my girl saout troop is makeing an hour glass and need to fill 1 liter bottle with sand. the box of sand that i got is 3 pounds. I need to know how many more boxes to buy.
138024
Measure the weight of 1 cup of sand and figure it from there.\n\n :-)
308699
If I'm reading your half pound rule right, the weights (with the metal included) would be:\n1, 3, 9, 27, and 81.\nIn other words, you should put in the following amounts of sand:\n0.5, 2.5, 8.5, 26.5, and 80.5.\n\nHere is the reason for it, though I'm not really sure how to write things like summation and subscripts here.  I will write a subscript in a square bracket.\n\nObviously we can use all possible sums of these weights to balance out our desired weight on the other side of the scale - weights of 1, 3, and 7 give values of 1, 3, 7, 1+3=7, 1+7=8, 1+3+7=11.  But we can also use all possible differences.  For example, if you had an 8 pound weight and a 3 pound weight, you could measure 5 pounds by putting the 3 and 8 pound weights on opposite sides.  Then you could measure some 5 pound object by putting it on the 3 pound side.\n\nLet x[0]...x[n] be the value of n weights (metal included).  In this case n=4 because we go from 0-4.  For simplicity in writing, define y[i] = x[0]+...+x[i] for any i.  The maximum amount we can balance on the scale using the weights x[0]...x[n] is y[n], by putting all weights on one side.  We need to find x[0]...x[n] such that for a given m up to y[n], we can add or subtract members of {x[0]...x[n]} to reach m.\n\nNow we prove this by induction.\nWe propose that x[n+1] = 2*y[n]+1 satisfies our desired property.\nDefine x[0]=1.  Then y[0]=1, so x[1]=2*1+1=3.  This satisfies our property because y[1]=3+1=4.  Clearly 1 and 3 can be weighed.  3-1=2 and 3+1=4, so we can reach all numbers up to 4=y[1].\nNow suppose you have x[0]...x[n] with this property.  So any number up to y[n] can be made by sums and differences of x[0]...x[n].\nNow we include x[n+1]=2*y[n]+1.\nYou can still get from 1 to y[n] by just not using weight #n+1, since the property holds true for x[n].\nBy putting weight #n+1 on the lighter side of any of the y[n] combinations, you can get\nall values from x[n+1] down to x[n+1]-y[n] = 2*y[n]+1-y[n] = y[n]+1.\nBy putting weight #n+1 on the heavier side of any of the y[n] combinations, you can get\nall values from x[n+1] up to x[n+1]+y[n] = x[n+1]+x[n]+x[n-1]+...+x[0] = y[n+1].\nSo with our choice of x[n+1] we can get all values from 1 to y[n+1].\nIn other words, our property holds true for x[n+1].\nBy induction, the property holds true for all x[n].\nUsing the formula to calculate the first 5 values, you get 1, 3, 9, 27, 81 - the powers of 3.
110011
let' see  one ton is 2000 pounds and there are 16 ounces to the pound so 2000 x 16 = ???? c'mon, you can do it, I have faith in you.
367876
(1) a total of (2+3)x=435\nso x= 435/5 =87\n\nso boys = 2x =174\nn girls = 3x = 261\n\n\n(2) a total of(5+3+1)x=27\nso x=27/9 =3\n\nso gravel = 5x = 15\n     sand = 3x = 9\n     cement = 1x = 3\n\n\nhave a nice day...
247751
Your local supermarket gets deliveries at least once a week. Just ask one of their employees on what day and at what time their deliveries arrive. They won't usually save boxes for you, but you can walk in and get as many as you like before they cut them down for recycling.\n\n\nOffice Depot has great boxes for larger items. They receive their reams of paper in sturdy boxes with lids, and these are always clean. Our local Office Depot receives paper deliveries twice a week. The store manager even agreed to hold the boxes for a few hours. We just need to go in on the correct day. Since these are such handy boxes, you may have to be the first to get to the store to beat others with the same idea.\n\n\nCheck the free ads in the Recycler. Companies go out of business all the time, and they would usually prefer to give their supply of boxes away than to pay someone to haul them away. We found a vitamin store that was closing down and got hundreds of small and tiny boxes. These are not only great for mailing small items, but they are perfect for double boxing a set of smaller breakables.\n\n\nLiquor stores have the sturdiest boxes since bottles of liquor and wine are so heavy. If you sell bottles, these may even come with cardboard slots in them already to keep bottles from hitting each other when you send out several to one buyer.\n\n\nDumpsters behind stores sometimes have good empty boxes. Sometimes you are lucky enough to find the boxes piled outside the dumpster. Either way, you should check them to be sure that they are clean and dry (a box will be weakened forever once it gets wet).\n\n\nThe most obvious is for sellers to BUY on www.Ebay.com. Your package will come in a box, and many will come full of styrofoam peanuts or bubblewrap also. Just be sure that the box is sturdy before using it again. And, don't use bubblewrap as padding for delicate items if the bubbles are popped.\n\n\nGet inventive, you may find even more good suppliers of boxes.
