) H 0,B279,B:A,0A,191b HASH MARKS HA279S:H,B2H,B2: HA0S:H,B2H,B2: VB191S:A2,VA2,V: VB0S:A2,VA2,V: : ACTUAL GRAPHICS  H0 POSITION THE ORIGIN : S IS THE SCALE FACTOR ] DETAIL INCREASES WITH S 21:"AFTER THE BOTTOM HALF OF THE GRAPH IS ","FINISHED >> TYPE ANY KEY!" I13000:::22 "THERE IS A HASH MARK (/) ON THE AXIS FOR EACH UNIT OF DIS":, Y14:"I N H I - R E S":I13000:{ Z:23:"THERE ARE 280 HORIZONTAL DOTS. HOW MANY DOTS/UNIT DO YOU WANT? ";S _ d21:"CHANGE THE LOCATION OF THE ORIGIN BY","MOVING THE CURSOR WITH THE L-R-U-D KEYS","F=FINISHED!" x620: RE **) ** ** H *************************N 2:] <NUM(X)Xp FDEM(X)(X2)v K: P DEF FN DEN(X)=1>> IF YOU HAVE A NON-RATIONAL GRAPH U: V: X12:6:"G R AP H I N G F U N C T I O N - GRAPHING HGR FUNCTIONS L *************************k ** ** ** GRAPHING RATIONAL ** ** FUNCTIONS **** ** ** (C) 1982 ** ** MICRO SOFTWA   155635 ~X155ĺ"END":# 10) }: 耲 OHM'S LAW Q 進 BY MIKE REINHARTf ꀲ & CHRIS OBERTH 뀲 9-15-78 12:58:35 AMF :"VOLTAGE------ E=";E D8 N"CURRENT------ I=";I> SX X"RESISTANCE--- R=";R^ ]x b"WATTS-------- W=";W l: v"HIT 'RETURN' KEY TO DO ANOTHER PROBLEM" w"HIT 'ESC' KEY TO EXIT" {X(16384):X128635 }16368,0:X141XϭI0FLAGFLAG1% R0FLAGFLAG1: W0FLAGFLAG1J "FLAG2500S ,180n E0I0REI:WEI E0R0IER:WEI E0W0IWE:REI I0R0EIR:WEI I0W0EWI:REI &R0W0E(RW):IER 0E 'R' FOR RESISTANCE"/ "TYPE 'W' FOR WATTS"5 L "THEN THE VALUE "T X$ X$"E"X$"I"X$"R"X$"W"ĺ"WRONG KEY":180 X$"E"Ą"E=";E X$"I"Ą"I=";I X$"R"Ą"R=";R X$"W"Ą"W=";W FLAG0 E0FLAGFLAG1   936&FLAG0:E0:I0:R0:W0.17>"OHMS LAW"D(L213d<"BY MIKE REINHART"jZd"----------------------------------------"n"LIST ANY TWO VARIABLES"sx"TYPE 'E' FOR VOLTAGE""TYPE 'I' FOR CURRENT" "TYP    WIDTH20((TT$)2):WIDTH0ĺTT$:6x}WIDTH:TT$:R}23:12:"[ESC] TO END"w}24:8);"[SPACE] TO CONTINUE ";}"[ ]";:29:ZZ$:ZZ$(27)ZZ$(3)ĉ::}ZZ$(32)ı}868:32230T340',:"INCORRECT ENTRY"::250?} * TITLE PAGE * M}16368,0o }6:TT$"ELECTRONICS FORMULAS"}32100:9:TT$"BY BILL BLUE"}32100}16:"(PLACED IN THE PUBLIC DOMAIN- SEPTEMBER 1978)"(}32200:d} * PRINT CENTER * $n}(NQRP)(NEPP)-B(ERPQ)(NEPP)J:"X =";M;"Y + (";B;")"u:"IF THE ";B1$;" IS ";:110:A1(A$)(A$)ĺ::A11:270A1123451460"THE ";A1$;" IS ";A1MB;:A2A1MB:A2(A2)ĺ" OR ";((A1MB).5);::1520"Y = ";M;" X + (";B;")"An:"IF THE ";A1$;" IS ";:110:A1(A$)[x(A$)ĺ::A11:270nA1123451460"THE ";B1$;" IS ";A1MB;A2A1MBA2(A2)ĺ" 0R ";((A1MB).5);::1390WW1UW2(W2)U01360M0:X(A$)D:"WHAT IS VALUE #";N1;" FOR THE ";B1$;:110:Y(A$)[X0Y0AZ01280q(X0Y0N212802X0Y01360<NN1:AZAZ1FRRX:OOXX:EEYY:PPY:QQXY:1290PM(NQRP)(NORR)ZB(OPRQ)(NORR)d:"SECONDS"140:F1ĺ:1180'250X:"NAME OF FIRST SET OF DATA ";:110:A1$A$"NAME OF SECOND SET OF DATA ";:110::B1$A$(A1$)(B1$)1250N0:R0:Q0:S0:O0:P0:W0:E0:AZ0  :"WHAT IS VALUE #";N1;" FOR THE ";A1$;:11160~" OHMS"&140:F1ĺ:1070/250^"ENTER CAPACITANCE IN MFD.";:110:C(A$)"ENTER RESISTANCE IN OHMS";:110:R(A$)G6.28318:Q1(GCR):QQ1E06:M(GCR)1E06"3DB DOWN POINT = ";Q;" HZ":"TIME CONSTANT = ";M;" E":"C/R=ALL":58R0:S0:T110:A(T)0:T:T110mB"ENTER R";T;:110:A(T)(A$):(A(T)0)(T1)ī1090~LA(T)01120VRR1A(T):SSA(T):T`R1R::"PARALLEL RT =";R;:R2ĺ" OHM":1140j" OHMS"t"SERIES RT = ";S;:S2ĺ" OHM":1M0A(0)A(T)A(T)MA(T)A(0)980VNN1A(T):M1A(0)N:M1M:M10000000ī1060|" ","RX= ";M;:M2ĺ" OHM":1030" OHMS"T140:F1ĺ:940250$"NO MORE RESISTANCE NEEDED":1040."ENTER RESISTANCE VALUES ONE AT A TIM1ĺ:890250<:"THE POWER IS ";P;:P2ĺ" WATT":L" WATTS":"ENTER TOTAL RESISTANCE DESIRED (RT)";:110:A(0)(A$)A(0)0940M0:N0:R0:S0T110:A(T)0:T:T110"ENTER R";T;:110:A(T)(A$):A(T)01040$ 130@*"(1) P=(I*I)*R"::"(2) P=E*I"::"(3) P=(E*E)/R":e4110:F(A$):F830,860,890:800>170:190:P(II)R:920:140HF1ĺ:830R250\150:170:PEI:920:140fF1ĺ:860p250z150:190:P(EE)R:920:140 F:F(A$):F690,720,750:660:150:190:IER:780:140JF1ĺ:690S250t190:210:I(PR):780:140F1:750250210:150:IPE:780:140F1ĺ:750250 :"THE CURRENT IS ";I;:I2ĺ" AMP":" AMPS":EIR:650:140:F1ĺ:560(D250FN210:170:EPI:650:140UXF1ĺ:59_b250l190:210:E(PR):650:140vF1ĺ:620250:"THE VOLTAGE IS "E:130"(1) I=E/R"::"(2) I=SQ.RT.(P/R)"::"(3) I=P/E":110:450 2501 150:210:R(EE)P:510:140A F1ĺ:480J 250x :"THE RESISTANCE IS ";R;:R2ĺ" OHM": " OHMS": 130 "(1) E=I*R"::"(2) E=P/I"::"(3) E=SQ.RT.(P*R) ": &110:F(A$):F560,590,620:5300170:190:A11230,390,530,660,800,940,1070,1180,1250,1570:340? 130x "(1) R=E/I"::"(2) R=P/(I*I)"::"(3) R=(E*E)/P": 110:F(A$):F420,450,480:390 150:170:REI:510:230 F1ĺ:420 250 210:170:RP(II):510:140 F1A SCALING": @A10ĺ:"WHAT WOULD YOU LIKE NOTES ON? ";:110:A1(A$):A1110000,11000,12000,13000,14000,15000,16000,17000,18000,19000:230 J T"WHAT WOULD YOU LIKE TO SOLVE FOR ";:110 ^A1(A$):(A$)340 h(A$)57340 r15806 |;:110:P((A$)):P0ī380! ' 0 340= ::A11[ :"AT ANY TIME, 