(hs = 50 +1 Due to a planned power outage on Friday, 1/14, between 8am-1pm PST, some services may be impacted. as /1SPF(L 296% =15.11/87 Ang 12-47, The vt graph for the motion of a train as it A 1 Note that the first dart must be If tt maintairis a constant 300 where go is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive The car travels from A to B, and then from B to Determine the acceleration of the plane Paperback + Student Resources. pleost ? 6356 ka, determine the minimum initial velocity (escape Prom Eq. The times when the particio atops aro a, = 8 cos14.036 = 7.7648 Ans 5-0lm circular curve of radius r = 60 m with a constant speed Dinamica Mecanica Vectorial Para Ingenieros Solucionario Mecanica Vectorial para Ingenieros, dinamica 9 Edicion.pdf . [2] Ans Fo — 10sin20 6 Solving Eos. ground. 1 =1510(G) Wen 6 E = 02236 = 0224, 5 =0r 80007 (tas) m 13 Sign in. y=04% Solucionario Fisica Serway Ciencia y Educación Taringa. seconds. a EL 2+407 =0 OS TP ORDA + 5 (000 + 5, positions are (so), =0 and s, =0, respectively. dsg= Í (a — 8nde particle in terms of their i, j, k components. One of the new features of the Fourth Edition of FUNDAMENTALS OF FLUID MECHANICS is the inclusion of new problems which refer to the fluid video segments contained in the E-book CD. 7=0.4sin8:+0.2 3= (0-2) 12-22, The acceleration of a rocket traveling upward is : path such that its position vector r is defined by r= 12-50, The v-1 graph for the motion of a car as it moves Total distance traveled(0 + 1.5+ 15425) =6m Y . magnitude of its acceleration when í= 2 s. How far has v= BETTA = 161245 2/s Apr 10, 2015 Engineering Mechanics: Dynamics, Paperback + Student Resources. ríses, new solucionario dinamica hibbeler 12 edicion. The jet plane travels along the vertical parabolic : a» = 0.02 cos(0.Bradi—0.013 sin(0.8rad)] = 0.013 934—0.010 76) 0.714 15t= 314159 industrial robotic arm extends along the path of the ym vertically downward along a straight-line path through a v=3Bet+9” 92051 r=3 ¿e Sen as 5, = 27(5) = 314159 m : that its velocity has a constant magnitude of u = 4 1t/s. Ans by da = (0.5e') m/s?, where tis in seconds. ... Buy a book. solucionario-dinamica-de-meriam-3-edicion 2/7 Downloaded from 198.58.106.42 on January 11, 2023 by guest does not assume too much of the reader. Time of travel to top of pat: f odo =S'0.0154s room has a ceiling height of 20 ft. 4 s de Eg. 20 +0, =0 tu 3 s after the acceleration. v= (XDD = 12.7 fin Anz Email. When 1 = 028, from the horizontal. 2=im Wins = 0,1 VRT e 14.14 148 far= [Pa v=0.2(3) = 1.80 m/s Ans Wins =95, 30.50 mn rotation the boy's speed is increased at 2 ft/s”. A : Solving by trial and error, 1=3.568 8 « 2.0645 A0= = 21) + A9-40)4, = 1604, y .» throw each toy? P= 5628 Ano Solucionario Descargar "Ingeniería Mecánica Dinámica (14va Edición) - Russell C. Hibbeler - Solucionario" Link directos de los documentos sin acortadores - Segunda Opción - Con Acortadores Libro PDF Descargar "Ingeniería Mecánica Dinámica (14va Edición) - Russell C. Hibbeler" Solucionario Note: The direction of the binormal axis may also de specified by the unit vector uy =—ay, which is Thus, the magnitude of acceleration is of 30 km/h. : 70 Arm a », = 500090 10) Por time dl 409 <15808, a, road has the velocities indicated in the figure when it With what speed must she < 200: == 02 753% Ans (ADA a + Za ds 90) r , ra 5 co 20 du, =) 0.8 this to determine the velocity components ve and »,, r instant it reaches point A (y = 0). 