A man in a boat on a reservoir carries a large gold sphere. An observer is watching him from the shore. He drops the sphere to the bottom of the lake; it sinks to the bottom. After it sinks to the bottom, is the water level of the reservoir the same as, higher than, or lower than it was when the statue was in the boat?\n\n(source: John Shuttleworth)
120922
pretty sure it wouldnt make a BIG difference by ANY means, but it would make the water higher by a smigen....  i guess?  lol
494788
According to Boyles Law, P1V1 = P2V2\nlet p = rou, the density of pure water (assuming tt the lake is pure)[1000 kg/m^3]\nTake g to be 10N/kg, and atmospheric pressure [P(atm)] to be 1.013x10^5 Pa \n\n[P(pgh) + P(atm)] x 1.55 = P(atm) x New Volume\n(1000x10x120 + 1.013x10^5) x 1.55 = 1.013x10^5 x new volume.\ntherefore, \nnew volume = 1210130 x 1.55 / 10130\n                     = 185.16cm^3\n\nHope this helps =)\n\n**The Boyles Law which i stated above states that the volume of gas in an enclosed area is inversely proportionate to its pressure. Initially under the water, the pressure on the bubble is taken to be atmospheric pressure + the weight of the column of water above it. As it rises, the pressure gradually decreases to only atmospheric pressure when it is at the surface. As such its volume increases proportionately, and tt's how i found out the ans.
214272
During winter the water at the surface of lake or pond gets cooled.\n\nDue to cooling density of water increases; denser water\ngoes down, while the relatively hot water comes up. This is convection.\n\nBut water has a peculiar property that below 4 degree C it becomes less dense.\n\nTherefore when the surface temperature is 4degree and below, the water at the surface cannot go down. Hence convection of water is stopped. \n\nThe temperature increases relative to water surface.  If the temperature of the water surface reaches freezing temperature, ice is formed on the surface and the freezing takes place from top to bottom.\n\nThus lives of water living creatures are saved from freezing.
543727
Some people must listen to good logic and not blindly go through with the Mathematics. If the water rises, the boat will too. Boats float on water.
21971
It definitely is floating around.  I saw it for the first time probably two years ago and was really surprised to see it.  I believe it was on one of the discovery channels but I'm not sure.  Anyway, if I saw it on TV it's available on the internet.  Its a scene where he's carrying the kids through water, one under each arm, and a helicopter that was wired up above them leaned and fell on top of them.
182799
Not his original, but the image in the middle of the first link I'm listing is pretty close to Descartes' from De Meteorologica.  The second source has several modern takes on the image as well as a nice physical explanation of what is going on.
289362
The water level will go down when you throw the body in.  Try it in your sink.  Fill your sink with water and float a bowl (boat) in it.  Then, fill the bowl with silver wear (body and rocks) or anything more dense than water.  Mark where the water level is on the side of the sink.  Now, take the silver wear out, and put it in the water so it falls to the bottom of the sink.  You will notice that the water level went down.  When the silver was in the bowl, it displaced much more water than when it was at the bottom of the sink.  Therefore, the body will displace more water when it is inside the boat than when it falls into the water.
343761
The density of gold is 19.3*10^3 kg/m^3. So to calculate the statue's volume\n115 kg * (1 m^3/19.3*10^3 kg) = .00596 m^3. That mass of that volume of water \n.00596 m^3 * (1 kg/1*10^3 m^3) = 5.96 kg.\n\nThe weight of the statue: 115 kg*9.8 m/s^2 = 1127 Newtons\nThe weight of the 5.96 kg water: 5.96 kg*9.8 m/s^2 = 58.4 Newtons. This is the amount of bouyancy provided by the water.\n\nSo the tension in the cable while the statue is still under water is 1127 - 58.4 Newtons =1069 Newtons.\nAfter the statue is out of the water, tension is 1127 Newtons.
354185
Forget about the mass of the water.  When it was all over, the water had not absorbed or given up any heat, so all of the net exchange happened between the copper and the aluminum.  I am not going to look anything up for you, but you need to know the "specific heat" for each of the metals.  Since they all wound up at the 25C temperature of the water, and the heat gain of the aluminum equaled the heat loss of the copper, find out what quantity of aluminum would absorb the heat given up by 210 g of copper as it changed from 80C to 25C.  Set the problem up like this:\nSpHtCu x 210 g x (80C-25C) = SpHtAl x Mal x (25C-5C) where\nSpHtCu = the specific heat of copper, \nSpHtAl =  the specific heat of aluminum and\nMal = the mass of the aluminum block\nDivide both side of the equation by SpHtAl and by 20C and you will have a number on the left side of the = sign and just Mal on the right side of the = sign.  At that point, you have your answer.
192916
If the pressure is 0.76m of Hg, it is equivalent to 10.3 m of water. There is a vacuum in the tube above this water height.\n\nHowever, by “vacuum” we mean inside this space, the pressure is low. There will be vapour of water which will have some pressure.\n\nEven in mercury barometer, there will be mercury vapour inside the “Torricelli’s vacuum”.  \n\nThis is the reason that ordinary pump which sucks water from a well or sump is not able to lift water to a height more than 10m from the surface of the water.