'END'": "(1) OHMS":"(2) VOLTS":"(3) AMPS":"(4) WATTS" "(5) PAR. RES. SOLVE RX" ""(6) PAR. RES. SOLVE RT" ,"(7) R/C TIME CONSTANT" 6"(8) DAT"OR ANOTHER (2)";:110:F(A$)::W "ENTER VOLTAGE ";:110:E((A$))::E0ī380] "ENTER CURRENT IN AMPS";:110:I((A$))::I0ī380 "ENTER RESISTANCE IN OHMS ";:110:R((A$))::R0ī380  "ENTER POWER IN WATTS " ELECTRONICS FORMULAS1 BY BILL BLUE H ADDED TO LIBRARYZ JULY 1980 :936:16298,0:TT$(30):32000dD$(4):250nA$:A$"END"ė:x:"CHOOSE YOUR FORMULA":::# ::"WOULD YOU LIKE THE SAME FORMULA (1)":         13,1517:17,1415:13,1819:19,2215:13,1822:27,2413:13,1524:24,271515,1727:27,2418:16);"MATH SERIES":N03000:N::15,1727:27,2418:16);"MATH SERIES":N03000:N::A::8"========================================">L21)"BY"Rk15)"HOWARD JENSEN"q12)"PIONEER HIGH SCHOOL"N03000:NX::2:12,2911:11,2029:29,1220:20,1112:7:0,3937:18,1314:14,1713N(Z1$)13(Z1$)2710101(Z1$)27ė:5407wFLAG0:Z11(Z$):Z1$(Z$,Z1,1):(Z1$)48(Z1$)572020Z1Z1$"-"Z112005FLAG1:p(5)z"========================================":18)"DISTANCE";", THE DISTANCE IS 13."-1000::2x"IT'S BEEN FUN WORKING WITH YOU TODAY. LET'S DO IT AGAIN SOMETIME.":::&15);"GOOD-BYE, ";A$0(4);"RUN MENU":24:"PRESS TO GO ON - TO END";16368,0Z1$:ER IS WRONG, ";A$;".":"TRY AGAIN.":2A1:464g:2:"STILL WRONG, ";A$;".":"TRY ONCE MORE.":tA2:464z"I'M SORRY, ";A$;".":"THE ANSWER IS 13.""YOU NEED TO BE MORE CAREFUL ON THESE."510"THAT'S CORRECT, ";A$4600"CORRECT, ";A$;", THE DISTANCE IS 5."?1000::2v"WHAT IS THE DISTANCE FROM (10,3) TO (-5,9)?"Z$:Z$""Ģ(37):4652000:FLAG1Ģ(37):958:465R(Z$):R13500A1487A2494%:2:"THAT ANSW450I"THAT'S NOT RIGHT, ";A$;".":"YOU SHOULD HAVE DONE AS FOLLOWS:"Os5);"7 - 4 = 3......(3)^2 = 9"y5);"(-2)-2 = -4......(-4)^2 = 16"12);"9 + 16 = 2512);"SQR(25) = 5.":"SO THE DISTANCE IS 5."OORDINATES AND SQUARE."Ie3);"3. ADD 2 ANSWERS AND FIND SQUARE ROOT."Oh^m1000::2"SEE IF YOU CAN FIND THE DISTANCE BETWEENTHE POINTS (4,2) AND (7,-2)."Z$:Z$""Ģ(37):4102000:FLAG1Ģ(37):958:410T(Z$):T5===================="RIT3);"SQR((X(2)-X(1))^2 + (Y(2)-Y(1))^2)"QY:Z"========================================"\:^"IN WORDS, THE FORMULA SAYS TO:"_`3);"1. SUBTRACT X COORDINATES AND SQUARE."c3);"2. SUBTRACT Y-COORDINATES OF THE FIRST POINT AND X(2) AND Y(2) ARE THE SECOND POINTS."yE:"PUSH WHEN YOU ARE SURE OF THIS";F1005GH(2)J"WATCH CAREFULLY, ";A$;".":"I AM