There was a farmer with a 40 lbs rock used to measure wheat. He lent the rock to his neighbor, and the neighbor came back the next day with good news and bad news. The bad news was that he broke the rock. The good news was that with the four pieces it was broken into, he could measure any weight from 1-40 lbs to the exactness of a pound. how much did each of the four pieces weigh?
206397
dghose is right.  \n\nYou may have to place the rocks on either side of the balance to determine the weight, but you can do it.\n\nFor example, to weigh one pound of wheat, put the one pound rock on the other side of the balance.  To measure two pounds, put the three pound weight on one side, the one pound on the other, and add wheat to the one pound side until it balances.  Three pounds, easy enough.  Four pounds,  put the one and three pound rocks on one side.  Etc., Etc., Etc.
512161
You haven't given me any information with which to solve this problem.  How big are the blocks?  How much do the blocks weigh?  How far apart are they?  How much resistance is there when I push the first block?   How fast is the first block going when it is pushed?  Without all this I cannot figure out how much momentum is lost between the time the first block is pushed and when it hits the second block or if the second block will have enough momentum to move at all or make the third block move if it contacts it.\n\n\nBy the way, you cannot solve this without knowing how much each block weighs, how much momentum it has to begin with, how much friction there is (to reduce momentum), and how far each block has to travel before it contacts the next one.    You need to have the variables to create a formula. \n\nGeeze, don't you hate it when idiots try to act smart????
610949
First convert to improper fraction which will be 25/8,\nThen divide by 10 which will be 5/16 lbs each.
138024
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)
412015
we have F= G*M*m/r^2  and F=weight\n\nwhen M=M and r= r/2 then F=G*M*m/ (r/2)^2\n\ntherefore F(new)= 4*F and hence new weight is 4 times.\n\nthis is due to change in the value of g(acceleration due to gravity)\n\nsince F=mg= G*M*m/r^2   and hence only g is effected\n\nand hence weight also changes mass remainig same.
420411
Part a)\n\nVf^2 = Vi^2 + 2 * a * d\na = (Vf^2 - Vi^2) / (2 * d) = -4,489/162 = -26.7 m/s^2\nm = 220*10^3 / 9.81 = 22,426 kg\nF = m*a = -598,777 N\n\nW = F * d = 50,297,245 J\n\nPart b)\n\nF (from above) = 598,777 N\n\nI hope this helps!
356702
This is too freaky. A problem that goes out of its way to EMPHASIZE weight not mass would give the weight in Newtons. Gram is a unit of mass, not weight.
519771
Of cause, help.\nthe force of gravity is mM/R^2 = 550 N\nThis is for R = 3400km\n\nIn case we take the object up for 12 km, the distance from the centre of the planet will be 3412 N.\n\nand its weight will be mM/(R+12km)^2 = Wn\nWe can find  Wn   just from these two equations:\nfrom the first equation we shall express \nmM = 550(N)xR^2\nthan  Wn = 550 (N) x R^2/(R+12)^2 = 550(N)x3400^2/3412^2 = \n550 (N) x 0.996483^2 = 550(N)x0.9929783 = 546.138 N\nThat is the answer
387044
an ore sample weighs 15N (newton) in air. when the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 10.80 N. find the volume and the total density of the sample.. \n\n\nApparent loss in weight=(15 -10.8 )N\nMass of water displaced=(1.50-1.08)kg\n                                        =0.42 kg\nVolume=0.42 l(since cm3= l)\n1 kg= 1 l (litre)\n\nTherefore,volume =0.42 litre\n                             =21/50000 m3\n\n\nsince density equals to mass over volume;\ndensity=0.42 (divide by) 21/50000\n           =100000 kg m-3\n\n\n\n\n\nThis is called Archimedes principle\n\nBelow is a table that will help you to change units from Wikipedia. \n\n\n\nName of unitSymbolDefinitionRelation to SI units\n\ncubic metre (SI unit)\nm³≡ 1 m × 1 m × 