1)investigate the motion of magnet as it rolls down an inclined plane2) how to meaure speed of sound in liquid
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1) Assuming there are no strong EM fields near the magnet, it should behave exactly as a similar object would due to gravity. The gravitational field of the earth is too weak to produce a measurable effect for this experiment.\n\n2) You need a very long liquid filled container with a speaker on one side and a microphone on the other. Then using a computer to produce and receive the sounds, measure the time taken for the sound to traverse the distance. For accurate results the water needs be still.
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Waves are formed by putting directional kinetic energy into water and the surface tension of a liquid acts like a jump rope when you whip it really fast.
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the acceleration of the center of mass as the body rolls an an inclined plane is given by,\n A=g*sin(angle)/[1+I/MR^2]        (for proof see any dynamics boook)\n I=0.5MR^2\n  =0.06\ni.e.  A=3.27 \ntherefore the speed of the CM after travelling 6m is\n v=sqrt(2AL)\nv=6.26\nangular velcity W=v/R\n                W=6.26/0.1\n                W=62.6 rad/s
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Try: allaboutphysics.com I think that might help.
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Hello, \n\nThis problem is a bit more involved than you might initially think. There are 4 major considerations you must take into account. First, the acceleration is dependant on the angle of the hill at any given time. Therefore, acceleration is NOT constant, rather fluctuating with the incline. Second, the wind resistance is NOT constant by any means. It increases exponentially with the speed of the sled. Third, the force of the kinetic friction is not constant either. While the coefficient of friction does increase somewhat with speed (you could use an average for this), you cannot neglect the increase in friction when the sled goes around a corner. As you know, the friction is based on the normal force of the sled. At 90mph, the centripetal acceleration around a curve is V^2/r (V=velocity, r is radius of corner) which becomes quite considerable, even though steel on ice only has a kinetic coefficient between .01 - .05. \n\nNow that I have bored you out of your mind, the full equation would look something like this (I’m writing everything longhand so you know what I’m talking about)\nSum of the forces = mass x gravity x sin(angle of hill) – [(coefficient of friction) x mass x gravity x cos(angle of hill) – mass x velocity ^2 / radius of curve] – (Force of air drag)\n\nNotice the term included to account for curves: – mass x velocity ^2 / radius of curve\nDrops to zero when they are going in a straight line (r = infinity). \n\nTo find your acceleration (at a specific point in time), you simply divide everything by the mass of the sled and there you go!\n\nPS. don't take physics advice from dental students ;P
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Use the right hand rule, also known as the rule of thumb. If you have a wire (insulated!) with a current flowing through it, grab the wire in your right hand, with your thumb pointing in the direction that the current is flowing. Your fingers will curl around the wire in the direction of the magnetic field.\n\nIf the current alternates (periodically reverse direction) then the magnetic field also alternates. Envision it as a cylindrical zone that has the wire down its axis, and that the cylinder expands and contracts as the current of electrons increases or decreases. \n\nNow, suppose we wrap our wire around a coil form to make a coil. There is also a right hand rule, or rule of thumb, for coils. If you hold the coil (again, use insulated wire!) with your fingers wrapped around it pointing in the direction that the electron current flows, your thumb will point in the direction of the north pole of the coil.\n\nHow does that work? Remember, we are imagining the magnetic field as a cylindrical region around the wire. But the surface of the cylinder is rotating. (We'll stick with current in only one direction for the time being. It's easier to follow.) So think of the flow of the magnetic field as rotating around the wire. \n\nNow, if we wrap the wire into a coil, the rotation of the segements of the magnetic field add up in the same direction. They are working together. This is why the field is stronger inside the coil.\n\nWe should also think of the magnetic field as composed of individual lines of force that repel one another. That is why, when they exit one end of the coil and return to the other, they spread out in space. \n\nWe can see this if we hold a magnet under a sheet of paper or aluminum foil and sprinkle iron filings on top. They cluster most around the poles where the field lines are concentrated-- the field is strongest-- and more thinly around the outside of the magnet, where the field lines spread out, because the field is weaker there. This holds true for any magnet, not just electromagnets.\n\nI hope this is clear enough to help. It is a little tough trying to understand things we can't see.\n\n19 NOV 06, 1816 hrs, GMT.