437825
Piston 1 (in the figure linked) has a diameter of 3.0 in. Piston 2 has a diammeter of 1.5 in. In the absence of friction, determine the force necessary to support the 500 lb weight.\n\nAnswer: The force required at piston 1 = 125 lb. This is because the pressure in piston 1 = pressure in piston 2\ni.e. 125 /one quarter area = 500/ area.\nArea of piston 2 = pi x [3.0/2]^2  sq. in.       Area of piston 1 = pi x [1.5/2]^2  sq. in\nUse the principle of moments to find force F:\n\nie. 125 lb x 2.0   =   F x (2 + 10)\nHence F = [125 x 2] / (2 + 10)\n              =  125 / 6    lb force
532011
1.3-.9=.4kg for half a bottle of oil no bottle so 1 bottle of oil with no bottle should wiegh .8kg so the bottle should wiegh.5kg
518259
You expect to end up with 1L of 120 proof hootch.  Therefor you have an ending percentage of 60% alcohol and 40% water.\nSo the volume of alcohol is .6 L and water is .4 L.\nLet X be the quantity of Bacardi 151 and Y be the quantity of cheap rum.\nYou are then working to figure out X liters * (.755 Alcohol ) and Y liters * (.4 alcohol ) = .6 L alcohol \nIn algebra-speak, that is X (.755) + Y(.4) = .6\nYou also know that X + Y = 1 liter.  Taking this equation and solving for Y we have Y = 1 - X.\nWe can now substitute all occurances of Y in the equation with (1-X)  Therefor\nX (.755) + (1-X)(.4) = .6 and we have a single variable to solve for.\n.755X + .4  - .4X = .6 liters\n.355X = .2 liters\nX = .2/.355 \nX = .56338 liters of Bacardi 151\nThere for Y = .43662 liters of cheap rum.\n--------------------\nAdditional info.\n--------------------\nOK, Shirt   Sorry you got lost.  But I skipped a step.\nIn my solution, this is what happened between\n.755X + .4  - .4X = .6 liters and .355X = .2 liters\n\n.755X + .4  - .4X = .6  \nSubtract .4 from both sides\n.755X  - .4X = .2\nSubtract .4X from .755X\n.355X = .2\n\nI like steph's solution too.\nHe doesnt show all the steps either when solving for A.\n(0.755A + 0.4(1-A))/1 = 0.6.\nGet rid of the parenthesis by multiplying .4 \n(0.755A + 0.4 - 0.4A)/1 = 0.6.\nDivision by 1 doesn't change anything, So you can ditch the parenthesis.\n0.755A + 0.4 - 0.4A = 0.6.\nSubtract 0.4A from 0.755A\n.355A + 0.4 = 0.6\n.355A = 0.2\n\nI assumed you were using rum because Bacardi is the only one that make a 151 proof liquor.   Everclear is 190 or 95%.  That will change your quantities alot.  Use the same equation, but replace the 0.755 with 0.90.\nI made my own version of Goldshlagers which is more than 100 proof.  As you are doing, I too used Everclear to raise the proof.  It ended up tasting better than the goldshlagers.  Disolve a dozen Atomic Fireballs into a liter of vodka and add everclear.
94433
Get a GOOD mortar and pestle.  They are online, but you can find them at WM and any good kitchen store. If you are only grinding small amounts, get a smaller one.  I have a huge lava rock one and it's too heavy to lift easily for dumping out what's ground, and it's scary moving it over my counter!
