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Physics Question


LukasMcNeal

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I think im being retarded or just lost the plot.

The question is :

Calculate the apparent weight of 12kg of iron immersed in water.

I am given density of iron : 7800kgm^-3 and density of water 1000kgm^-3

Surely I need to know the volume of the piece of iron?

Ive been staring at my physics notes for ages and ive gone mindblank.

I know it involves archimedes but I swear I need the volume?

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Yeah, I think you need to work out the volume from mass and density then the buoyancy is equal to the weight of water displaced (I think, been a looooooong time since I did this stuff!). You've then got gravity pulling the iron down and buoyancy pushing it up giving the apparent weight of the iron.

That's what I'd be doing, anyway.

Edit: for shapes and stuff you would usually do length x breadth x height or whatever but if given any two of mass, volume and density you can work out the third since they are simply related.

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v=st

Are you thinking of v=s/t? In which instance v = velocity, s = distance and t = time?

As for the original question, basically what monkeyseemonkeydo said. It's boyancy is equivilent to the weight of water it's displaced, and it's volume is its mass divided by its density.

So that makes it's volume 12/7800 = 0.001538m^3, which makes it's boyancy 0.001538*1000 = 1.5385kg, giving a final apparent imersed weight of 12 - 1.5385 = 10.4615kg.

Edited by RobinJI
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Are you sure about that?... ;)

haha, hadn't spotted that, it's because I didn't round during the calculations, I just rounded each number as I wrote it down, so it's written to 4 sygnificant figures, but calculated using the true values, the answer's right anyway!

Your answer looks good anyway Lucas (Y)

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haha, hadn't spotted that, it's because I didn't round during the calculations, I just rounded each number as I wrote it down, so it's written to 4 sygnificant figures, but calculated using the true values, the answer's right anyway!

Sorry, I was meaning that the weight wouldn't be in kg's :P.

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