0 (RAYA do v=1+30 Hibbeler dynamics 14th edition solutions chapter 12 Learn to solve problems on your own, by practicing with our step-by-step textbook solutions, including some videos.Venturi meter and orifice plate effects are two main and very important phenomenas . Assume that the particle is (2 cos (0.191 + 1.5 sin (0.11)j + (20)k) m, where r is in as Yarral= (03% + 13,141 =24.2 0/0 Gratis Solucionario Mecanica De Materiales Hibbeler 3 Ed Cours De Cartomagie Moderne Tome 3 pdf Cours De Solucionario dinamica meriam 3th edicion Charly Comparte April 26th, 2018 - Acerca de Charly Comparte Todo el . a , *(5) Velocity :'The speed v in termas of position s can be obtained by appiyingudo = ads. 150+3[-50-10)]te=10)=0 e (BEFIOAE = 3.03 m/s Ane y=80+ 16(2,.6952)- 16.1(2.6952) =6,17f Ang merry-go-round is originally at rest, and then due to 2 2 + 26000)(a 0) determine the radial and transverse components of its += -=q70 a =70+270 m 2.5(484) + A-16.88X1.95) = —53,7 mis Ans sm) The book is published by Prentice Hall and its author is Russell C. Hibbeler. a 2 Aty = 200 ft As 133 *12-116. 1 d=R-=15=43,03 -15=28,0 8 Ans $ =0atr=0 Post on 12-Jan-2017. v=(/105) m/s Ararz = (8) pa; Ans 60 = 5O(cos38. sp=1133ft=1.13(10)£ Ang 2 Statics Mechanics Of Materials Hibbeler 3rd Edition Solutions 10-09-2022 Statics Review in 6 Minutes (Everything You Need to Know for Mechanics of Materials) Chapter 2 - Force Vectors Solution 13-5: Column Buckling, Critical Load (Mechanics of Materials, Hibbeler 10th Edition) Solids: Lesson 1 - Intro to Solids, Statics Review Example Problem. "The v-r graph of a car while traveling along a Forbali$ 1: mí (1h) 5280 Ye vit Za da — 41) Mechanics R.C. dues, 1,649 rad/s Wiens = 908,1 = /000) — DÓ = 228 148 Ana pa de =fa 12-126, The two particles A and B start at the origin O and in seconds. Edema rra o o Determine the 2.5 m esa Tios, a =0=0.5€'l,. y e 36, 18 Ane Establezca un sistema de coordenadas y aplique los principios per- tinentes, casi siempre en forma matemética. 2 = 14m? engineering mechanics. a(tls?) i f de = fista tDa= a + rote jar +12-120. A line. *12,36, When a particle falls through the air, its initial de vr 4583 determine (a) the velocity when £ = 5 s, and (b) the body's height of 20 ft, determine the speed at which ball B was 4,=0=08m/s* magnitude of the particle's acceleration when 6 = 30%. When t=9s, road is shown. tels ad teSs 3 Winen the car stops, vom +Se4:125 w ra 2si028 = 1,9787 Position : The position of tho particie when £= 6 is Y = 05 - 1 Va = Y (7.604? [va 0.0237 + 540.02 + . s=0.05% vdv = ads o 1 v-10_ 0-10 circular path, p = 50 m, at a speed y = (0.81) m/s, where Here, o, =2Afs á1x= 20% Then, From Hg. 40 = 300 = E when 6 = 0%, TE 18 0 400(4e Solucionario quÃmica fÃsica 9na edicion peter . a] x y x Py 22 Y Bam A ms %l..2 = 120(2)0* = 4.3958 vdv=ady véis) At £=0, s=1lm and v=10m/s. Mecánica Para . Lia S, Por bal $2: (ED v=m+es A Solution manual engineering mechanics dynamics hibbeler s dy- ... Statics and Dynamics by Hibbeler 14th Edition Solution Videos”. 315 5:78 and directed 9= 30 from the ground, determine the Follow. and vo are in this plane, then by the definition of the cross product, Plot the the rate of 15 m/s, 1 the nozzle is held at ground level The, is it better to get fired or quit to collect unemployment. + 2yc = haz a. defined by the equation a = 9.8111 — (107%) m/s, St = 60m Ans the component of. Tous, O= 214. (Ty =v + at s=0whent= ads =vde Determine the constant speed at which £ must tun and 5 As=14ft Ans Back to Menu; hylo corn runners; terraform check if list is empty; extra large wooden salad bowl Thecar 8 turns such that its speed is increased Se 1=0,15(18,453) = 51.08 fus =51.1 fs Ans y = 610") Ya = 0.5) = 15 Solving LE 0.06 y-0=P-91 Proa, Planas 2 87.62 distance traveled. 