how can i make a 3-d model of newton's third law of motion?
457731
I'd suggest a rocket but there are always going to be some people who say it's pushing against the air so you'll spend time explaining how a rocket really works. How about something that actually does push on the air, a fan-propelled car? Easy to make, easy to run (but not if you have to make quantitative measurements) Fan pushes air backwards, air pushes fan and car forwards.
27041
Newtons law is called the Laws of Motions. There are three laws for it. The first law is known as the  "law of inertia" states that if the forces acting upon an object are balanced, then the acceleration of that object will be 0 m/s/s. Basically if objects are not at equilibrium they do not accelerate, they only accelerate if there is a net or unbalanced force acting upon it. The second law is mainly about the behavior of objects. That all existing forces are not balance. It states thatacceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The third law is known as " for every action there is an equal and opposite reaction " this basically states that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.
159509
Start with a toy car sitting on a table.  Point out that the car is not moving. Point out that no unbalanced external forces act on the car, hence it remains at rest, i.e. doesn't accelerate.  This demonstrates Newton's First law.  You can also demonstrate the other part of Newton's First Law, about objects in motion staying in motion unless acted upon by an outside force, by pushing the car and catching it a short distance later, before the force of friction had much opportunity to act on it.  Or you could use a marble, which is less affected by friction.\n\nFor Newton's Second law, use two cars, one more massive than the other.  You can arrange a weight and pulley sytem to pull the cars along if you want, or you could just tell your audience about the forces you are using as you push the cars.  Push 1 car lightly and see how fast it goes.  Then push it harder and see if it goes faster.  Next, push a more massive car and a less massive car with the same force, and see which ends up going faster.  The cars that went faster were accelerated more as they were pushed.  Thus, you "showed" that the acceleration of an object is directly related to the force accelerating it, and inversely related to the mass of the object.  This yields Force is proportional to mass x acceleration, (Newton's Second Law)\n\nFor the Third Law, have your audience imagine what it feels like to push the car.  It is impossible to push on something without it pushing back on you.  And the amount of the push-back is equal to the amount you push.  If they slam their hand against the car very hard, they will feel more force applied to their hand than if they gently push it.  And of course the push-back is in the opposite direction of the push.  If they push against the car to the right, the car pushes back to the left.  Hence, for every action, i.e. applied force,  there is an equal and opposite reaction force.
865909
Use clay and then you'll be able to form the 3d shapes
345363
Try this website:\n\nhttp://www.screwgravity.com/
806553
C'mon, kid.  Go and read your book and do your homework.  I can understand if you are having problems with one problem, but you are posting this so that others will do your learning for you.  This is simple Newtonian physics.  Three laws you have to work with.  Basic, basic concepts.  If you can't handle this, then might as well give up your dreams of becoming a rocket scientist where you have to deal with special relativities and zero-gravity physics.  \n\nNow get moving!
610467
Go to your local physics professor. If you are correct, (s)he will no doubt help you to get it published. If you are incorrect, (s)he should be willing to show you where your math is off.
700654
Newton set down the 3 laws of motion\n\n1. An object in motion tends to stay in motion in a straight line until it is acted on by a force.(inertia)\n2. Force = mass x acceleration\n3. Every force has an equal and opposite force (jet propulsion)\n\nNewton also discovered that light was made of all the colors of the rainbow. He split light with a prism
392123
http://www.pemberley.com/janeinfo/pptopic2.html#protofem3
1121613
Check this one\nhttp://science.nasa.gov/Realtime/jtrack/3d/JTrack3D.html
746185
I can only answer 1 and 3 for you, cause i dont know the others, but here it goes:\n\n             1.) Think of this problem in terms of a right triangle. Draw a horizontal line. This line is the tidal force of 3* (10^3) N. Then draw a line on the end of the line perpendicular to the horizontal one. This line is your wind force of 6* (10^3) N. Connect the two ends of the lines with another line, to make a complete right angle triangle. This is where it gets a little tricky. The angle formed when the slanted line and the horizontal line meet is the angle you are trying to find. To find it, use the tan function, which is defined as opposite side's value (6*10^3 N) over the adjacent side's value (3*10^3 N). Lets use x as a variable for the unknown angle. So...\n\ntanx=(6*10^3) / (3*10^3)\nUse inverse of tan now (on calculators, you usually have to press shift or 2nd, then the tan key.) So...\n\nInverse tan ( (6*10^3) / (3*10^3) ) = x\n\ndo that in your calculator, and the value you get is the angle.\n\nFor the net force just subtract (3* 10^3) from (6*10^3) and the result will be your net force.\n\n3.) I dont know if u know this or not, but there is a formula that is part of Newton's second law. It is F=m*a.\n"F" being the Force, while "m" is the mass, and "a" is the acceleration. Just plug in the values you are given so that it comes out....\n\n390=270*a\n\n390/270=a ("a" is in meters per second squared)\n\nHope that helps
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
483925
You may want to try Centripedal Force i.e. Turnings and slopes..\n\nBut it has some stuff to do with Physics as well. \n\nAlso you may want to find the maximum and minumum speeds at turnings. Include Frictional Force and try to minimize it using Calculus. \n\nTake Simple Circuits like circles, oval, parabola etc. Then take their equations and use calculus to find area under the curve etc. etc.\n\nWell thatz all comes to my mind for now!! :) BEST OF LUCK..
47010
1.  Hurl it spinning into space.  The vehicle will spin around its center of gravity, and its center of gravity will travel in a straight line.\n\n2.  Balance it on a see-saw three ways: upright, on its side, and on one end.  Each balance will define a middle plane that cuts through the vehicle.  The intersection of the three planes is the center of gravity.
954062
put an arm perpindicular on the shaft, put another arm pinned to the first, and slide the second through a sleeve.\n\njust like an old fasioned steam engine turned linear movement (piston) into rotary(drive wheel
987754
Actually the only way it would be possible is if you use Elliptic or Riemannian geometry, and would require a manifold.  A representation of such a construct would be similar to drawing a 4 dimensional shape in 3 or even 2 dimensions.\n\nEuclidian geometry doesn't allow triangles with a sum of all angles to be greater than 180 degrees.
912141
Here are two (see refs.). Note that the second one is a rotating 3-D model.