using a balance scale 3 times or less find the odd weighted ball and be able to identify the unkown ball and weather it is heavier or lighte then the other pool balls
131734
1.  Weigh all 12 balls together.\n2.  Weight 11 balls.\n3.  Remove a ball from the 11 and weigh the 10.\n\nNow for some calculation.\nweights 1,2,3 --> w1, w2, w3\n\nif w1-w2 = w2-w3 --> you have the weight of a normal ball.  To find the weight of the odd ball, multiply the normal weight by 11 and subtract from the weight of 12 balls.\n\nif w1-w2 does not equal w2-w3 -->  one of the two differences is the weight of the odd ball.  You have to assume one of these differences is the weight of a normal ball and analyze.\n  \nTake the larger difference and multiply by 11.  If the difference between this value and the combined weight is equivalent to the one of the differences from the previous paragraph, then the weight of the oddball is the "smaller difference" we havent analyzed yet.  The oddball is thus lighter than the rest.  \n\nIf the subtraction from the previous step isn't equivalent to the smaller difference, then the weight of the oddball is the one you've assumed to be the normal ball.  The oddball is thus heavier than the rest.\n\nOriginal solution, may be kind of complicated, but hey it's orginal, LOL\n\nTo answer below: How would that work if the odd ball is LIGHTER not heavier?
563722
Here's one: Measure the density of air.\n\nTake a flask, weigh it.\nVacuum out all the air, then weight it again.
512832
You have 8 rungs that are nine inches apart with the last being 9 inches from the bottom of the pool so I would draw it on a piece of paper I I I I I I I I I as nine rungs and count the spaces in between the rungs (lines). I get 8 spaces of nine inches. So then 8 spaces times nine inches is 72 inches. Now we need to convert inches to feet so we divide 72 by 12 inches per foot. That works out to 6 feet. if the pool fills at 1 foot per hour then it will take 6 hours to fill the pool to the top rung. There is an algebraic approach but that isn't what they'd be looking for in 5'th grade. Teach you son to draw a image and label it when working with word problems. Its a great habit and eventually he'll learn to visualize without the paper. Cheers:0
231401
http://www.weightchart.org/\ncheck this out
178794
IF we know the coin is heavy and we have 5 scale usage , we can find out the heavy coin amist 243 coins.\n\nFor the first question, with 3 scale usage and when u dont know is the coin is heavier or liter\nNumber the coins A-L. \n\nBegin by weighing four against four. If they balance, then weigh any of the remaining three against any three of the good coins. \n\nIf they balance then the odd one is the remaining coin and can be \nidentified (whether it's lighter or heavier) in the final weighing. \n\nIf three against three do not balance, take the three containing the odd \ncoin and weigh any one against another. \n\nThus: \n\nIf the first weighing of four against four does not produce a balance, then \nthe second weighing involves three against three with the coins switched \nbetween the two pans and a good coin introduced. \n\nSo: \n\nIf A+B+C+D > E+F+G+H \n\nWe try A+B+E against C+F+J \n\nIf A+B+E + C+F+J then either D is heavier or G or H is lighter, so weigh \nG against H. \n\nIf A+B+E > C+F+J then either F is lighter or A or B is heavier, so we \nweigh A against B. \n\nIf A+B+E < C+F+J then either E is lighter or C is heavier, so weigh \neither against a good coin, eg. K against E. \n\nAn alternative second weighing is A+B+E against C+D+F, which follows \nsimilar lines to the above.
616173
Zero out all scales. You start will the scale with the largest ballast (big heavy thing). Move it larger by notches until the scale tips the other way, then move it back one notch. Repeat this on each successively smaller scale. Then your weight is the sum of each scale.\nSo 150 + 10 + 5 + 3 would be 168.
255788
Divide into two equal groups of 4.\nOne will be heavier than the other.  Take that group off the scale.\nDivide the remaining four into two groups of two.\nOne will be heavier than the other.  Take that group off the scale.\nDivide the remaining group into two groups of one.\nThe lighter one is the answer.\n\nEDIT: This can be accomplished in just 2 weighings.\nSplit the 8 into two groups of 6 and 2.\nDivide the group with 6 into two groups of three.\nCase 1: If they weigh the same then the lighter one is in the group of two.  Weigh the two to see which one is lighter.\nCase2: If they don't weigh the same, take the lighter group of three and weigh one on each side.  If they are the same then you know which one is the lighter one.  If they are different then you still know which one is the lighter one.