493290
Too easy as written- he can do it with ONLY ONE 40 litre can!\nFill the 5\ndump into the 4, so theres 1 left in the 5\ndump the 1 from the 5 into the 40\ndiscard the milk in the 4 and repeat the process, dumping the 1 into the 40.\nnow theres 2 in the 40.\ndiscard the milk in the 4\nfill the 5\ndump into the 4, so theres 1 left in the 5.\ndiscard the milk in the 4 & dump the 1 from the 5 into the 4\nfill the 5 and dump into the 4, so theres 2 left in the 5.\ndiscard the milk in the 4 and dump the 2 from the 40 into the 4.
413026
A) a grain of sand.\nThink about it this way: one thousand milligrams= 1 gram\n                                          454 grams= 1 pound\nso 454 THOUSAND miligrams=1 pound\n\nhow much would each of those items weigh if you had 454 thousand of them?
513634
First question\n\n1000m = 1km\n60s = 1 min\n60 min = 1 hr\n8m/s * 60 (to convert it to mins) = 480m/min * 60 (to convert it to hrs) = 28 800 m/hr / 1000 (to convert it to km/hr) = 28.8 km/hr\nTherefore she can skate 28.8 km in an hour.\n\nSecond question\na) The question is asking you how many times bigger the volume is. Volume = l *w *h. The first box is 2x4x2 which gives a volume of 16 units^3. If you double it the box becomes 4x8x4 which gives a volume of 128 units^3\n\nTo find how many times more the second box will hold, divide the second volume by the first. 128/16= 8 times more candy.\n\nb) The question is asking you how much bigger is the surface area of the box. There are 6 faces to the box, each with an area that has to be added together to get the total surface area. If the box is 2x4x2 that means there are 4 faces of 2x2 and 2 faces of 4x2. So you would find the area of the first face (2x2 = 4), multiply it by the number of identical faces (4 * 4 = 16) and then take the area of the second face (4x2=8) and multiply it by the number of identical faces (8 * 2 = 16). That leaves you with two answers, 16 and 16. Add them together to get your surface area of 32 units^2.\n\nRepeat the same steps with the dimensions doubled (4x8x4) and you'll get a surface area of 128 units^2. Divide the second surface area by the first (128/32) and you'll see you need 4 times more cardboard for the second box.
477757
First solve for the force of friction between the block and the table using the formula,\n\nf=uR, where f is the force of friction, u the coefficient of static friction, and R is the force normal to the table. R therefore is the weight of the block equal to 31g.\n\nf=0.425*31*9.8\n =129.1N\n\nThe weight of the sand added to the bucket plus the weight of the bucket itself should be just slightly equal to 129.1N so that the system will begin to move. At that point, however, the block will begin to accelerate because the force of friction acting on it will no longer be static friction but kinetic friction, which is lower in magnitude. Thus there will be a net force equal to 129.1 minus the force of kinetic friction on the block. Thus,\n\nNet force, F=129.1-0.320*31*9.8\n                 =129.1-97.2\n                 =31.9N\n\nF=ma, where F is our net force of 31.9N, m the total mass of block plus mass of sand and bucket, and a is the acceleration of the system.\n\nMass of sand and bucket=129.1/9.8\n                                    =13.2kg\nSubstitute known values:\n\n31.9=(31+13.2)a\na=31.9/44.2\n  =0.72m/s^2
22096
3 on one side, 3 on the other.\n\nIF they all balance, then the 2 left over has the lighter stone, just weigh and find out the lighter stone.\n\nIF they do NOT balance, then take the 3 that's lighter, take TWO stones from that group and weigh.\n\nIF they are balanced, the remaining stone from the 3 is the lighter stone.\n\nIf there's one that's lighter, then that's the stone you want.\n\nOnly two weighings in either case!
554747
Let's see....\n1 liter=1000 ml\n1000ml=4 cups water\n1 cup=8 oz\n 32 oz=1 liter\n32x3=96 oz\n\nSpace out the water. Try dividing it into three 1 liter bottles and carry one with you while at work, school,...Good luck cuz that's alot of water....
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.
190545
u shud check this site becoz if u know the density and type of milk and also the temperature at which the milk is it is easy to calculate the wt of one litre of milk in kgs