Two swimmers start simultaneously from the opposite ends of a pool with speeds v1 and v2  respectively.....They meet first time t1 seconds after,reach the ends of the pool and the second time t2 seconds after the first meeting. whats the ratio of t1 and t2 ?\n\na)v1:v2\nb)v2:v1\nc)1:3\nd)Data Insufficient\n\n\nhow to solve this?
214928
I believe what you are saying is they meet twice. this is actually a simple 'trick' question but:\nvelocity  x time = distance\nthey first meet after t1 and we have:\nv1xt1 + v2xt1 = length of pool\nAfter they have met they swim to the end of the pool(or at least one of them does since we don't know what the velocities are and one may be super slow) and turn around and swim back and meet again, but in any case we have either:\nv1xt2 + v2t2 = twice the length of the pool and since their relative velocities do not change we have t1/t2 = 1/2, or one is super slow in which case he is caught before 2 pool lengths are covered so "Data insuficient"
668266
this can be a problem to you if you don't know anything about simultaneous equation or if u don't know how to form simultaneous equation. another thing is u should at  least have an  idea about chemistry. look\n  \nm1v1+m2v2=MV.............1\nm1+m2=total volume after dilution=150Lt..............2\n now\n              take m1 is the initial conc of saline soln=32%=0.32\n                   ''  v1                      volume                 =?\n                 m2 is the conc of water =0\n                 v2 is the volume of water =?\n               M is the final conc of the soln = 20%=0.20\n                V is the final volume of the soln obtained after mixing=150Lt\n\nnow\n             0.32v1+0*v2=0.2*150......................1\n\n             v1+v2=150.................2\nfrom eqn 1\n                      0.32v1=30\n           so v1= 93.75Lt's\n   so the remaining amount is of water\n                v2=150-93.75=56.25lts\nso you need 56.25litres of water and 93.75litres of 32%saline to prepare the solution you need.\n\nif you have these solution just try to mix at that proportion then u will tell me.
297281
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.
395797
let,Pump A and pump B can themselves empty the pond in x and y hours respectively\nTherefor in 1 hour pump A can empty 1/x part of the pone,and\n               in 1 hour pump B can empty 1/y part of the pond.\nBy the problem\n10/x +6/y=1.........(eqn. 1)\n6/x+12/y=1...........(eqn.2)\nmultiplying equn 1 by 2 and subtracting it from 2,we get\n-14/x=-1\n=>-x=-14=>x=14\nPutting the value of x in eqn 1 we get\n10/14+6/y=1=>6/y=4/14   =>4y=84    =>y=21\nThe pump A can empty it in 14 hours and pump B can empty it in 21 hours
934
You're really close! The answer is actually 1:03:11. You were off one minute because you forgot to borrow a minute from the 34 so you could subtract the seconds.
533582
Use PV = n RT, RT = constant. Therefore n = PV * constant in the three cases.\nWe have \nH2 :  P = 372 torr V = 2 liter  n_H2=  744 * constant\nN2:  P = p torr V = 1 liter  n_N2 = p * constant\nCombined: P = 303 torr V=3 liter  n_combined = 909 * constant\n\nTherefore  744 + p = 909 => p = 165 torr
497285
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
115909
come on. all these questions are almost EXACTLY the same. \np1v1/t1 = p2v2/t2 = nr = constant.\n(0.998atm)(2.1L) / (36+273K) = (0.9atm)v2 / (28+273K)\nand then solve for v2. and u get\nv2 = 2.268L\n\n3. p1v1 = p2v2 = nr/t = constant\n(1.08atm)(14.57mL) = (1.43atm)v2\nv2 = 11.0039mL.
