Recent content by 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

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.

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 ??

Homework Statement if an object to be coated in a .20mm thick layer with a surface area of 2.0x10^2 cm^2, what is the mass of the silver required? (density of silver is 10.49 g/cm^3) Homework Equations d=m/v , surface area formula? not sure which one The Attempt at a Solution i...

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?

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...

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 =...