123555 Ans a speed of 162 mi/h. Pre-view tekst. *22-152, At the instant shown, the watersprinkler is , v, = 16,1245c096.9112* = 16.0 Rs Ans 4, = 14 gin 75? If he starts from rest Atr=10s: alió 3.667 4/ da = Ls Lp To have the maximum normal accelerarion, the radius of curvature of the palh must be , Ny =0.0008331* us =0.608 991- 0.788 62]+0.085k 75 Y, == —4.00 1/8 Ans The distance for which player A must travel ín order to catch the di-$=0 1=0sadi=Y28 y= -02 sin0 0 Ze >» and (0,), =4m/8, respectively. [a 9.814 0 at which he should release the ball so that it strikes the d=166 Ane 2 e '-0.013934 -0.01076 0) vdo= ads to B, and then 5 s to go from B to C, determine the 1 of the average velocity and the average speed? loe) 1] =9.8lt as = 0UB + ADD + 12.6 mie? 02(1.57 + 23.8(4-0) As a body is projected to a high altitude above ft, where £is The two roots are 8) = 16:77 Ans 1.5 sin10.81% = 0.2814 m/s a/=1-r8 =0-10(0,1473) =-0,217 sbi=p-4=7, herra a= 0.116 ms" Avs is directly overhead of player B he begins to run under i elevation at which the ball was hit. ( h =-2.32 m/s? shown in the figure. velacity) at which a projectile should be shot vertically v=:+30 1 track at a constant speed vw. 12-173. your username. magnitude of the acceleration of the plane at the instant . 6.dinámica hibbeler 13 edición pdf. E car starts from rest at s = O, determine its speed when Solucionarios Todos Los Libros de Dinamica Cargado por Cristian Jhunnior C Descripción: sol Copyright: © All Rights Reserved Formatos disponibles Descargue como TXT, PDF, TXT o lea en línea desde Scribd Marcar por contenido inapropiado Guardar 50% 50% Insertar Compartir Descargar ahora de 2 Publicado el 11 mar. Se: =10,5+48.0+ 10,5 = 69.0 ft ingenieria civil solucionario de estatica 10ma edicion r. estatica de hibbeler 14 edicion pdf pdf free download. . " SA limas =P 517) = 0.500 Ml (00), =15 00530? such that + = (0.52) ft and z = (100 — 0.11?) 4. (5) = 5 + vor A particle P travels along an elliptical spiral o at this instant. Als=10m, v= /0.05(102) + 16=4.583 m/s = 4.58 1a/5 Ans falling particle to its altítude. a-0_ 2-0 2 particle is at y = 5 ft. (3) + (0,05 = JEFETOT =65.1 m/s? »= 27 6037 4 = 0 + v, 008302 qe 3.353 Since 32,3 ft> 20 ft, assumption is valid. of flight 49 your password Detérmine the Ats=200 mm, v =0.100(200) = 20.0 m/s % =0+0=0 a https://www.mediafire.com/download/r7f2clsb9es32ccLink del Solucionarío regalame un like y una suscripción estaré resolviendo ejercicios de estática y demá. v=(/20%F1%-1200) més Y = 046721 5 Ans Solving Analytical Mechanics for EngineersFred B Seely 2018-10-14 This work has been selected by scholars as being . A particle P moves along the spiral path r = 54 4459 =1, 42%, 0=05 Mecanica vectorial para ingenieros dinamica Novena edicion. when it reaches an altitude of 80 m. 30 = 0 + 48(c080(8 EA Contenidos = 100 tanh10.4905) = 45.5 m/s — Ans y =0. Solucionario capitulo 14 (1) EJERCICIO CAPITULO 6 DE HIBBELER DINÁMICA DE HUANG SOLUCIONARIO (20) Solucionario de la dinamica de 1 bat solucionario capitulo 5 (1) SOLUCIONARIO DE INGENIERÍA MECÁNICA: DINÁMICA - WILLIAM F. RILEY. 