suppose u have a 100kg tomato having 99% water & 1% solid part. After one day the water is reduced to 98% .Now what is the weight of tomatos. I give the answer 50kg ,plz let me know am i right & if not plz explain the right answer\nThanks in Advance
126179
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.....
471613
um, i may be wrong(...)\n\n1. i dont get what the first one is saying...like what it means(?)\n2. well, first you'd tell the whole glass to STEP AWAY from the glass and orger any goofy silly kid to please get out or else they get seriously cut!\n    then with maybe the teachers help(or actually just ask the teacher to do it) with a paper or anything else thats hard and prevents you getting hurt..)\netc...continue from there...\n3.(water is a liquid).....(? is a solid). the mass.....OH YEASH!!!!\nwith liquids you pour a certain amount in a graduating cylinder and y-oh sorry ^^; that to measure something else...um....\nim sooo sorry !!! of course....\nMASS is measured by (with solids) with a triple beam balance.\ni guess one would measure mass with liquids like this: you er...put it in a gasllon,.....mass that gallon....mass the water(with the gallon) and when you get the mass you subtract it from the mass of the gallon(without the water)....\nbut then that way there'd be no difference......\n\n\nim horrible...im really sorry....
188215
In your example, the concentration of the solution weight/weight would be (presuming the 100 ml is water, thus 100g) would be 10.7% (12 g solute/112 g total solution).
676982
I think that's dalton's law.  Whoever came up with it, the pressure is additive.  So convert so everything is in the same units, multiply by the percent, add, and then compare to the answers.  I bet it's more fun than you predict.  Whooo!  Conversion factors!  I don't want to take that kind of enjoyment away, so get cracking!
79985
somethings wrong here dude. you cant get something more concentrated. like just think about it, the only way you could make it more concentrated is to take otu water from it, which isnt what youre supposed to do here.\ni think maybe you got the question backwards. from 100mL of 0.92M NaCl, make 0.39M solution. it doesnt matter, its same thing. which is \n0.92 x 0.1 = 0.39 x v2      \nv2 = 236mL. so basically you take your 100mL and add 136mL of water to it  so the new volume is 236mL. and the new concentration would be 0.39M
148007
hey man, dont worry ill help. first of all our given:\nh=280m\nsg=1.03\ngc=9.7m/s^2\n\nreq'd Pressure(absolute)\n----------------------------------------------\nsol'n:\nfor DENSITY of subs:\nformula\nSpecific Gravity=density of subs/density of water\ndensity of subs=1.03 x 1000(density for water)\ndensity of subs=1030 kg/m^3\n\nFOR PRESSURE GAGE:\nP(gage)=Vh\nP(gage)=density x gravitational constant x height\nP(gage)=1030 x 9.7 x 280\nP(gage)=2,797,480 Pa\n\nFOR PRESSURE ABSOLUTE:\nP(abs)=P(gage) - P(atm)\n\nwhy minus? bec. the submarine is pressurized, meaning its vacuumed\n\nP(abs)=2,797,480 Pa - 1atm(101.325x10^3 Pa)\nP(abs)=2,696,155 Pa\n\nconvert Pa to Kpa\n=2,696,480 /1000\n=2,696.155 Kpa\n\nconvert to bar\n=2696.155kPa x (1.01325 bar / 101.325 kPa)\n=26.96155 bar\n\ndont mind the others, they dont even know how to answer this simple question. They just want to earn points!
415050
You haven't included enough information to have anyone else answer this question.  Specifically missing is the density of your sand.\nTo help you as much as possible, you have to calculate the quantity of sand that is equivalent to 900 ml of water (900 grams).\n\nGood luck.
396569
When I was in 9th grade we did a plant project. I bought two small plants that were the same. One of the plants I left it in the sun watered it everyday and took good care of it. The other we put it in a dark place and also watered it everyday and also took good care of it. I did it for two weeks. Everyday I would make an observation of the lenght of the plant and the leaves it dropped and wrote it down. Then on a posterboard I did 4 large graphs. Two of the graphs were the length it grew each day for each plant. And the other two I graphed how many leaves would fall out of each plant. And when I presented it in front of the class I brought both plants and my conclusion of the project was the affect the sunlight has on a plant. It really hopes that works. By the way I got an A+ on it.
