How much water does a 100 year old tree drink / hold and how many sappling trees equal that amount?
487526
Sounds like a Math/Science type question to me. Why are you posting it here?\n\nBesides, I have the the slightest idea.
225446
First, we must assume that no one gets out alive.  We won't allow enough water to return from the desert after saving the flower.  \n\nWater consumption rate is p((q+.2)n/p+1)/20 where \np - number of people still going\nq - amount of water in each bottle\nn - number of bottles \nSolving this for 10 people with 10 bottles you get a rate of 2.1 L/km.  If instead you load all 10 bottles on one person, you get 1.65 L/km so it's best to do that. If she drops a bottle whenever it's empty she can cover 138.6km. Calculate Int(20dv(30+2+1),0,3)+ Int(20dv(27+1.8+1),0,3)+ Int(20dv(24+1.6+1),0,3)+ Int(20dv(21+1.4+1),0,3)+ Int(20dv(18+1.2+1),0,3)+ Int(20dv(15+1.0+1),0,3)+ Int(20dv(12+0.8+1),0,3)+ Int(20dv(9+0.6+1),0,3)+ Int(20dv(6+0.4+1),0,3)+ Int(20dv(3+.2+1),0,3)\n\nIf each person is only allowed to carry one bottle, then they need to ditch people as quickly as possible. When the initial 10 people have 2.7L left, one person should top up the others' bottles then sit down while the nine carry on with full bottles.  They continue in this manner leaving people behind whenever one can bring the others all back to 3L.  The last person accumulates 56.04km before running out of water. \nCalculate 10*Int(20dv(3+0.2+1),0,.3)+ 9*Int(20dv(3+0.2+1),0,.33)+ 8*Int(20dv(3+0.2+1),0,.37)+ 7*Int(20dv(3+0.2+1),0,.42)+ 6*Int(20dv(3+0.2+1),0,.5)+ 5*Int(20dv(3+0.2+1),0,.6)+ 4*Int(20dv(3+0.2+1),0,.75)+ 3*Int(20dv(3+0.2+1),0,1)+ 2*Int(20dv(3+0.2+1),0,1.5)+ 1*Int(20dv(3+0.2+1),0,3)
213685
see this:\nhttp://answers.yahoo.com/question/index;_ylt=Asf07LGr3t0fySCHqllplZDsy6IX?qid=1006032408384\n\n\nA team of 10 women is to go through desert to save a flower. We know that the more the weight of the equipments more water is needed to travel certain distance. The formula for water consumption is dx = 20dv/(m+1), where dx is differential displacement in kilometers, m is the weight of load on person in kilograms and dv is the amount of water (in litters) needed to travel dx. So a walker should continuously drink water for any small movement. They have ten bottles containing 3 litters of water and each empty bottle weights 0.2 kilogram. The flower need one drop of water to be saved. What is the minimum distance for flower from the base, that guaranty the team will not be successful? (Density of water is 1000 kilogram per cubic meter)
278197
none, nil, nada, zip\n\nalcohol dehydrates and plants need water to grow.
757835
Sap is the common word, but there are two different fluid transporting systems in trees: Xylem and phloem.\n\nThe xylem is the principle tissue and its liquid is called the xylem fluid (sap).
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
340687
http://www.rain-tree.com/facts.htm\nread it and u will get a answer
287974
At Least 64 OZ. (equvalent to 8 glasses of water a day) if not more their is a hydration calculator on the weather channels website they base it on location weight and how much activity you do a day hope this helps!! the link below will take you to the hydration calculator
230223
AGE OF THE TREE the rings inside 1 per year \nhurbs  dont know
470626
Well when I was at camp as a kid what we did was wrap a bag around a tree limb.  You know like the big garbage bags.  during a day a good amount of water collects.  Another way is to place a plastic sheet over a thing of salt water....during the day the salt water evaporates onto the plastic and runs down.  That's drinkable too....need to make sure to catch the run off though.
354060
0.2 gal. or 25.6 fluid oz.
302446
I agree with the first answer:  Here is something that might help you with maths:  When you have a problem such as this make a 'rough estimate' of the answer.  In your problem almost all the tomatoes were water, nearly 100kg, the water was reduced by 1% - that's very little, it being 1/100 th. So you nearly have the same weight as you started with.  It cannot be 50kg as that would have meant that half the water 50% would have evaporated.  I hope you follow. Go back to my first point, the rough estimate should have been 'nearly 100kg but a little less'.  I hope I have made sense and that this gives you a good idea as to how you should proceed in the future.  Always work out a 'nearly right' answer first.....
337551
It depends.  Wood floats on water, as does ice.  Rocks sink.
56978
water, water and more water
58837
Using 6,525,486,603 as the world population, (http://www.census.gov/ipc/www/popclockworld.html\n\nand 150 lbs as an average mass per person (my guess)\nyou get total mass of people on the planet of:\n489,411,495 tons. This would equal the weight of 122 billion gallons of fresh water. \n\nLake Mead, near Las Vegas, Nevada, holds about 9,300 billion gallons of water. The entire human population would fit into only about 1.3% of the lake. \n\nWithout people, the mass of the earth would be changed by a very very insignificant amount.
1120275
8 to 10 glasses of water a day. Roughly 1/2 gallon
488001
It`s about 45-50 % !\nBut If you realy means world as the earth !\nMay be there are more trees on the other unknown live planets !\nYou know 50 % is the percentage of 1990 and i guess that now in 2006 it should be 45 % , and sadly i should say if this tree termination method continues 1n 2025 you only can find tree in the museums !\nTry to hold them !\nTry to grow the trees !\nThey are our fathers !