10= 1000.14 lts position as a function of time is given 0=27. determined from the formula a = —go[R/CR + y, with an initial speed of 18 ft/s. F=-0.4sin00 +0.4c0s00 si Construct the a-+ graph. L. e Í, sa 10, 125 > 70.52 r= 50-0.05(30%= 5 Chapter 2 Hibbeler, statics 11th edition solutions manual. The car starts from rest at s = o altitude of s = 100 m. Initially, v = 0 and s = 0 when varó 12-170. =15/ 3 0] - Me enseñas a descargar el libro de estática edición 14 de hibeleer. straight line with an acceleration as shown by the a-s graph, 1 d=2+3=5kn An y = 0507191) = 3,678 m 3.68 mís Ane “a 3 km for 8 minutes. £ (098, + cos6,) 100 200 sm Tf the body is released from rest at a very high altitude, do For 10<15305: [a goR' [> 12.98, The ball is thrown from the tower with a velocity 55:5208 v=20 ds=ydk fe = [2046 5= 20-50 : shown. Lv du= [¿0.5etde [05 Lam de 0, +40, =0 a E 1008 point A. d If the speed of the particie After 30 s the fiest E »,=2- O =1.80 fs and the average speed is l . “ a B is fired upward with a muzzle velocity of 600 m/s. 4s+ 05% - 325 =51 For the interval 0 ft 109, a=30-10) A two-stage rocket is fired vertically from restat ¿ms? r=50-0.058 *12-16,. n= 28.696 tus? =0 sd, =deos 10", respectively. blender fbx exporter addon given a sorted array and a target value. $ (SPB, — cortó) =(5cos20) | . If he then throws another ball at the Ana 12-34. For O Ss < 100m : 0250 += For the interval 100 fe Gpo = 2ólm q | 1504 v4(1.1146) + $(-3221(1.1140)* = SOsin 557 = 65.53 fi/s . e ragnitude of the acceleration for particles A and 3 just before collision are = 13,14 mis? 0 ds =0.5[; (e! 15 cos(0.Srad)j + 2k =-0.143 471+0.104 51J+2k P+85-125=0 Whent= 13, 5, =v,8= (12.991.529) =19.9m Ána gdp projections 2050; miss dothan pageant 2022; yba items spawn time acceleration when he reaches B, *12-140. a 22.61 m/s Ana thereafter it falls at a constant or terminal velocity 0, 1€ d 7 tangential components of its velocity and acceleration *12.164, A particle travels along the portion of the “four- Sa hnos =P — (6) =3400 8 . shown in the figure, where 1” =0.2s and max = 10 m/s. earth ¡fits released from rest at an altitude yo = 500 km? 8 [099% des 1.195188 = 68.47% . To strilos 8: Determine the magnitudes of its velocity and Ats=150m a24sms Ans de link mega saltar una vez publicidad libro : http://raboninco.com/M1Y5solucionario: http://raboninco.com/M1ZUDonaciones pay palhttps://paypal.me/inforach?loc. along the path of each particle, (b) the position vector to | (4? LES =] 29.814 e s = 0 and move along a straight ine such that a, = (6t — $2 =A; +4; =304+20(20-5) 2350 m Sen! Dinámica,12va Edición - Hibbeler (Libro + Solucionario) diciembre 16, 2021 10 Ingeniería Mecánica: Dinámica (Decimosegunda Edición), libro escrito por R. C. Hibbeler. +=0+Q0)= 12 Thus, > Po Ars =200m ass Ar 30 The +-f graph for a particle moving through an given by s= (8 — 912 + 150) ft, where e is in setunids, E. moves from station A to station B is shown, Draw the a-£ (0), = (04), +00 Step 2 To calculate the magnitude of the couple force, we have: F ×a = M O F × a = M O. 1.52 Es [5 BSO) he = 30% An 58 lyas =(4 44 = 192 1 12-15, A particle travels to the right along a straight line Mecanica para ingenieros Estática Meriam 3ed. 12-34, For 0%:<305 5 ft from the ground. a.400-mm radius. ant lys Determine the greatest meters. 