http://session.masteringphysics.com/problemAsset/1009449/3/yf_Figure_26_53.jpg
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Use the node voltage method. Label each node with a voltage, then apply Kirchoff's current law at as many nodes as you need to generate enough equations to solve for any unknowns you have introduced.\n\nLet the node at the bottom of the circuit be ground, and then try to figure out the voltages at the other nodes relative to ground. Clearly the voltage at the node on the top will be 14V (I'll just call this V1) but no other voltages are apparent. So call the voltage at the node between R1, R3, and R4 Vx and the voltage at the node between R2, R3, and R5 Vy.\n\nNow that we have labeled all the node voltages, you can figure out every single current in the circuit in terms of V1, Vx, and Vy (using Ohms Law, the voltage drop across a resistor is equal to the resistance times the current in the direction of the voltage drop). For example, we know the current going down through R1 is (V1-Vx)/R1, the current going right through R3 is (Vx-Vy)/R3, etc. Now we just have to solve for the unknowns Vx and Vy.\n\nKirchoff's current law (KCL) says that the sum of the currents leaving a node is equal to zero. If we apply this law to the node with voltage Vx and the node with voltage Vy, we'll get a pair of equations that we can use to solve for Vx and Vy.\n\nApplying KCL at the node with voltage Vx should give you the following equation:\n\n(Vx-V1)/R1 + (Vx-Vy)/R3 + Vx/R4 = 0\n\nThe first term is the current going up through R1 (we already figured out the current going down through R1, but we want currents that are leaving node Vx so we had to flip the sign), the second term is the current going right through R3, and the final term is the current going down through R4.\n\nI'll leave it to you to figure out the equation for node Vy. Once you have that you'll have two equations and two unknowns and you can solve for everything. Once you have Vx and Vy, you can easily figure out a numerical value for any current in the circuit.\n\nBTW, the node-voltage method is pretty powerful and can solve most linear circuits. Check out the link below, it was written by an old EE prof of mine who is very thorough.
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http://cwx.prenhall.com/bookbind/pubbooks/hillchem3/medialib/media_portfolio/text_images/CH08/FG08_17.JPG
359606
I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
510305
http://hyperphysics.phy-astr.gsu.edu/hbase/class/phscilab/dens.html\n\ntry this site
636658
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
572034
would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
503761
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
453548
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
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F = ma\n\nF = m(Δv/Δt)\n\nF = 38000(17-0/20)\n\nF = 38000(17/20)\n\nF = 3.2 x 10^4 N
48071
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
413210
Your answer key is wrong, or you are not reading the right answer.\nFrom the data you presented above y = x^3 is OK.
339467
Consider a force diagram at the point that the 499 N load is applied.  You have 499 N in the negative "y" direction, and two equal Tension forces (call them T), each at an angle to the horizontal.  Thus, the tension forces have each an "x" and "y" component.  One tension force goes to the tree, the other to the car.  You've got the angle from the horizontal from trig.\n\nNow do a force balance on those three forces, in the "y" direction. There is no acceleration, so sum of forces = ma = 0\n\nT*sin(4.76) + T*sin(4.76) - 499 = 0\n\nOnly one unknown here, solving.......\n\nT = 3006.67 N\n\nThat's a lot of force!\n\nCheers,
547100
18\n\nhttp://mudandmuck.com/str2/tri.gif\nThe angles in the picture are derived working from left to right (From angle A to the right)\n\n\n\n<DAF = <DFA = X\n<ADF = 180 - 2X\n<FDE = 180 - <ADF = 2X\n<DEF = <FDE = 2X\n<DFE = 180 - 4X\n<EFC = 180 - (180-4X+X) = 3X\n<ECF = <EFC = 3X\n<ECB = 90 - <ECF = 90-3X\n<CEB=<BCE (complement to <BAC) = 90-X \n\n90 - 3x + 2(90 - x) = 180\n5x = 90\nx = 18
291422
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.
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http://static.flickr.com/66/177746990_d467577ab5_m.jpg
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Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
97778
Howstuffworks has good diagrams and explanations that may help you.  Good luck.