1 m\nlitre\nL≡ 1 dm³\nlambda\nλ≡ 1 mm³= 1 μL\ndrop (metric)\n ≡ 1/20 mL= 0.05 mL\nminim (Imperial)\nmin≡ 1/480 fl oz (Imp) = 1/60 fl dr (Imp)≈ 0.059 193 880 208 333 mL\nminim (U.S.)min≡ 1/480 US fl oz = 1/60 US fl dr= 0.061 611 519 921 875 mL\ndrop (U.S.) (alt)gtt≡ 1/456 US fl oz≈ 0.064 854 231 mL\ndrop (Imperial) (alt)gtt≡ 1/1824 gi (Imp)≈ 0.077 886 684 mL\ndrop (U.S.)gtt≡ 1/360 US fl oz= 0.082 148 693 229 1(6) mL\ndrop (medical) ≡ 1/12 mL= 0.08(3) mL\ndrop (Imperial)gtt≡ 1/288 fl oz (Imp)= 0.098 656 467 013 (8) mL\ndash (U.S.) ≡ 1/96 US fl oz = ½ US pinch= 0.308 057 599 609 375 mL\ndash (Imperial) ≡ 1/384 gi (Imp) = ½ pinch (Imp)= 0.369 961 751 302 08(3) mL\npinch (U.S.) ≡ 1/48 US fl oz = 1/8 US tsp= 0.616 115 199 218 75 mL\npinch (Imperial) ≡ 1/192 gi (Imp) = 1/8 tsp (Imp)= 0.739 923 502 604 1(6) mL\nfluid scruple (Imperial)\nfl s≡ 1/24 fl oz (Imp)= 1.183 877 604 1(6) mL\nfluid drachm (Imperial)\nfl dr≡ 1/8 fl oz (Imp)= 3.551 632 812 5 mL\nfluid dram (U.S.); U.S. fluidramfl dr≡ 1/8 US fl oz= 3.696 691 195 312 5 mL\nteaspoon (Canadian)\ntsp≡ 1/6 fl oz (Imp)≈ 4.735 510 416 667 mL\nteaspoon (U.S.)tsp≡ 1/6 US fl oz= 4.928 921 595 mL\nteaspoon (metric) ≡ 5 mL\nteaspoon (Imperial)tsp≡ 1/24 gi (Imp)= 5.919 388 020 8(3) mL\ndessertspoon (Imperial)\n ≡ 1/12 gi (Imp)= 11.838 776 041 (6) mL\ntablespoon (Canadian)\ntbsp≡ ½ fl oz (Imp)= 14.206 531 25 mL\ntablespoon (U.S.)tbsp≡ ½ US fl oz= 14.786 764 782 5 mL\ntablespoon (metric) ≡ 15 mL\ncubic inch\ncu in≡ 1 in³= 16.387 064 mL\ntablespoon (Imperial)tbsp≡ 5/8 fl oz (Imp)= 17.758 164 062 5 mL\npony ≡ 3/4 US fl oz= 22.180 147 171 875 mL\nfluid ounce (Imperial)\nfl oz (Imp)≡ 1/160 gal (Imp)= 28.413 062 5 mL\nfluid ounce (U.S.)fl oz (US)≡ 1/128 gal (US)= 29.573 529 562 5 mL\nshot ≡ 1 US fl oz= 29.573 529 562 5 mL\njigger ≡ 1 ½ US fl oz= 44.360 294 343 75 mL\ngill (U.S.)\ngi (US)≡ 4 US fl oz= 118.294 118 25 mL\ngill (Imperial); Noggingi (Imp); nog≡ 5 fl oz (Imp)= 142.065 312 5 mL\ncup (Canadian)\nc (CA)≡ 8 fl oz (Imp)= 227.3045 mL\ncup (U.S.)c (US)≡8 US fl oz ≡ 1/16 gal (US)= 236.588 236 5 mL\ncup (metric)c≡ 250 mL\nbreakfast cup ≡ 10 fl oz (Imp)= 284.130 625 mL\npint (U.S. fluid)\npt (US fl)≡ 1/8 gal (US)= 473.176 473 mL\npint (U.S. dry)pt (US dry)≡ 1/4 bu (US lvl)= 0.550 610 471 357 5 L\npint (Imperial)pt (Imp)≡ 1/8 gal (Imp)= 0.568 261 25 L\nfifth ≡ 1/5 US gal= 0.757 082 356 8 L\nquart (U.S. fluid)\nqt (US)≡ 1/4 gal (US fl)= 0.946 352 946 L\nquart (U.S. dry)qt (US)≡ 1/32 bu (US lvl) = ¼ gal (US dry)= 1.101 220 942 715 L\nquart (Imperial)qt (Imp)≡ ¼ gal (Imp)= 1.136 522 5 L\npottle; quartern ≡ ½ gal (Imp) = 80 fl oz (Imp)= 2.273 045 L\nboard-foot\nfbm≡ 144 cu in= 2.359 737 216 L\ngallon (U.S. fluid; Wine)\ngal (US)≡ 231 cu in= 3.785 411 784 L\ngallon (U.S. dry)gal (US)≡ 1/8 bu (US lvl)= 4.404 883 770 86 L\ngallon (Imperial)gal (Imp)≡= 4.546 09 L\nbeer gallonbeer gal≡ 282 cu in= 4.621 152 048 L\npeck (U.S. dry)\npk≡ ¼ US lvl bu= 8.809 767 541 72 L\npeck (Imperial)\npk≡ 2 gal (Imp)= 9.092 18 L\nbucket (Imperial)\nbkt≡ 4 gal (Imp)= 18.184 36 L\ntimber foot\n ≡ 1 cu ft= 28.316 846 592
50817
If you start with the times from A to B is t1 and the time from B to C is t2, then the distance from A to C is t1*10 + t2*50, and the total time is t1+t2, so the average speed is:\n\n    (t1*10 + t2*50)/(t1+t2)\n\nThat's the weighted mean of 10 and 50, with weights t1 and t2.\n\nIn general, to average a bunch of averages, you need to compute a weighted average.  For example, if you have ten rocks averaging one pound apiece, and I have five rocks averaging four pounds apiece, then to determine the average of all the rocks, you would take the weighted average of 1 (with weight ten) and 4 (with weight 5.)  That is:\n\n    (1*10 + 4*5) / (10+5) = 30/15 = 2
142442
A "stone" is also 14 pounds, but no one uses that unit of measurement anymore.  It was the "Imperial" British system.\n\nOne stone is 6.350 kilograms.