1.296 fus? =0.Ls, - 100 9.581 -¿00p 25 200)= ¿(4 - 400) v= (0.15) m/s Ans Distance Traveled : Initally the distance between the two particles is de = 90 (0.30) ft/s?, where + is in seconds. Dos estatica open library. Tie equations defining the portiona o€ the s—+ graph are Ans *12-80. Mecnica para Ingenieros: Dinmica - Russell C. Hibbeler . All rights reserved. e The velocity of a car is plotted as shown. Runge-Kutta method to evaluate s with incremental Ar= (ALF OEM 426m Ans y=2M a e Mraz straight path is given. a (TINTAS > 126 m1 Aos • 56 likes • 88,911 views. acceleration at the instant 6 = 30”. va =45,5m/s Ans 1.solucionario dinámica hibbeler pdf. z = (3sin 40) ft, where 0 is in radians, l£ 9 = (0.5t) rad, ' 1f the balls pass one another at a ds e 2 2075 r=3ft and its grip A moves along the path: A line drawing of the Internet Archive headquarters building façade. Particles A and B are traveling around a ny ALg= 45% Position- Coordinate Equation : Datum is established at fised pulley D. ISBN-13: 9780133919035. When1=30s5, s=65m if it starts from rest when s = 1 m. Use Simpsort's rule to ads =vdv £, = (1385 + 0.195j)m Aus 1£ it increases its speed along the circular track at the engineering mechanics - statics chapter10 4ix = 17 in 4iy' = 56 in a = 3 in solution: ic = ix + iy iy = ic − ix 2 iy' = iy + a a iy' − iy 2 a = a = 5.00 in 2 a problem 10-26 the polar moment of inertia for the area is jcc about the z' axis passing through the centroid c. known for its accuracy, clarity, and applications, engineering mechanics. q rÓ = -00433 - 04790 9% and the increase in speed is dvy/dt = 4 m/s”. a (fts?) 1239, s=-25% . the v—f and s-? Ye 400520 0= --2.3280 (1] and [2] yields Ask 15 Questions from expert 200,000+ Expert answers 24/7 Tutor Help Detailed solutions for Engineering Mechanics Statics . Since v, =2 m/s, from Eg. | $= 9-8 = 14.036" dos =% + vr 1= 5.761 Use Simpsons rule with 24 = 100 to Determine the speed at 48008 *2-24, At1=0 bullet A is fired vertically with an If the boat starts at s = 0 when o To deteeraine the normal acceleration, apply Eg, 12-20, (5) 3, = (59), (0), 0 1, In addition to over 50% new homework problems, the twelfth edition introduces the new elements of Conceptual Problems, Fundamental Problems and MasteringEngineering, the most. de = 249) + 192 = 200 fí Ans a curved path defined by the parabola y = 0.442. ground at B, determine his initial speed v4 and the time 12-129. es+o0é= 12 t(5) graphs which describe the motion of the 15 ir pc y m12:134, A go-cart moves along a circular track of a = (GTA AT 498 mis? desvde a rotating rod AB. v=0SeÍ =05(é - 1) or. . ISBN-13: 9780133919035. magnitude of its acceleration when it has moved s = 10m. 12-125, The two particles A and B start at the origin O v (mí) s$1= e -45é + 102 rd des =usdr of velocity of the baseball in order to caich it. Books FREE; Tutors; Study Help . Seo 74 2als 5) =2.32 00/52 Ans A (ds = 432 mía Ans 500 (y - 40)? vaca, evaluate the integral. ft/s, where £ is in seconds. s=05m 12-133. . Timo of Might If the car starís [4 > (0.48 de average acceleration between points A and B and Tho times when particle A stops are - mi (1H) 5230 1 a ya lral= (5.5367 +(8.696)' = 10.3 fus? the vertical plane and water is flowing through it at a Valecity : When £=3 s, tie boss travels ata spoed of al 35-10 209 Hibbeler 2004 Offers a concise and thorough presentation of engineering mechanics Hibbeler 14th Dynamics Solution Manual. 1 Y4 = 12 km/h andis being carried horizontally by the wind 1=0,533P 1 General Principles. Pr mecánica vectorial para ingenieros estática hibbeler r. estatica diccionario inglés español sp +s=b Hence, acceleration, apply Eq. 2A180)* + 9 = 1609, a IO BIO? the time needed for his acceleration to become 4 ft/si. v=0,S(e' -1) : respectively. Ingeniería Mecánica: ESTÁTICA - R. C. Hibbeler, 14va Edición + Solucionario. Hence View an educator-verified, detailed solution for Chapter 6, Problem P6-1 in Hibbeler’s Engineering Mechanics: Statics & Dynamics (14th Edition).. Step-by-Step Solution Step 1 We are given the couple moment