485995
(P1V1)/T1 = (P2V2)/T2\n\nP1 = 1 atm\nV1 = 1 m3\nT1 = 0°C = 273 K\n\nP2 = 1 atm\nV2 = x m3\nT2 = 105°C = 378 K\n\nV2 = T2(P1V1)/(P2T1)\n     = 378[(1)(1)]/[(1)(273)]\n     = 1.38 m3\n\nOriginal density is 0.179 kg/m3\nnew desity (0.179kg/m3)/1.38\n\n=1.30 kg/m3
344253
Hello, I would be glad to help you with this:\n\n a=(v-c)/t\nat=v-c\n-c=at-v\nc=v-at \n\n I hope this helps and good luck with the rest of your course work ,Please Please remember me for best answer
508574
a) F = ma\n    F = 0.0412 kg * 9.8m/s^2\n    F = 0.404 N\n\nb) F = 0.0370 kg * 9.8 m/s^2\n    F = 0.363 N\n\nc) The difference in "weight" is equivalent the weight of the displaced volume of water...  So...\n   0.404N - 0.363N = V * 1000 kg/m^3 * 9.8 m/s^2\n   0.041N = V * 9800N/m^3\n   V = 4.18 * 10^-6 m^3   or  4.18 cm^3\n\nd) Oh Juliet you are my density.... (bad joke!)\n     Density = m/V\n     D = 0.0412 kg / 4.18  * 10^-6 m^3\n     D = 9848 kg/m^3
428707
Distance = Speed * Time.\nWe don't know the "distance" variable for either. We need to find the "time" at which the "distance" will be the same. Let distance = d\n\nWe want to find Thomas's time, so let Thomas's time = t.\nAlexis left 3 hours earlier, so her time is t + 3.\n\nThomas speed is 30, Alexis is 20. So lets fill this in.\nThomas:\nD = 30t\nAlexis:\nD = 20(t+3)\n\nThe distance will be the same, so the D in both equations is the same variable. A basic principle is that if a = b and b = c, then a = c.  Therefore, we can combine the equations for the two people as follows:\n30t = 20(t+3)\n30t = 20t + 60\n10t = 60\nt = 6 hrs\n\n\nThe same applies to the second.\nTimothy:\nD = 58t\nSavanah:\nD = 47(t+2)\n\n58t = 47(t+2)\n58t = 47t + 94\n11t = 94\nt = 8 6/11 OR 8.5454545454545\nt = 8 hours, 32 minutes, 43 seconds\n\n\n\nThe last one is slightly different.\nJet 1 Distance = 331t\nJet 2 Distance = 333t\nThe two distances aren't equal this time, but the two distances added up will equal 1,705, SO:\n331t + 333t = 1705\n664t = 1705\nt = 1705/664 = 2 377/664 = 2.56777108\nt = 2 hrs, 34 minutes, 4 seconds
391052
Yes you are right to subtract the water pressure from the total pressure.\n\nThen use p1v1/t1 = p2v2/t2\n\nSTP is p2v2/t2\n\nSTP = 1 atm and 0C or 273K\n\nSolve for V2.\n\nKeep units the same on both sides!
240330
P1V1/T1 = P2V2/T2\nPlug in your initial and final data and solve for x.
502362
The formula for a cylindrical volume is (pi)(radius)^2(depth)\nSo the total volume of water we are working with is:\n(Vtot) = (pi)(4)^2(12.5) = 200(pi)                  [eq1]\n\nThe volume of water in Tank1 and Tank2 at particular depths d1 and d2 are:\n(Vtank1) = (pi)16(d1)                                    [eq2]\n(Vtank2) = (pi)9(d2)                                      [eq3]\n\nWe also know that at any given moment...\n(Vtank1)+(Vtank2) = (Vtot)                            [eq4]\n\nWe want to know the volume in Tank2 when the water depth in both tanks are the same, so let d1 = d2 = d, and solve equation [eq4]  using [eq1], [eq2] and [eq3]:\n\n(pi)16(d) + (pi)9(d) =  200(pi)\n\n16(d) + 9(d) = 200\n\n25d = 200\n\nd = 8\n\nUsing [eq3], determine out how much water Tank2 has in it when its depth is 8.\n\n(Vtank2) = (pi)9(8) = 72(pi) =  226.1952 m^3\n\nIf the pump is running at 10 m^3/min, then Tank2 will have 226.1952 m^3 of water in it after 226.1952/10 minutes.\n\nAnswer:  22.6 minutes
246498
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.
274579
i haven't done this in a while.\n\nThe volume of the reservoir is equal to (1/2)*(height)*("base")*length.  the reservoir is a rectangular prism with the "base" being the distance across the surface of the water.  We are given the shape of the full volume so we can scale down to the level of the water to find what the volume in the reservoir at its current level.\n\nif you think about the cross-section of the volume, and use half of that forming a right triangle, we can determine the ratio of the surface (the "base") to the height (water depth).  In this case, it is (40/2)/(20*sqrt(3)) or 1/sqrt(3).  so ("base"/2)/(height) = 1/sqrt(3)  and then "base" = (2*height)/sqrt(3).\n\nsubstitute that into the volume equation so that\nvolume = [(height^2)*length]/sqrt(3) or literally,\nV = [(h^2)*L]/sqrt(3).  differentiate with respect to time,\n(dV/dt) = [2*h*(dh/dt)*L]/sqrt(3) and substitute the numbers in:\ndV/dt = 200 - 50 = 150\nh = 10*sqrt(3)\nL = 450\nand solve for dh/dt which is the rate of change of the water level in yds/hour.  these are odd units, you might want to change to inches/hour.\n\ni did it this way and got 1/60 yds/hr or 0.6 inches/hr.\n\nbut again it has been a long time....