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electromagnetic or magnetic fields. It is easiest to research and to do the practical of. Just place the magnets and use iron filings to measure it
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1.Kinetic energy at equilibrium point\n=1/2*m*v^2\n=1/2*.3.v^2\n=10J\nv^2=66.66\nv=8.16 m/sec\nAs the ladle is moving at this speed\nthe spring will come in to action\nHowever at the equilibrium position \nthe spring tension=zero\nHence at the instant the ladle is\n passing the equilibrium \nposition work done =zero\n2. At .1m tension force in the spring\n=500*.1=50N\nWe assume the velocity remains the same\n(strictly velocity will decrease due to deceleration\nimposed by the spring)\nWork done =50N*8.16 m/sec\n=408 N-m/sec\n=408watts
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See if it is attracted to a magnet?
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stick it in a tank of water then measure the water displaced.archemedes law
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Place the sand on a piece of paper, then place the magnet underneath, then the tilt the paper, that way the iron will stay where it is but won't just stick to the magnet!!
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This can be done with a density gradient column.  It is made of 2 fluids that are carefully mixed so that the density of the mixture varies from the top of the container to the bottom.  An object will then come to rest where its density matches the mixture.  This is actually a very accurate way of determining the density of materials.  However, over time the fluids in the column mix by diffusion and will eventually cause the suspended objects to sink or float.
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I'll answer the second part first. RIme taken to get to the other side is distance/speed = 2km/20kph = 6 mins. The downstream motion of water does not affect te perpendicular motion.\nHow far will it reach = distance covered downstream at the speed of the river = speed x time = 6 kph x 1/10 hr = 0.6 km.
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mass of the object\nforce of the magnet in a magnetic field\nthe object's own interference with the magnet
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1. Usually you have to solve these problems in terms of energy, where initial energy = final energy + friction (lost) energy. The initial and final energies are each the sum of the respective initial and final potential + kinetic energies. (Sometimes the problem doesn't explicitly give you an initial potential energy; it may just tell you that an object slides x meters downhill.) If you know the initial and/or final momentum and either the mass or the velocity, you can find the missing variable (velocity or mass) and then transform the momentum into kinetic energy. \n2. The coefficient of friction 'mu' of course affects the friction force. Sometimes you are asked to find mu so you have to work backwards. You may know the lost energy and the length over which the object encountered friction. Then you solve for friction force. Then, knowing the normal force as well (force perpendicular to the friction surface), you can find mu.\n3. Inclines with friction are a big favorite of the problem writers. Just remember that the gravitational force normal to the incline is g*m*cos(theta) and the gravitational force along the incline is g*m*sin(theta). If a ramp is tilted just enough to allow the object to slide, g*m*sin(theta) = friction force = mu*g*m*cos(theta), so mu = sin(theta)/cos(theta) = tan(theta). Or if you know mu and want to find theta, it's arctan(mu).\n4. Another fave is centripetal force on a horizontal turntable with a friction surface. We know the centripetal force is w^2*r*m (w being angular rate in rad/sec). And we know the normal force is g*m. So, if we rev up the turntable until the object slides off, or we move the object out from the center until it slides off, we know that, at the instant it lets go, w^2*r=g*mu, so we can find w or r. Easy, right?\nI hope this helps; if you have followup questions I'll be watching this question until it closes.
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when u will draw ur inclined plane and mass on it there r different forces acting in it \nfirst force will be directly perpendicular to the ground level i.e. Fg=mg\nsecond force acting is mgcos(theta) which is perpendicular to hypotenuse of the inclined and makes up Fn=mgcos(theta)\nthen another component parallel to the hyp. of the inclined \nwhich equals mgsin(theta) which goes down the ramp\nand  then u can\nequal the force going down with the frictional force\nmgsin(theta)-(mew)mgcos(theta)=ma\n(as we know frictional force is (mew)Fn=(mew)mgcos(theta) )\nsolve by putting values\nmew=0.146    ans
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The speed of sound: 340.29 m / s