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The statue has 274.4 joules of gravatational potential energy

relative to the fround.

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Q: A statue has a mass of 14 kg on a shelf it sits 2m from the fround what is the gravatational potential energy?
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Lindy is decorating her house She puts a statue that has a mass of 14 kg up on a shelf so it sits 2 m from the ground What is the gravitational potential energy of the statue?

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What instrument measures the density of a statue?

Let's review and then try to solve the problem. Density is the weight per unit of volume of something. (Maybe mass per unit of volume, if you like.) We'll need to find the weight of the thing, and then find its volume. Weighing the statute should be fairly simple. Find an appropriate scale and perform the measurement. But what about volume? How big is the statue? And it has an irregular shape so that its volume cannot be easily calculated, doesn't it? Yes, it does. I hope the statue is waterproof and can suvive a swim 'cause we're going to have to immerse it completely. About the easiest way to find the volume of an irregular object is to immerse it completely in water to see how much it displaces. Let's figure out how it works. Say the statue is a couple of feet high and we know someone who has an aquarium store. We drive over with our statue and "borrow" an aquarium that the statue will fit in. (It's empty.) We need a container large enough to hold the statue, and as small as possible beyond that so we can be as accurate as possible. Lay the statue in the bottom of an aquarium that it just fits into and add water to just cover it completely. Carefully mark the level. Now remove the statue and mark the level. (The level went down, obviously, and the amount of "water loss" in the tank is equal to the volume of the statue. Measure the difference in the levels, and then measure the inside dimensions of the aquarium. You will have three dimensions, or a "volume" to work with. The fish tank has length and width, and we've just discovered height. We'll have "x" cubic inches of water (or "x" cubic centimeters of water, if you prefer). We have a volume of water that was displaced, and (as stated) that volume of water is equal to the volume of the statue. We had its weight, and now that we have its volume we can find its density. So let's. Weightstatue / Volumestatue = Densitystatue Assume we found the statue weighed 8,000 grams or 8 kilograms (8 kg). And let's say it displaced 1,000 cubic centimeters of water (which turns out to be a liter). Plug the values into the formula and get an answer. 8,000 grams / 1,000 cubic centimeters = 8,000 grams per 1,000 cc Let's reduce this by dividing the top and bottom by 1,000. We'll get 8 grams per cc for the density of the statue. It tuns out that bronze (about 10% tin) has a density of about 8 grams per cc, so it may be that the statue is made of bronze. In any case, it's density is 8 grams per cubic centimeter, which is 8 times as dense as water. Note: We could have added water back to the aquarium to fill it back up to the level where it was with the statue in it and kept track of how much water we put back in. This would allow us to directly find the volume of the statue. The statue's volume is the same as that of the water we put back in to bring the level back up to where it was. This directly measures the volume of the water and, therefore, the volume of the statue. We could have also put the whole aquarium on a scale, filled it to just cover our statue and then removed the statue and weighed the tank and water. If we then added water back to the tank to fill it to the level it was with the statue in there and then weighed it again, we'd have two weights. One is the weight of the tank and water after removel of the statue and the second weight (the heavier one, obviously) is the one with the water "put back" to bring the level up to what it was. The difference in weights is the weight of the water that the statue displaced. We know that water weighs 1 gram per cubic centimeter, and we can calculate the volume of that weight of water to find the volume of the statue. (Whew!) Any we do it, the smaller a container we can find to completely immerse the statue, the more accurate the measurement. If we cannot immerse the statue to find its volume, the problem becomes much more difficult. There are ways of estimating volume, but they invlove some measuring and some tedious (but doable) calculation.


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