NOW GOING TO SHOW YOU THE FORMULA:"O:Q"====================R(100) = 10.""'310E,"EXCELLENT, ";A$;", 10 IS CORRECT."K1Z61000::2;"THE LAST PROBLEM IS AN EXAMPLE OF HOW TOUSE THE 'DISTANCE FORMULA' TO FIND THE DISTANCE BETWEEN TWO POINTS."G@:"REMEMBER X(1) AND Y(1) ARE THE C1Ģ(37):958:250!W(Z$):/ W10300{"NOPE! ";W;" IS NOT THE CORRECT ANSWER.":"WATCH THE FOLLOWING STEPS:"6);"9 - 3 = 6 , (6)^2 = 36"6);"2 - 8 = -8, (-8)^2 = 64"10);"36 + 64 = 100" 13);"SQ"16 + 9 = 25",20);"SQR(25) = 5."2;240l"FINE, ";A$;", 5 IS THE CORRECT":"ANSWER."r1000::2"HERE'S ANOTHER FOR YOU TO TRY:"5);"SQR((9-3)^2 + (2-10)^2) = ";Z$:Z$""Ģ(37):2502000:FLAG."+12);"SQR((4)^2 + (3)^2) = ?"FZ$:Z$""Ģ(37):207j2000:FLAG1Ģ(37):958:207xX(Z$):X5230A$;", THE ANSWER IS 5.":"YOU SHOULD HAVE DONE AS FOLLOWS:"12);"(4)^2 = 16 AND (3)^2=9" 18);SORRY, ";A$;", THE ANSWER IS 9.")200D:"REMEMBER, ";A$;",""WHEN YOU SQUARE A NEGATIVE NUMBER YOU GET A POSITIVE ANSWER.":"THE ANSWER IS 9."200"THAT'S RIGHT, ";A$;", (-3)^2 = 9."1000::2"TRY THIS ONE, ";A$;"3:130Z(Z$):!Z7150D"SORRY, ";A$;", SQR(49) = 7."M160o"GOOD, ";A$;", SQR(49) = 7."~1000::4"WHAT DOES (-3)^2 =";Z$:Z$""Ģ(37):1652000:FLAG1ė:4:165Y(Z$):Y9180Y9190 "L '^' MEANS RAISE TO A POWER AND 'X' IS THE POWER.":Vx"HERE ARE SOME EXAMPLES:"^}:u15);"SQR(9) = 3":17)"(9)^2 = 81"1000::3"NOW, ";A$;", WHAT DOES SQR(49) =";Z$:Z$""Ģ(37):958:1302000FLAG1ė:P"ANOTHER THING WHICH YOU WILL NOTICE ABOUT ME IS THE WAY I DO EXPONENTS.":Z"IF YOU WISH TO RAISE A NUMBER OR EXPRES-SION TO A POWER, YOU USE THE SHIFT KEY AND THE 'N' KEY AS FOLLOWS:"_:d15);"(---)^X"i:7n"WHERE THE SYMBO ABOUT ME.": 2"AS YOU CAN SEE, THERE IS NO SQUARE ROOT KEY ON MY KEYBOARD. SO, IN ORDER TO FINDSQUARE ROOTS, YOU MUST TYPE:" <(37)3:15);"SQR(-----)." >(37)3:"WHERE '-----' IS THE NUMBER YOU WANT TO FIND THE ROOT OF." A1000::2SAME ";& 'A$:A$""Ģ((37)):18:39G ((A$)11Ģ(37)2:958:38 )((37)3):"HI ";A$;",":"I'M GLAD YOU CAME TO WORK WITH ME TODAY.":1000::2 *"TODAY'S LESSON IS ON THE DISTANCE BETWEEN 2 POINTS. BUT FIRST YOU MUST LEARN A LITTLE****"  *PROGRAM MAY BE COPIED*@  *FOR EDUCATIONAL USE &*^ *MAY NOT BE SOLD.COPY-*| ! *RIGHT 1981 CUE,INC * " *********************** #A$"ABCDEFGHIJKLMNOPQRSTUVWXYZ" %7000::6000::6 &"HELLO, I'M APPLE."::"WHAT IS YOU N SOFTSWAP *2  * *P  * 333 MAIN STREET *n  * REDWOOD CITY *  * CA 94063 *  * (415) 363-5472 *  * *  * *  *******************DWARDS$ ***********************B * *`  * SAN MATEO COUNTY *~  * OFFICE OF EDUCATION *  * & *  * COMPUTER-USING *  * EDUCATORS *  * *  * ! DISTANCE JUL 21,1981. M PROGRAM HELPS USER UNDERl -STAND AND CALCULATE THE DISTANCE BETWEEN POINTS. APPLE II W/APPLESOFT PIONEER HIGH SCHOOL REVISIONS BY CHRIS E! ! ! ! !!!!!!!!!       ˠ PG K(LOCK U(NLOCK N(EW D(ELETE E(XIT ----------------------------------------VOL PG OF SECT-USED LEFT ---------------------------------------- ؅D$% 0 +DE ` A PCL~@E^E_EZE[`1 B A PC @`6798 `H|C h|C``2D3DEEiEiA)m3D3D +D 4D +D) 4Dm2D2Dm3D3Dة$eeELCؠ C82D2D3D3Dؠ 3D$C >B$E  ``L#BH$%  hWB\  `CATALOG\Y/N\LOAD\LOCK\UNLOCK\RUN\SAVE\DELETE\NOMONC\START\ LENGTH\,D\E  EmEB&BBNBmBJnBmBeeE``8EEPEE &B B00   5DEEE B ALT@ BG A CL~@Ъ2'EE &B B B" yD$ >B B A C }CL~@ :L@% E E >B yD AE/5 >Bs r ; >Ba ` `$$(@ @_E%E^EZE[E̍^ZE_[ 0E^ D C1  }CE)ȹE)ȹE) 5D E Bɠ 0   5DL@ M  EТ E XELiLLT@ E &B B B E A EL@ E"""""!!! Юh%L"%H  0L% "(I(ȱ(I(%ߥ% "'(I( % аh%L"A                          %H%' "( I(%h%L"%H& L% "(I(ȱ(I(%ߥ% "'(I( %   ( 2 F d                   x n d Z P F < 2 (                                      } x s n i d _ Z U       "HЅNi qȅN8!Oi qe"O8#N$O" [(ɠH(%h% "h($" [(ɠ!""#h"`ГЏ               #&)-           E:BLINK:70,1:NOISE:I:1 21(A$)2:A$;:7 `R j MACHINE CODE FOLLOWSk t DON'T CHANGE THIS:q ~| 770:x ʚhhXϊiZΐ`F eNȱ(eυON)NO,0N NON NO` ><]cwKHIT ANY KEY FOR CATALOG";:A$:@ :(4):"BRUN SUPER HELLO"F ` SP:70,NT:I1(A$)p (A$,I,1);{ NOISE I I1500:I  21(A$)2:CROSS:A$;:NOISE:2000:12:" ";: 70,9:NOISE:I18:70,2:NOIS ";:NOISE:OCA 9:9:" PHYSICS SOFTWARE";^ 11:9:" FOR"; 13:8:" APPLE ][ MICROCOMPUTERS"; I11500:I 34,16:TY 34,6:33,6:TY 32,34:TY:32,0 34,0:33,40:35,5:TY:7 :21::9:"4:" ";:I: 16:A$" ":5000:D 2000 70,2:24:3000:I815:I:8:" ";:I:24:3000 70,7:I13:NOISE:I:A$" PRESENTS:":NT3:11:21(A$)2:1000: 255:70,9:NOISE:CR:11:12:" 2000 Z255,64:70,4E dA$" EDUCATIONAL COMPUTING NETWORK"h nI123:I:A$:70,1:NOISE:Ir s2000 x70,4:NOISE:BLINK:NOISE:2000:NOISE:BLINK:NOISE }2000 :7:A$" ":5000 I716:I:7:" ";:3o 9 :A$"302:A5 B8 8D 00 03 A5 B9 8D 01 03 60 N D823G"nI1(A$):511I,((A$,I,1))128::72,0:144y60000(INIT(768)256(769)52NOISEINIT25:TYPEFALLINIT128:BLINKINIT765:CROSSBLINK35:OCROSSCROSS105<INITF11 P""" " " " " """"