Related Rate Problem?
966711
The volume of the water in the cup at any moment is:\n\nV = (1/3) pi [ (Rw)^2 + (1)Rw + (1)^2 ] \n   = (1/3) pi [ (Rw)^2 + Rw + 1 ] h\n\nRw = Radius of upper part of water\nh = height of water\n\nRw / h = 2 / 6\n\n6 Rw = 2 h\n\nRw = 1/3 h\n\nSubstitute in to the volume equation\n\nV = (1/3) pi [ (1/9) h^2 + (1/3) h + 1] h\n = (1/3) pi [ (1/9) h^3 + (1/3) h^2 + h ]\n\ndV / dt  = (1/3) pi [ (1/3) h^2 + (2/3) h + 1] (dh / dt)\n\nSolve for dh /dt  and plug in the values h=4 and dV/dt = 3
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
511464
I think there is something wrong with the problem.The given details are not sufficient.Any how if you find the answer for it please email me.Lemme make use of it.
596087
This is very straightforward. Which step are you having trouble with? The same goes for your other question.
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.
901904
I think here is a mistake:\n\n(1500 * 36) = 1000v\nv = 55.8 m/s\n\nyou have to put 500 instead 1500, because this is what we have after 1000 left us!\n\n(500*36)=1000v\nv=15 m/s
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
45955
Question 1 - 7 minutes.  Other question ??  Sorry..
268703
Use the overall rate equation: rate = k[A]^x[B]^y...\n1) define for rate1 and rate 2 with the appropriate concentrations.\n2) divide both sides (e.g. rate1/rate2 = ...\nNote that this cancels k\n3) Solve for x, y, ...  These are your respective orders.\n4) Add x, y, ...  This is your overall order.\n\nYou can get the value of k by:\n5) Plug in values of x, y, ... to either rate1 or rate2 equation.\n5) Solve for k.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
434195
That is the rate of change in an "instant" in time.  You calculate it by finding the slope of the curve at that point.  You do this by taking the first derivative at that point.  If it is a straight line, you can get it by doing the average rise over the average run:  Pick two points on the line. Call them x1,y1 and x2,y2.  You are probably looking at supply and demand so we'll do price and quantity, so call them Q1,P1 for the first point and Q2,P2 for the second.  The slope (instantaneous rate of change) is: (P2-P1)/(Q2-Q1).\nIf it's not a straight line, you can approximate it by picking one point to the left and one point to the right of the point you are trying to find.  Make these two points close to the point you are looking for, and equidistant from the point you are looking for.  Use these points in the above equation.
530622
Ok, first you want to see where they meet. To make this easier, advance train A for 5/6ths of an hour and see the problem in terms of when train B starts out. 65 x 5/6= 325/6\nkilometers. So the problem is the same only they start out at the same time over a distance of 600- 325/6 kilometer =\n545 5/6 kilometers.\n\nNow, they travel in the proportion of 65/105 kilometers per hour. So the distance till they meet will be in the same proportion. That is 545 5/6 x 65/105 = 212875/630 or about 338 kilometers for train A to travel. Since it travels at 65 kph, this means it travels for 5.2 hours from the new start time equal to B's which was 6:20 AM. Ends up being 11:32AM.
744023
- 1.24 rad/s ? at steady state\n\nw = v/r = (3.4 m/s) / (2.75 m)
141032
None. That's a hardware problem.
239222
no, reaction times are too slow at that temperature.  (I'm just guessing, but I'm a chemist.  You should be able to look up reaction energies, try Beilstein.)
839113
rate of change of demand or supply