93448
erm, bit hard to show you on the net. You rest the pool cue (tip end pointing away from you) between your thumb and finger of your non-writing hand, hold the cue with the other hand near the other end of the cue, slide the cue back and forth a bit to get used to it, then hit the white with the tip in the centre of the ball.\n\nGood luck
351188
Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)
248752
Two\n\nWeigh ABC against DEF\n\nIf the balance tilts, take the lighter group and weigh one ball against a second one. If it tilts take the lighter ball, If it doesn't tilt, it must be the third ball of that group of 3 balls.\n\nIf ABC weigh the same as DEF, weigh G against H, take the lighter one if it tilts. take I if G and H weigh the same as one another.\n\nTwo weighings suffice.
279773
Make a tree that shows all possibilities you have.\nOn the first level there is the choice of the box with 0.5 probability each. On the next level, box A, you have two choices again, with the prob. is 3/7 to find a red ball, 4/7 to find a white ball, etc.\n\nWhen you have finished to tree, use it to make a list and\ncalculate the prob. for picking red-red, red-white, etc.
196227
in 9 ball you have to hit the lowest numbered ball 1st or it is a ball in hand fall.if you scratch.if you dont hit a rail after you contact the lowest number ball with either the cue ball or the object ball. in 8 ball you have ball in hand if...you dont hit one of your balls and a rail.you scratch.
177807
Seperate balls in 3 packs of 3.\n\nFirst weigh pack 1 against pack 2. If both are the same, then your heavy ball is in pack 3. If one weighs more, you ball is in that pack.\n\nSecond you weigh 2 balls in the heavy pack. If both balls are the same, it's the 3rd ball. If one weighs more, that's your ball.\n\nToodles.
168315
catching soft ground balls with bare hands...\nand...have them get a partner, have them on two knees and give each pair about 8 balls, have one flip 8 balls quickly to the other and tell the one recieving the balls he can only catch with one hand...and have them both do right hand,then left hand...two sets\n..one more..\nhave them get on one knee sitting up straight and have them play catch from short/regualar distances
535379
You must show not tell. Take a bowl filled to the top with water and put it on a plate to catch overflow. The put objects in the water and compare the overflow for sinking and floating objects of the same size. If you have a scale weigh the objects and the water if not just hold them and judge weight. Then explain density.
184557
Divide them into 2 groups: 6 and 2\nWeigh the group of 6 placing 3 on each side.\nIf they are equal, then the odd coin is in the group of two.  Weigh each coin to determine the heavier one.\nIf they are not equal, then take the 3 coins from the heavy side.\nTake two of them and place one on each side.\nIf they are equal then the third coin is the heavy one.\nIf they are unequal, then you know which one is heavier.
134910
Answer: 1,3,9 and 27 kg\n\nWay of solving:\n\n1) 4 pieces are very small number to weigh such number (40 of different weights. So I just tried to understand a most efficient way of weighing different weights using a small number of pieces.\n\n2) Firstly, I understand the most efficient set of pieces when we put some thing we are weighing on one scale and pieces on another. There are 15 (2^4-1) ways of taking different non-empty subset from set of 4 pieces. So, in most effective set all those subsets must have different weights. Eady to see, that pieces which are powers of 2: 1,2,4,8 kg make such a set (15 different weighings from 1 to 15 kg). \n\n3) Next I guessed that it is possible to put pieces not only on another but also at the same scale as we put some thing we are weighing. So, the weight of little piece can be not only added to the big piece, but also subtract from it, giving more possible combinations of weights.\nThen I guessed that some subset1 without smallest piece (ss1), ss1 + smallest piece (ss1+), another subset without smallest piece (ss2), ss2 with smallest piece on another scale (ss2-) must be consecutive numbers. So, ss1 = ss2 + 3. So, if smallest piece is x kg, the next piece must be 3 * 1st=3x kg.\nThe same situation is with other pieces: 3rd = 3 * 2nd = 9x kg, 4th = 3* 3rd = 27x kg.\nThis set of pieces makes possible 40 different weights.\nAlso, we know that sum of weights of those pieces is 40 kg.\nSo, x + 3x + 9x + 27x = 40kg\nx = 1kg\n\nAnd the final answer is: 1 kg, 3 kg, 9 kg and 27 kg.