Anyone Smart? in physics?
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Assume that the incident rays are incident normally on the flat side (one of the sides forming the right angle) of the prism. The side opposite to the top angle is the base. When any light is incident perpendicular onto a transparent medium, it is simply transmitted without undergoing any refraction. So, the ray incident on the flat surface normally (means perpendicularly) is transmitted to the slant face containing 65* angle. Construct a line perpendicular to this slant face at the point of incidence. You will find that the angle between this line and the light rays is 25*, also known as the Angle of Incidence. Using\nRefractive Index = (Sine of the Angle of Incidence/Sine of the Angle of Refraction), we can find the Angle of Refraction for each colour depending upon its Refractive Index. The Angle of Incidence is the same 25* for both the rays. The problem is that the Refractive Indices given are for light travelling from air medium into glass medium, but, here, the light rays are travelling from the inside of the glass medium to the outside air medium. So, reverse the Angles of Incidence and Refraction so that it looks as if the rays are travelling from the air medium into the glass medium. So, now, the <i is to be found out and <r = 25*\nU = (sin <i / sin <r), sin <i = (U x sin <r). Calculating this, <i for red = 44.62* and <i for violet = 45.857* These are the answers. <i is the Angle of Incidence and <r is the Angle of Refraction.\n\nStill doubts? Visit www.pcmhelp.netfirms.com
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thermodynamics?  heat transfer?  physics?
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www.physicsclassroom.com
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websites aren't that trustworthy for accurate scientific advice. I'd recommend finding someone to ask in person... Wilkipedia is good for some basic things, but anything beyond Physics 101 is a crap shoot.  \n\ngood luck
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Way to go Arbitrage!  Oh to funny guy- yeah we get that E=MC2  IS a formula- but if you really want to "get it" Check out Maxwells equations. It's a great time to be a science nerd with all this info pouring in now! Figure it out & go up one on Albert!
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I don't know but, I suggest you put this in the "Physics" category if you want the best possible answer.
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My magic 8 ball says no. \n\nSo does my physics professor.
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Have you tried asking this question under the heading Science & Mathematics > Physics? You might have a better audience there.
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plz ask in physics category
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ok i fell asleep in physics, but...\n\n1/1837 th of the motion would be transferred to the atom.\n\nand 1836/1837ths would be the speed of the bounced electron.
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http://theory.uwinnipeg.ca/physics/
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The last poster needs to have his dynamic physics retaught, or stop trying to sound like an engineer.\n\nThe first mistake, is identifying newtons = force x distance.  This is actually a measure of work or energy and measured in joules.  The second mistake identifying force as being measured in joules.  Force is measured in Newtons.  Since it was Isaac Newton in the second law of motion that defined Force.\n\nYour question cannot really be answered without more information.  What is the direction of motion, horizontally or vertically?  \n\nVerically?  Then 150 lbs of force.  Pounds were once used to measure force.  If you want to use metric, convert the 150 lbs to kilograms, multiply by 9.8 and then you must exert at least this much force in Newtons to overcome the force of gravity.  150 lbs is 68.04 kg.\n\nHorizontally, what type of surface?  Determine the coefficient of friction, and multiply it by the mass in kilograms.  Then this is the amount of force you need to exert to overcome the force of friction.  There are two types of friction, static and dynamic.  In other words it takes more force to move an object at rest, but once the object is in motion it takes less force to keep it move (opposing the dynamic force of friction).\n\nWatts by the way is a measurement of power, which is energy derived by time, Power = dE/dt.  Joules per second.
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try the following. they are really helpful.\n\nwww.tutor4physics.com\nwww.physicsclassroom.com
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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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Only if you show your work and do not expect them to do the question for you; only to provide hints and guides.\nhttp://www.physicsforums.com
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u can consult ur physics book
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http://www.physlink.com/education/askexperts/ae25.cfm