All balls look exactly the same, but One ball is slightly heavier.  You can use the scale two times, but thats it.  How do you find the heavy ball?
132282
I think I have the answer.  Make three groups of three balls.  Now, take any two groups and place one group on one side, and the other group on the other side.  If the balance indicates that the groups are equal, then set aside those balls.  Take the remaining group of three balls and place any one on one side, and another on the other side, leaving the third on the table.  If both balls are even, then the one on the table is the heaviest.  Now for the different scenarios.  Going back one step, if the scale indicates that one side is heavier, then that ball is heavier, and you've found the heaviest.  When placing the groups of three balls on the scale, basically the same procedure is used.  If one group is heavier, take that group and set aside all other groups.  Then place one ball on each side and find if the scale is balanced.  If it is, then the one on the table is the heaviest etc.
383253
Analyze the forces at the top of the loop:\n\nThe force of gravity pulling the ball downward,\nthe upward centrifigal force keeping the ball on the track,\nand the force of the track pushing the ball downward.\n\nThe answer to the question is when the centrifigal force is exactly equal to the weight, so the force of the track equals zero.\nmg=mw^2R\nw is the angular velocity\nw=sqrt(g/R)\n\nOnce you know the angular velocity, compute the equivalent linear velocity by mutlipying by R\n\nthen use mgh=1/2mv^2\nto compute h\n\nWhere is point Q?\n\nj
275241
You only need to perform ONE weighing.\n\nFirst: take 1 match from the first box, 2 matches from the 2nd box, 3 matches from the 3rd box, ..., and 10 matches from the last box.\n\nWeigh the 55 selected matches. If the matches are all 10 grams, the total weight will be 550 grams. However, this is not true because some of them weigh 9 grams. For each 9-gram match, the total weight will go down by 1 gram.\n\nSo if the total weight is 549 grams, then the first box has the 9-gram matches.\n\nIf the total weight is 548 grams, then the 2nd box has the 9-gram matches.\n\n...\n\nIf the total weight is 540 grams, then the 10th box has the 9-gram matches.\n\nThus: Given this setup, the nth box has the 9-gram matches where n = (550g - total weight)
334586
As it is certain that the first one is red and the second one is blue, we can ignore the fact that they were even mentioned. (It would be different if you phrased it as find the probability of the first ball being red, the second being blue and the third being yellow...)\n\nP(Yellow) = 1/5
90201
I believe the result is the same, assuming replacement.  This is because every event (drawing a ball) is independent of every other and it does not matter if you look at all 1000 results or some 100 of them.\nGood question.
351527
Hi!  You are right.\nIf they have equal masses and are elastic then  b) it is going to happen. Why? well Vf= 2*Vcm-Vi. Where all are vectors and :\nVf= final speed;   Vcm=center of mass speed ;  Vi=initial speed.\nThus 0=2*Vcm-Vi => Vcm=V/2=(m1V1+m2V2)/(m1+m2)=\n=m1*Vi/(m1+m2)  => m1=m2.    \n  " a)" isn't possible because the total momentum must be preserved.\n          Hope it helps!
19495
That sounds like a question for a magic eight ball.\n\nOr a coin.
256077
5\nbring it higher because at the first 20 feet it won't break\n5 because you only said two balls the same color  hit the wrong number damnit
312876
Divide all 9 balls into 3 groups of 3 balls.\n\n1) weigh any 2 groups:\n         - if they are equal, the crooked ball is in the 3rd group\n         - if not, you have found the group where the ball is\n\n2) repeat the same experiment with that group until you find the ball
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.
553487
Choose the heaviest that you think you can handle and control.....it needs to be as heavy as possible......the correct roll(spin or English) on the ball is very important but even then it needs to be heavy enough to hit the "King Pin" which is the 5 and not the 1....the 1 or head pin is almost 3  feet in front of the 5 so with out the right roll and weight it will deflect and deny you a strike or a good count and often times leave you a split....recommend a 16 pound ball but no lighter then 14.5......good luck on the tour and I will be expecting my 1% commission!!
