Looking for concise explanation of tricycle physics puzzler?
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The simplest explanation I know is in terms of the instant center.  We all think of a wheel as rotating about its axis.  That is true, relative to the tricycle frame.  However, relative to the ground, the axle moves forward.  The point where the tire touches the ground is stationary.  It does not move with respect to the ground.  An instant later, another part of the tire is touching the ground and again, for an instant, that point on the tire is stationary.  \n\nOn any rotating body, the spot that does not move (it may rotatebut not translate) is the center.  At any instant, the part of the tire touching the ground is not moving and is, by definition, the instant center.  The wheel is rotating about this instant center.  SInce the pedal is above the instant center, your forward horizontal force rotates the wheel forward.\n\nYou can prove this all to yourself by trying a few things with the string on the pedal.  Instead of pulling horizontally, raise the angle of the string.  When the line of action of the string passes through the contact point between the tire and the road, you will find that pulling on the string will not rotate the wheel.  If you pull hard enough the trike will move but the front wheel just skids.  Since the line of action of the applied force is passing through the center of rotation, there is no torque to spin the wheel.\n\nRaise the string higher and the trike will roll backwards because the force is acting on the other side of the center creating a torque in the opposite direction causing the wheel to turn the other way.   \n\nIf you don't have a trike handy, you can do this with a spool and a string.  Just wrap the string sround the spool with the end coming off the bottom of the spool.
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http://www.geoblox.com/freemodels.htm\nScroll all the way down for a model of a Phoenician trireme.
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Try those...\n\nhttp://www.eng.abdn.ac.uk/~eng624/paper2004b.pdf#search='Dynamic%20Characteristics%20of%20a%20Jeffcott%20Rotor'\n\nhttp://www.dyrobes.com/Chapter3_Excerpt.pdf#search='Dynamic%20Characteristics%20of%20a%20Jeffcott%20Rotor'\n\nhttp://www-personal.engin.umich.edu/~shihang/p32.pdf#search='Dynamic%20Characteristics%20of%20a%20Jeffcott%20Rotor'
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there are some solved physics problems on \nhttp://www.dctech.com/physics/help/\nbut I don't know if it will help you out.
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So I'm assuming there is no static or kinetic friction in this problem.\n\nWell, since the cart experiences a force in the horizontal direction, then: \nF = ma\na = F/m\na = 11N/26kg = 11/26 m/s^2\n\nHow far will it move in 2 seconds starting from rest?  What do we know?\n\na = 11/26 m/s^2\nt = 2 secs\nVelocity initial = 0 m/s (since it starts from rest)\nSo what is d (for distance although I know some schools use X)?\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/26)(2)^2\nd = 11/52*4\nd = 11/13 m\n\n30 N child in the cart before he begins to push it?\n\nNewtons is just another way of saying weight but in order to find the acceleration we need mass and not weight.\nWeight (in N) = mg\nm = weight/g\nm = 30/10\nm = 3kg  <-------- this is the mass of the child\n\nRepeat as above with little modification.\na = F/m\na = 11N/(26kg + 3kg) = 11/29 m/s^2\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/29)(2)^2\nd = 11/58*4\nd = 44/58 m\n\nGood luck!
300027
connect the dots games. They make puzzle books. Try a newstand.
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Actually it is a bot more complicated than those answers would show.\n\nThe spinning wheels cause a gyroscopic effect which makes it easier to balance.
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1.\nPhases:\nf1 = (t/T1)*2*pi, where T1 = 1 sec\nf2 = (t/T2)*2*pi, T2 = 1 sec\nIn phase: f1 = f2 + 2*pi*n, where n is integer (0, +-1, +-2, ...)\n(t/T1)*2*pi = (t/T2)*2*pi + 2*pi*n\nt(1/T1 + 1/T2) = n\nT2 = 0.98 = 98/100 = 49/50, 1/T2 = 50/49\nt(49/49 - 50/49) = n\n-t/49 = n\nt = 49 seconds, n = -1\nNumber of oscillations if second, shorter pendulum (shorter pendulum has shorter period): N = t/T2 = 49/(49/50) = 50\n\n2.\nCoordinate: y(t) = 3*sin(20*pi*t)\nSpeed: v(t) = y'(t) = 20*pi*3*cos(20*pi*t)\nMean position: y(t) = 0. Take t0 = 0.\nMaximum displacement: y(t1) = amplitude = 3, speed = v(t1) = 0\nv(t1) = 0\ncos(20*pi*t1) = 0\n20*pi*t1 = pi/2\nt1 = 1/(2*20) = 1/40
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Howstuffworks has good diagrams and explanations that may help you.  Good luck.
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Hey! I just finished learning about active transport! The teacher confused me soooo bad! I had to have a friend in my class explain it to me. I would suggest going to google and typing in "active transport" and clicking on the news tab on the top. good luck with your project!
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Try these links:\n\nhttp://cse.ssl.berkeley.edu/segwayed/lessons/light/emspectrum.html\n\nhttp://search.yahoo.com/bin/search?fr=ybr_sbc&p=electromagnetic%20visible%20graph
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http://www.worldwidewords.org/qa/qa-thr1.htm\n\nhttp://www.bartleby.com/59/4/threesheetst.html
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Check out Three Ball Charlie:\n\nhttp://sideshowworld.com/ats3ball.html
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http://www.ask.com/web?q=where+can+i+find+the+solutions+manual+%5B6th+edition+hibbeler%5D+for+mechanics+of+materials%3F+&qsrc=0&o=0
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trillian... u can get it from download.com
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http://www.puzzlemaker.com/CrissCrossSetupForm.html
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I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.