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are these objects made of same or different metals? assume that calculated densities are accurate to 1.00 %

2007-08-17 10:46:35 · 7 answers · asked by mannu 2 in Science & Mathematics Chemistry

7 answers

3 cm * 3 cm * 3 cm = 27 cm^3
140.4 g / 27 cm^3 = 5.21 g/cm^3

4/3 * pi * (1.42 cm)^3 = 11.99 cm^3
61.6 g / 11.99 cm^3 = 5.14 g/cm^3

(5.21 - 5.14) / 5.14 * 100% = 1.36%
(5.21 - 5.14) / 5.21 * 100% = 1.34%

The % different is over 1.00 %. Well it is in the range. if 5.21 is 1% lower, 5.14 is 1% higher, then they are probably from the same material.

2007-08-17 10:53:24 · answer #1 · answered by Carborane 6 · 0 1

The way to tell is to compare their densities. We'll assume that the objects are solid, not hollow and made of only the one metal (e.g. not the top half steel and the bottom gold).

The density of an object is its mass divided by its volume. It's a matter of geometry to determine the volume of those shapes. (This is mainly a maths question, not a chemistry question.)

Cube: volume = lenth of side cubed = 3^3 = 27 cubic centimetres. So density = 140.4 / 27 g/cm^3. You can do the division on a calculator.

Sphere: volume = 4 pi radius^3 / 3
= 4 * 3.1416 * 1.42 * 1.42 * 1.42 / 3
= v (you can do it on a calculator)
Density = 61.6 / v.

If the two densities are the same to within 1% of each other, then you assume that they're the same metal. If not, then you assume they're differrent.

1% means you take the bigger number (call it d), work out 1% of it (d/100), and both add and subtract that from the number (d - d/100, d + d/100). If the smaller number lies in that range, then they're the the same to within 1% accuracy.

2007-08-17 11:12:05 · answer #2 · answered by Raichu 6 · 0 0

To determine if two objects are made of the same material, their densities have to be the same.

So here we need to find the density of the cube and the sphere. To do this we need to know how to get the density. Here's the formula:

D = M/V, where D is the density, M is the mass, and V is the volume.

So let's start with the cube:

Mass - 140.4 g
Volume - 3.00^3 >>> To get the volume of a cube, you take the edge length to the third power.

Now we plug it into the formula we had:

D = (140.4)/(27.00) = 5.2 g/cm^3


Now we do the same for the sphere:

Mass - 61.6 g
Volume - (1.42)^3 * (pi) * 4/3 >>> again that's how you find the volume of a sphere

And we do this formula thing again:

D = (61.6)/((1.42)^3 * (pi) * 4/3), which is about 5.14 g/cm^3

Personally that's close enough for me. (5.2 and 5.14?)

2007-08-17 10:58:31 · answer #3 · answered by Anonymous · 0 0

Density = Mass / Volume

If the two objects have the same density (to within 1.00%) you can assume it is the same.

Calculate the volume of each and use the above formula.

A cube has the volume of the side x side x side.

Look up the formula for the volume of a sphere based on the radius..

The calculations are just plug into the formulas

2007-08-17 10:54:33 · answer #4 · answered by reb1240 7 · 0 0

a super circle is a circle that is going throughout the time of the middle of the sector. The radius of a super circle is the comparable because of the fact the radius of the sector. The formula for the area of a circle is: A = pi * r^2 the two aspects are: A1 = 3^2 pi = 9 pi A2 = 5^2 pi = 25 pi The ratio between the two is 9 pi / 25 pi. The pi cancels out, so the respond is: 9/25 a speedy thank you to do it is purely sq. the two radii and write them as a ratio: 9:25

2016-12-12 05:15:54 · answer #5 · answered by ? 4 · 0 0

3 cm * 3 cm * 3 cm = 27 cm^3
140.4 g / 27 cm^3 = 5.21 g/cm^3

4/3 * pi * (1.42 cm)^3 = 11.99 cm^3
61.6 g / 11.99 cm^3 = 5.14 g/cm^3

If the 'true' density was supposed to be 5.17g/cm^3, then the two measurements are within 1% of the 'true' values, so they COULD be the same material.

However, the fact that two objects have similar densities doesn't automatically make them the same material.

2007-08-17 10:59:10 · answer #6 · answered by tinfoil666 3 · 0 0

dont think so...volumes are different, so different metals

2007-08-17 10:55:09 · answer #7 · answered by SmartHealth 1 · 0 1

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