if  the bucket of a tipping-bucket rain gauge has emptied 10 times how much rain has fallen? explain?
539633
Often it tips for every tenth of inch of rain, but this value varies and sometimes it can be adjusted.\nIf this is the case (tipping for every 1/10 inch of rain), your answer is 1 inch.
932537
The volume of rainfall can be calculated as hectare-mm or acre-inches in an area or watershed.\n\nThe equation is:\nrainfall volume = hectares in watershed x mm of rain\nrainfall volume = acres in watershed x inches of rain
946895
Let the height be h (inches), radius r.\n\nNow by proportion, h = 5r at every level.\n\nHence volume is given by\n\nV = (1/3)pi*r^2*h which is\n       (5/3)*pi*r^3\n\n"Rate at which water is drained" means find dV/dt\n\ndV/dt = 5*pi*r^2*(dr/dt),\n\nand dr/dt = (1/5)*(dh/dt)\n               = -2/5\n\nYou can finish from there.
861406
(1000kg is not a ton, it is called a tonne or metric ton)\n\nThe object of 1000kg will have a force of acceleration on it of about 9.8m/s² during its fall of 5m.\n\nTo find the kinetic energy Ek that it will have at the end of its fall you just need to find its gravitational potential energy at the beginning of the fall.  By the end of the fall all the gravitational potential energy will have turned into kinetic energy.\n\nUg=(m)(g)(h)\n\nUg is grav. potential energy, m is mass, g is acceleration due to gravity, h is height\n\nUg=(1000)(9.8)(5)\nUg=49,000 Joules\n\nSo our Ek is also 49,000 Joules.\n\nThe object will land on the earth with 49,000 Joules of kinetic energy.\n\nNear the earth's surface, the object will have a force towards the earth of \nf=ma\nf=(1000kg)(9.8m/s²)\nf=98,000N\n\nAt the time of impact the composition of the earth will determine how much force the 1 tonne mass imparts on the earth.  If it lands somewhere soft where it can dig in far its kinetic energy will be eaten more slowly because it is pushing on the earth with a smaller force, if it lands somewhere very hard then the Ek will be eaten up very quickly as it will decelerate very quickly by imparting a large force on the earth.
328342
Simply d1/ d2 = weight of metal / loss of weight in water = 10/ 2 = 5.\n\nd1 = 5x10^3 Kg/ m^3\n  \n\nDetails\n\nThe reduction in downward force is 2N.\n\nThis is due to the upward force given by equal volume of water. \n\nIf V is the volume and its mass is V.  10^3 ( density of water is 10^3 kg/m^3).\n\nThe weight of this volume of water is V.  10^3 g This is equal to 2N. \n\nThe weight of the metal is V. d . g ( d is the density of the metal) = 10N.\n\nTherefore,  V.  d . g/ V.  10^3 g ==10/ 2 = 5\n\nd = 5 x10^3 kg/m^3.
11917
measure the height of the water in bucket or sink you're putting the model in BEFORE you put the model in.  Then put the model in and remeasure the water height.  Use the height and the dimensions of the vessel (bucket, sink, etc.) to calculate the volume of water that's displaced.  Using a perfectly circular or square vessel will be ideal.
1044964
The weight of water equals the weight of the cloud, since the cloud is nothing but water. If it loses water by raining, the cloud loses that amount of weight.
742159
So lets roll...\n\nmathematically, set it up as follows:\n\nLet x = the number of kg of water @ 82degC that must be added\n\nEnergy = m * c * T\n(m = mass, c = specific heat of water, T = tempearature)\n\nNeglecting convective and radiation heat losses:\n\n (x kg)*c*(82 C) + (0.2kg)*c*(14 C) = (x + 0.2 kg) *c*(36 C)\n\ndividing by c and re-arranging:\n\n82x + 2.8 = 36x + 7.2\n\nx = 0.096 kg
750229
the rain is measured in a rain gauge.  The wind is measured by a anemoter(i could have spelled it wrong), and the temperature is measured by a thermometer.  None of that stuff is expensive.  I bought something with a rain gauge and a thermometer all in one and i paid like 10 dollars.  I am not sure how much an anometer costs though.  I hoped i helped ya.
554775
Put it in a graduated cylinder, record the water level, drop the rock in, and note the new water level. Subtract.\n\nYay science.