486205
Fg = mg = (0.20 kg)(9.80662 m/s^2) = 1.961324 N\nFc = (mv^2)/r = mrω^2\nFc = (0.20 kg)(0.60 m)((2π rad/s)^2\nFc = 4.73741 N\nFt = √( 4.73741^2 + 1.961324) @ arctan(1.961324/4.73741)\nFt ≈ 5.127 N @ 22.5°\n\n[The vertical component of Ft must offset the weight of the ball.]
290502
And I see that you are no exception, as you HAVE confused them!!\n\nIn the S.I. system, mass is kg, and weight is N (that is Newtons)\n\nIn the Imperial system, mass is slugs and weight is pounds.\n\n1 slug is equal to 14.6 kg.\n1 lb. is qual to 4.5 N\n\nPeople generally confuse them because the authorities decided  to equate 1 kg to 2.2 lb.  Not that it makes that much of a difference if these are used for commercial and not technical/scientific purposes, since gravity is basically the same at all habitable places on Earth.
604774
The explanation is really simple, really!\n\nOkay, take a look at the mystic ball chart right now. Take a look at the numbers 0, 9, 18, 27, 36, 45, 54, 63, 72, and 81. They are all the same symbol, right? And they are all multiples of 9! Now, before doing anything, just click on the mystic ball. You'll notice that the symbol that pops up is the same symbol that the numbers mentioned above show. Click on the blue link to try it all again. You should get another sheet with different symbols. Now, look at the very same numbers that you looked at before. They all share the same symbol, but now it is a different one, right?\n\nThis is the trick. No matter what number you choose from 0 to 99, the final answer will always be a multiple of 9. Try it out yourself! Go through each number and figure out what the final number of each number would be. Figure out what the final symbol would be, too. You will see that no matter what number you choose, the same symbol will keep popping up.\n\nIf you want the equation, it would be 10a + b - (a+b). 'a' represents the first digit of the number you chose, and 'b' represents the second digit of the number you chose. '10a + b' represents the number as a whole (for example, let's choose the number 32. a = 3, b = 2. 10(3) + 2 = 32. Make sense?). the '- (a+b)' part represents the time when you add the digits together and subtract it from the original number to get your final answer. After reducing the equation, you get 9a, which is why the answers all come out to be a multiple of 9.\n\nHopefully this makes sense!
620429
Most of them got it right.\n\nThis is another way to think about the result.\n\nFirst player = T or H doesn't matter\n\nSecond Player = Must be that same as the First Player\n                        =  only 2 possibility (H or T)\n                        = 1/2\nThird Player = Must be the same as First Player\n                    = again, only 2 possibility ( H or T )\n                    = 1/2\n\nIn order for the Second Player and Third Player to have the same result as First Plaer = 1/2 x 1/2\n                                 = 1/4.
418796
If they are thrown with the same MAGNITUDE of velocity, and you are either not counting air resistance or are assuming that they all have the same drag parameters, then the MAGNITUDES of the final velocities will all be the same.  The actual velocities of the ones thrown up and down will be exactly the same while the one thrown horizontally will have a horizontal component an a lesser vertical component.  Still, all magnitudes will be the same.\n\nIf however, they are thrown at different speeds, then it will be the one that is thrown the hardest If you are counting air resistance and the balls have different drag parameters, then you have not provided enough info to tell.\n\nAnd if by final velocity you mean after they have come to rest on the ground, then obviously they all have a velocity of 0.  Typically this is not referred to as final though.
450067
according to the conservation of momentum the second ball has the same momentum, though it depends on the angle at which the first ball collides with the second\n\nalso if you put English on the cue ball\n\nif it is a solid shot, the cue ball can stop and transfer all momentum to the second ball\n\nhope i could help