93094
Actually I had something to help you but x=0 which makes it pretty difficult.\n\nI guess you could try to talk about tomatoes or something.  Say Tom (X) has 1/2 the amount of tomatoes that Jill has, plus four.  Or in other words, if you take Jill's amount of tomatoes, multiply it by two, and take Tom's amount and multiply it by three, there total amount of tomatoes will equal 12.  How many tomatoes does Jill have?  How many does Tom have?
473376
There is a buoyancy force on the ball, and an equal reaction on the water bucket. Let B be the ball's weight, so buoyancy is B/s, where s is the unit-less ratio of the ball density to water density. Equilibrium requires that \n\nW + (B/s) = B - (B/s)\n\nHence B = W/(1 - (2/s)).\n\n\nPS: kaksi_guy says it doesn't matter that the ball is inside the bucket. He doesn't say why, and I think he's wrong.
387314
Let x = weight of crate, and y = weight of grapefruit.\n\nYour equations are:\n\nx + 36y = 4 kg  ← the weight of the crate to begin with\n\nx + 24y = 3 kg ← the weight after 12 grapefruit are removed\n\nSo, subtract the second equation from the first to solve for y:\n\nx + 36y = 4 kg\nx + 24y = 3 kg\n------------------------\n0x + 12y = 1 kg\n\ny = 1/12 kg\n\nNow substitute this value for y into either of your original two equations, to solve for x:\n\nx + 36y = 4 kg\n\nx + 36(1/12 kg) = 4 kg\n\nx + 3 kg = 4 kg\n\nx = 1 kg\n\n~ ~ ~ ~ ~ \n\nSo the crate weighs x = 1 kg\nAnd each grapefruit weighs y = 1/12 kg\n\nHope that helped!\n\n~ ♥ ~
439200
Tom baked 4 batches of fairy cakes. Each batch consisted of x cakes. He put 3 cherries on top of each cake in the first batch, but couldn't resist taking off and eating 5 of them. He did the same again with the second batch, then the third, then the fourth! Meanwhile, Sarah baked 5 batches of fairy cakes, again with x cakes in each batch. She put 2 cherries on each cake and an extra two on the final one. She repeated this procedure with each of the batches. She managed to resist eating any at all! Tom and Sarah discovered that Tom's cakes had exactly the same number of cherries in total as Sarah's. Set up an equation for this.\n\nThen: \n\nNumber of cherries on Tom's cakes = 4(3x-5)\nNumber of cherries on Sarah's cakes=5(2x+4)\n\nBut number of cherries is equal so:\n\n4(3x-5)=5(2x+4)\n\nPhew!! I think that works!! Lol!
16730
Let the mass of the girl be m\nLet her net volume be v\ndensity in air = d\n\nSince 97% of her body is beneath water, so volume equal to 97m/100 is displaced and weight of 97m/100 mass is balanced by buoyant force. \n\nSo, here mass density remains the same = d = m/v\n\nBut her weight density = (mg- [97mg/100])/v\n                       =  3mg/100v\n                       =  3% of weight density in air
387724
I have actually sort of answered this question already.\n\n1:10 dilution of the first step will give you 1/10th the amount of cells = 50,000 cells\n\nThe next dilution will get you 1/10th again = 5,000 cells.\n\nThanks for the 20 points... by picking the Best Answer please.
512376
It will vary greatly depending on how the vegetable was grown and harvested.  The best way to find out (perchance) would be to weigh it, then dehydrate it and weigh it again, noting the difference.
361914
Its a trick Q...\n1. If floating on Surface, then 0 (since ur ballon is massless, although theoretically, nothing is massless..)\n2. If forced to stay inside water, then it would be equal to the weight of water displaced (i think ur Q is phrased wrongly, it is 1000cm3 displaced water ( or else the answer depends on the volume of the balloon).. Or simply, it is equal to the force used to keep the balloon inside water.
108175
Let H be the nunber of hens. Assume rate of food consumption per hen is equal for all hens. Then, if X days feed left at decision point and r is feed consumption per hen per day:\n\nSell: (H-75)*r*(X+20) is feed left\n\nBuy:  (H+100)*r*(X-15) is feed left\n\nSo (H-75)(X+20)= (H+100)(X-15)\n\nHe has to know how much feed he has to say what he says. So if we know X we can find H