Find Meters to Fill Copper Cylinder with 8.60lbs of Covelite

[geoti8]

Homework Statement

copper can be drawn into wires. how many meters of 34-gauge wire (diameter 6.304 x 10 -3 in) can be produced in copper from 8.60 pounds of covelite, an ore that is 60% copper by mass. (hint: treat the volume as a cylinder, V of cylinder= pie r squared, density of copper = 8.95 g/cm3

Homework Equations

V of cylinder = pie r (squared)

The Attempt at a Solution

I don't know where to start, is the 60% mass of the covelite ther to confuse me? or is it relevant?

 

[rock.freak667]

if there are 8.60 lbs of covelite and 60% of this is copper, how many lbs if copper is in it?

Now that you have the mass, what is the volume, given the density and the found mass? (Note that your mass should be in grams)

also I think your equation should be V= πr²h, as πr² alone gives units of m² which is used for area. (h=height of the cylinder)

 

[geoti8]

ok..I'm following you up until then.. does that mean that for the 34-gauge wire i use the 6.304 x10^-3 and that's the diameter? so for the (vol = pie r^2) i would use (6.304x10^-3 / 2) for the radius?

so I used the V = m/d formula and got 255.66...what do i do with that now?

i'm thinking myself to death here. should have never take 5 years off school to go to the military.

 

[geoti8]

ok, πr2 alone gives units of m2 ...does that mean that i just take the square root of my m2? because I'm looking for length in meters?

 

[rock.freak667]

ok..I'm following you up until then.. does that mean that for the 34-gauge wire i use the 6.304 x10^-3 and that's the diameter? so for the (vol = pie r^2) i would use (6.304x10^-3 / 2) for the radius?

so I used the V = m/d formula and got 255.66...what do i do with that now?

i'm thinking myself to death here. should have never take 5 years off school to go to the military.

worry not! We will get through here.

Right now, so you have V=255.66 cm³ (assuming you converted your lbs to grams)

so now you know that V=πr²h. (r is half your diameter, so convert that to cm!)

so put πr²h=255.66 and solve for h

 

[geoti8]

ok so i plug my vol, and radius into the equation:

255.66 = pi (6.304x10^-3 /2)^2 x h ... and get...

255.66h = (3.14)(.003152)^2 ...

h = .000031196/255.66 ...

1.22 x 10^-7 is my m2? (was that in cm? or m?) do i need to get the square root...

3.493 x 10^-4 is that in cm or meters ??

 

[rock.freak667]

ok so i plug my vol, and radius into the equation:

255.66 = pi (6.304x10^-3 /2)^2 x h ... and get...

255.66h = (3.14)(.003152)^2 ...

h = .000031196/255.66 ...

1.22 x 10^-7 is my m2? (was that in cm? or m?) do i need to get the square root...

3.493 x 10^-4 is that in cm or meters ??

convert the 6.304x10^-3 inches into cm, then plug it into the equation

 

[geoti8]

ok so apparently the answer was 1.88 x 10^12 or ^14

and i got

1.875 x 10^14 or ^12

as soon as i saw i got it incorrect i stopped caring, FML.

 

[geoti8]

thanks for your help though, you must be a teacher, professor..or something along those lines to give me step by step help like that

 

[rock.freak667]

ok so apparently the answer was 1.88 x 10^12 or ^14

and i got

1.875 x 10^14 or ^12

as soon as i saw i got it incorrect i stopped caring, fml.

1.875 ≈ 1.88 how'd you get it wrong?


(also, I am a student like you )