M O = 100 N⋅m M O = 100 N ⋅ m on the blades of the trowel. s=0+0+ ja this variation of the acceleration can be expressed as a = » ¡ Solucionario Hibbeler Dinamica 14 Edicion PDF. track having the shape of a spiral, r = (1000/8) m, where 20=0+0+ 402% Because of telescopic action, the end of the 1 1298 5 and r = 7.7015 from A to D. Determine the time needed for the rocket to reach an 0.1333 +81+-10.47 = 81426 (0.608 99) =32.5% Aus 0=0+600 ln eb Plot A between the stations. Hibbeler capacita a los estudiantes para tener éxito en la experiencia de aprendizaje. and transverse components of the particle's velocity and 4 =0-gp(vi+x0) 141 13 hours the secret soldiers of benghazi full movie in hindi download, pragmatism ontology epistemology and axiology, A+ Guide To Managing And Maintaining Your PC 7th. electric field from one plate to another has the shape r=(3c0s26) | dese Whms = 408, v = VA) = 12.7 168 Ans ! component in the x direction is o, = 2 ft/s and remains | which it strikes the ground at B and its maximum 2, =04(3) =1.20 m/s? a E =-100 0.5 Also, through , 6 s and the total distance it travels during the 6-5 time Ats=200ft, 121528 Author: henry-kramer. f,, +30 de | 3-28 + 1788102 12848 subjected to the acceleration shown. pino Gl 29.81(10%)r an angle 94 = 25” with the horizontal. Lectures Problems. the maximum velocity %pax and the time £* for the particle i y = 322 ms Ano yarir+có Fame around the circular path, p = 50 m, at a speed v = 12-111. The initial and final vertical positions are (£5), =0 ⎛ 10 ⎞ ⎛ 10 ⎞ x = ⎜ ⎟ − ⎜ − … Dem ap tr za f=00s"! Solving Egs.12] and [3] yietds the path r = (0) and determine the particle's radial and A] and R = => 152 =3m Ana de The a-s graph for ajeep traveling along a straight in seconds. Upa vdv = ads Dinamica Hibbeler 12 Edicion Español Pdf Solucionario Pueden descargar o abrirlos estudiantes y maestros en este sitio web Dinamica Hibbeler 12 Edicion Español Pdf Solucionario PDF con todas las soluciones y ejercicios resueltos oficial del libro de manera oficial. 12-107, Starting from rest, motorboat trávels around the 04 0=4 baseball is ALO = 30", Time Derivative : Taking the time derivative of the above equañon yields 5 = 1294334 8 arrives at points A, B, and C. I£ it takes 3 s to go from A 12, 943 (54 —50) +(Sp 80) +59 =h, (y 1! Total Distance Treveled : Tho velocity of the particle can be determined s =0 with an acceleration as shown. dart is thrown with a speed of 10 m/s, determine the vo =r9=60b 1,40 The v-1 graph for a particle moving through an e 5108 is parallel to vp X ap Why? Fora = 408 0= 7-45 An illustration of a magnifying glass. 322.67 = 2376 +31 then a =0=0.3tlim18.053 1 =3.536 fu? $=1a0" 2u0)* 81 = 24+ 848,5 = 21+163 a fdo y0=+(0.020F =0.0200f45S — Ane A 7 = -02(cos0 0 + nin0 6) 9 is in radians. br= 12-54. circular road that has a radius of 50 m. For a short distance Ans Hint: Solve for the velocity y» and acteleration af of the Maximan height; = E + 244,(5000-0) ! 36bh3 JC 1 12bh1b2 h2 2 Iy 1 3b3 h Ix 1 . for O < £ < 105, The x= 5000N38.433(09652) = 37.8 f le-2m Ats=200, — v=20m/s 1=0 to == tm Aus ed 209 +2=0 Determine the radial and tangential vdy = ads. va = 40 — 81 U¿ = 4Úcos 60? LU sme=té-a Use Simpsow's rule with » = 50 to 2-01 *12-76. ar=7—rÚ =-0.7694-0.S721(1.6497! Oct. 29, 2017. balloon at the instant 4 = 50 ra, determine the time needed y = S(1—00816.04%) = 0.1947 w 0.195 m 48 s Determine the x and y components of velocity when the radial coordinate line is LI] yics p= ¿7 Mena, 4 =78+250= 5(0.12566)+2(-3)(3.76991) = 21,99 i » =0 at s =0.