871293
Imagine you had an umbrella.  If you walked in the rain with it... you'd likely not get wet at all.  Now if you ran with it, you would... same principle applies to _HOW_ wet you get without one.\n\nI personally run when it's raining because I don't like cold rain on the back of my neck... the best of both worlds is to run with you torso bent at a 45 degree angle.  The amount of water landing on your body is about the same, but the chinese water torture doesn't last as long.\n\nHow wet does a sponge sphere get when getting from point A to point B at different speeds?  It really depends on the density and velocity of the rain and the surface area of the sphere.  Assuming constant density and gravitational velocity with no wind factor... the wet amount is inversely proportional to the velocity.  \n\nBest case scenario, sphere moving at the speed of light would collect all droplets "suspended" in mid air in its path.  Worst case scenario, speed of sphere at near zero, would approach "infinitely" wet... right??\n\nNow a sponge swells up as it gets wet... but that's a different issue we don't wanna talk about :)
6661
It depends on how wet the snow is. The US Geological Survey says a rule of thumb is 10 inches of snow equal an inch of rain. But wet heavy snow my be 4-5 inches  and light fluffy powder may be more than 10 inches.
724828
Hmmm, I get 0.033 Joules, not 0.33 Joules:\n\nW = F * r\nW = m * a * r\nW = (3.35e-5 kg) * (9.8 m/s^2) * (100m) = 0.033 Joules\n\nAs for the air resistance, well, the equation we did above was just the work done by gravity.  The total force is actually zero because the velocity is constant.  Therefore, there must an equal and opposite force preventing the  raindrop from accelerating...the air resistance.  \n\nF(total) = 0\nF(total) = F(grav) - F(air resistance)\ntherefore,\nF(grav) = F(air resistance)\n\n...note that the forces are the same over the whole length of the fall, but opposite in direction.  Since work is the dot product of force and distance (i.e. direction matters), the work done by air resistance will be equal to the work done by gravity but of opposite sign...-0.033 Joules.\n\nLet's be clear on this: The total work done on the raindrop is zero because it's not accelerating...but the gravitational force *is* doing work, and the air resistance is doing an equal amount of negative work.
257846
I don't understand why you are only concerned about 1/4 inch of rain, but a good method for accurate measurements is collecting rain in a coffee can placed at a location with as few obstructions as possible and where little if any will bounce off the ground or other objects into the can. The diameter of these cans is normally about 3.9 inches.\n\nThe contents can then be poured into a tube of much smaller diameter - for example 1 inch. The amount in the smaller tube corresponds to that collected in the larger can as follows:\n\nLa = Sa (Sd/Ld)^2\n\nLd: diameter of coffee can\nLa: amount collected in coffee can (what we want)\nSd: diameter of tube\nSa: amount in tube\n\nI.e., the ratio of squares of the diameters provides the required conversion from the amount in the tube to that in the can.\n\nExample: Suppose the coffee can has a diameter of 3.92 inches and the small tube has a diameter of .95 inches. You collect an amount in the can that is very difficult to measure directly, but when you pour its contents into the tube, you find it is not that difficult to measure as 5.22 inches. Helps to have a clear tube with accurate markings, and for large amounts it may have to be emptied and filled several times (best to wipe any amount sticking on the inside when that is done). Anyway for this situation:\n\nLd = 3.92\nSd = .95\nSa = 5.22\n\nThus\n\nLa = (5.22)(.95/3.92)^2 = .307\n\nslightly greater than 1/4 inch :)\n\nNote that (Ld/Sd)^2 = 17.02, so pouring into the small tube increases the resolution of the measurement by a factor of about 17.\n\nThat is essentially what the Long-term gauge shown here does:\n\nhttp://www.robertwhite.com/cgi-local/shop.pl?type=item&item=longtermrg&categ=rw_rainfall&nav_bar=weather\n\nThe advantage of that being that the smaller tube inside is marked with the conversion factor already accounted for - the amount can be read directly in 1/100's of an inch.\n\nSo if you want to save some money, you can probably construct something very similar using a coffee can &/or funnel and some sort of smaller tube for it to go into. I have the gauge, and can say it is designed very well and is accurate. Measurements with it and the coffee can as described almost always agree within 2% and often are virtually identical.
543679
Typical raindrops fall at a rate of 14mph and large raindrops fall at 20mph. So about 4 minutes if a cloud a mile high.
894074
Assuming the drop of water is a sphere or close to one...\n\nvolume of a Sphere = (4/3) * pi * radius^3\n\nTo find the volume of a drop, measure out a small volume of water into a graduated cylinder.\nUse a dropper to take drops and deposit them into another container.  Count the number of drops needed to empty the measured water.\nvolume of drop = measured volume / number of drops
706188
It means, if you look at billions of days with weather conditions just like this day, 20% of them would be days with rain.\n\nI'm not a pro.\n\nBTW, when they give people stats and predictions, wherein it rained 40 out of a hundred times that the weather person said there was a 40% chance of rain, people tend to think the weather person was a lousy predictor.\n\nEven though the stats suggest they were a perfect predictor.\n\nFolks just don't understand this stuff, do they?