87
The answer is 9 stones (or maybe more). \n\nAfter adding a look-alike to 9 stones, there will be 10. Name them ABCDEFGHIJ.\n\n"ABC<DEF" means "the stones ABC as a group weigh less than the stones DEF."\n\nWeighing 1. Weigh ABC vs. DEF.\n\nIf ABC<DEF:\n--Weighing 2. Weigh ABC vs. GHI.\n----If ABC<GHI, the look-alike must be light and among ABC.\n----Weighing 3. Weigh A vs. B.\n------If A<B, the look-alike is A.\n------If A=B, the look-alike is C.\n------If A>B, the look-alike is B.\n----If ABC=GHI, the look-alike must be heavy and among DEF.\n----Weighing 3. Weigh D vs. E.\n------If D<E, the look-alike is E.\n------If D=E, the look-alike is F.\n------If D>E, the look-alike is D.\n\nIf ABC=DEF, the look-alike must be among GHIJ.\n--Weighing 2. Weigh G vs. H.\n----If G=H, the look-alike candidates are I and J.\n----Otherwise, the look-alike candidates are G and H.\n----Weighing 3. Weigh one of the two candidates against A.\n------If they differ in weight, the look-alike is that candidate, otherwise it is the other candidate.\n\nIf ABC>DEF:\n--Weighing 2. Weigh ABC vs. GHI.\n----If ABC>GHI, the look-alike must be heavy and among ABC.\n----Weighing 3. Weigh A vs. B.\n------If A<B, the look-alike is B.\n------If A=B, the look-alike is C.\n------If A>B, the look-alike is A.\n----If ABC=GHI, the look-alike must be light and among DEF.\n----Weighing 3. Weigh D vs. E.\n------If D<E, the look-alike is D.\n------If D=E, the look-alike is F.\n------If D>E, the look-alike is E.
660738
4 weighings. First weigh 6 and 6. If one is heavy you take the heavy side. 3 and 3. Pick the heavy side. 1 and 1, if equal it is the third.\n\nThe 4th weighing is one of the coins may be light and when you weigh 3 vs 3 they might be the same and you would have to pick the other pile of 6.
426497
F = GmM/r^2; where G is a constant, m and M are the two masses, and r is the distance between them.\n\nm + M = 4 kg; so m = 4 - M\n\nF = G(4 - M)M/(.25)^2 = 2.5*10^-10; where G = 6.67 × 10−11 N m^2 kg^−2 \n\nUse the above to solve for M, then m = 4 - M to find m.  \n\nAnd there you have it, you do the math.
361639
1 pound = 4.4475 newtons
364708
1. friction\n2. W=mg  , he has a mass of 100Kg. Weight is the force=newtons, g=pull of gravity N=100 x 9.8m/sec^\n98Newtons\n3. correct
383019
1)\nF = mg = kx\n8 * 9.8 = k * 0.1\nk = 784 N/m\n\n2)\nelastic PE = (1/2) * kx^2\nelastic PE = (1/2) * 784 * 0.4^2\nelastic PE = 62.72 N-m\n\n3)\nAt maximum height, all elastic PE stored in the spring becomes the gravitational PE in the stone.  Therefore, the gravitational PE equals...\n62.72 N-m \n\n4)\nelastic PE = gravitational PE = mgh\n62.72 = 8 * 9.8 * h\nh = 0.8 m = 80 cm (from release pt)