what would be the weight of a aluminium metal with these dimensions?

Assume a Aluminium Cube with length of 15 inch and 1.14 inch width and thickness of about 1 inch.

15 inch = 0.381 meter
1.14 inch = 0.028956 meter
1 inch = 0.0254 meter

so the total volume = 0.381 X 0.028956 X 0.0254 = 0.000280218 meter³.

so how much would a 0.000280218 meter³ volume of aluminium weigh exactly??

the Density of Aluminium at room temp is 2.70 gm/cm³

1 cm³ = 0.000001 m³

so a 2.70 gm/cm³ = 2.70 gm/0.000001 m³

and to find how many 0.000001 m³ in a 0.000280218 m³,i divide

0.000280218/0.000001 = 280.218 m³ in a 0.000280218 meter³ volume

so 280.218 * 2.70 grams = 756.58 gm = 0.756 Kg...

am i doing it a harder way?and simplified way to do this? thank you

Seriously?

Should we ignore the initial error that the dimensions are not a cube??? If not, then this is a trick question and the answer is zero because a cube of 2m x 3m x 4m does not exist in our universe.

Otherwise...

Google can tell you the density of Al: 2.70 gcm⁻³, and 1 m³ = 10⁶ cm³

That gives a mass of 24 x 10⁶ x 2.7 = 6.48 x10⁸ g = 6.48 x10⁵ kg

Assuming we are weighing it on the Earth, at standard 9.81ms^-2

weight = 6.48x10⁵ x 9.81 = 6.36x10⁶ N

In other words, the answer is either very heavy or zero!

FoxyRick said:

Seriously?

Should we ignore the initial error that the dimensions are not a cube??? If not, then this is a trick question and the answer is zero because a cube of 2m x 3m x 4m does not exist in our universe.

Otherwise...

Google can tell you the density of Al: 2.70 gcm⁻³, and 1 m³ = 10⁶ cm³

That gives a mass of 24 x 10⁶ x 2.7 = 6.48 x10⁸ g = 6.48 x10⁵ kg

Assuming we are weighing it on the Earth, at standard 9.81ms^-2

weight = 6.48x10⁵ x 9.81 = 6.36x10⁶ N

In other words, the answer is either very heavy or zero!

Click to expand...

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oh sorry,not a cube,a rectangular shape it is.as all the sides are not symmetrical

LOL, yes, how did I get power 8 at the start??? Fat fingers, that's how!

In my defence I was doing it quickly because I thought it was going to be a trick question anyway.

I hope no one orders their aluminium based on my calculations...

FoxyRick said:

LOL, yes, how did I get power 8 at the start??? Fat fingers, that's how!

In my defence I was doing it quickly because I thought it was going to be a trick question anyway.

I hope no one orders their aluminium based on my calculations...

Click to expand...

you helped me though,thank you

Your procedure is correct, only you don´t need that steps every time, just use a unit conversion constant.

If your data are in inches and g/cm3 . Then, for alluminium: W = k * l * w * h = 0.044 [Kg/in3] * 15 [in] * 1.14 [in] * 1 [in] = 0.756Kg

where k = 2.7 * 0.254 ^ 3 = 0.044[Kg/in3]