130730
No monster in that lake....but a nice tourist attraction\n\nhttp://ourworld.compuserve.com/homepages/lesj/ness.htm
217390
Everyone says that The surface tension causes the bubble to be spherical. But how? and Why? \n\nLets explain surface tension first- \nIn an air-water interface, the molecules that are close to the surface are attracted by fewer molecules of the liquid than the ones that are NOT close to the surface. The former thus, has less attractive potential energy than the latter, right?\n\nNow, for any matter (solid or liquid), attractive potential energy results in stability & repulsive energy results in instablity (common sense). \n\nSo, it is energetically unfavourable to have a surface for a liquid (or solid) as it results in loss of attractive potential energy.\n\nNow, for that reason, a bubble wants to have the lowest possible surface area to remain stable. Now, using some high level mathematics, someone found out that a sphere has the lowest possible surface area for a given volume. (it is better than a cube atleast).\n\nHence why we can see that the water bubbles and drops tend to be spherical. Due to the effect of gravity, although the drops appear oval in shape but, in the absense of gravity, all the liguid-air surfaces tend to be spherical.\n\nI hope it helped.\nHave a nice day!
558995
The symbol on Lana Lang's back is the symbol for transference. Water is the flowing element, thus qualifying as a link for transference (in this case it was for Countess Isabel to be able to transfer her soul into Lana's body). \n\nI believe the symbol shows up again when Lionel and Clark switch bodies? I don't remember exactly but I believe it's on the stone Lionel used. I'll have to go back and watch the episode which is oddly called Transference.\n\nI'm a walking textbook when it comes to Smallville, so if you need further help contact me: bubblegumbrainiac@sbcglobal.net....\n\nGood luck!
515463
All decent explanations so far; I hope I can add to what has already been posted.\n\nLiquid and Ice most definitely can exist at the same temperature and Ice can be and often does get colder than 0C.\n \nFirst of all the Triple Point of water is the pressure and temperature at which Ice, Liquid Water and Water Vapor exist at the same time in equilibrium.  Vapor is the right word, not steam although you can argue that the only time that Water Vapor and Steam are the same thing is at a very low pressure.  The Triple point of water only happens at a near perfect vacuum, 0.0063 ATM to be precise.\n\nAdding Salt to water will lower the freezing point, but then it isn't water anymore, it's salt water.  Even so, Liquid Salt Water and Ice Salt Water can also exist at the same place and time at the same temperature, slightly below 0C.  But let's stick to water.\n\nIce can easily be lower than 0 degrees.  Consider a bucket of water, liquid water sitting outside and the air temperature is -10C.  Little by Little, as the liquid reaches 0.0098 C, liquid starts to become ice.  The ice crystals form in the bucket, around the walls of the bucket and on the surface first.  The air is still -10C so the bucket, the liquid and the ice will keep getting colder.  The ice on the surface will drop to slightly below 0C but very slowly since it is still absorbing the heat of the not-yet-frozen water.  Eventually, the entire bucket of water will be frozen and will be just below 0C.  First the walls of the bucket will reach -10C (same as the air tempertaure).  The ice near the walls of the bucket and on the surface will reach -10C first, continually absorbing the heat energy from the relatively warmer ice near the center of the bucket.  Until finally, the entire bucket of ice is steady at -10C.  That is until the sun comes out or the day gets warmer.\n\n\nAnd by the Way, dry ice is not water, it's frozen CO2 (Carbon Dioxide)\n\nOne more thing, at a near perfect vacuum, you can actually cool liquid water to less than 0C without it freezing.  Never tried that but I did accidently freeze a beer last night...
536971
The weights of the liquids inside 2 branches must be equal, therefore:\n\na) m_water + m_oil = m_mercury (the amount of mercury that raised in the other branch to keep the balance)\n\nm_water = D_water * V_water = D_water * h_water * Area\nm_oil = D_oil * V_oil = D_oil * h_oil * Area\nm_mercury = D_mercury * V_mercury = D_mercury * h_mercury * Area\n\nSubstitute this in the balance of weights above, we have\nD_water * h_water * Area + D_oil * h_oil * Area  = D_mercury * h_mercury * Area\n<-> h_mercury = (D_water * h_water + D_oil * h_oil) / D_mercury\n\nb) To bring mercury's surface two the same level again, the weight of the to-be-added water must be the same with the weight of both oil and water in the other branch.\n\nYou can do the same as above to find:\nh_water2 = (D_water * h_water + D_oil * h_oil) / D_water
599919
I USED to go to the RIVER and I noticed that during the WINTER, when the Hydo-electric Establishment let the river down, so's people could fix-up their piers, the bushe's that used to be under the river water would start growing.  Interesting, cause all most of SPRING and SUMMER, they were under the water!!!!  Just something to think about, can't explain how they stayed alive, take a Botanist I guess!!!! Hope this helps........
66315
Seems like a pretty interesting riddle. I am so sorry that it didn't get the response it deserved. I think you should have placed it in the "Jokes and Riddles" sub-category under the category "Entertainment & Music". It would have garnered an immense response there.\n\nAnyways, let me try here-\n4 is golden,eh? I think here 4 somehow refers to "silence"\nNow, 5 is silence. Ummm.... what should be 5? I give up.\n\nGood Riddle. Try posting it in the "Jokes & Riddles" category.