# Simple volume density problem

#### groundknifer

[SOLVED] Simple volume density problem..

Hello, my college is some how affiliated with utexas so we gotta do all our
physics probs online and submit them to utexas... the problem is that if i
get 51.12 and the correct answer is 50.3 (tolerance 1%) then i get it
wrong!!!!!!

So heres 1 problem were i keep on getting same answer but according to
website its wrong:
http://f0rk.com/~dmitri/prob.jpg [Broken]

can someone gimme correct answer please, i keep on getting 25.59 and its
wrong!!!@#!@#

THANKS!!!!!!!!!!!!!!!!!!!!!!!!!!

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#### enigma

Staff Emeritus
Gold Member
Well, I'm getting a different answer.

Can you post your work so we can see where the problem is?

#### groundknifer

Volume_top_part=1.07m x .00517m x .168m = .000929m^3
Volume_middle_part=1.07 x .00517m x .276m = .001527m^3
Volume_bottom_part=Volume_top_part

V_total=.00398m^3

density of iron is 7560 kg/m^3
density=mass/volume

so,
7560kg/m^3=mass/.00398m^3
mass=30.11

is that correct?

#### groundknifer

ok i found my mistake.. i got 25.162 (thats correct i think)

also i ran into a problem while finding the amount of iron atoms in that mass...

my equation is:
atomic mass (iron): 55.85

25.162 x ((6.022x10^23)/55.85) = 2.713E+23

#### enigma

Staff Emeritus
Gold Member
Yeah, you were counting the intersection of the T's twice. The trick is to count the top and bottom, and then use h-2d instead of h for the height of the center piece.

For the next part, the atomic mass* Avogadros number is number per gram, and you are working with kilograms.

Always keep the units in your equations. If in doubt, make sure they all cancel. That is the #A1 "gotcha" for all intro level science classes.

Welcome to Physicsforums, BTW

#### Integral

Staff Emeritus
Gold Member
Originally posted by groundknifer
ok i found my mistake.. i got 25.162 (thats correct i think)

also i ran into a problem while finding the amount of iron atoms in that mass...

my equation is:
atomic mass (iron): 55.85

25.162 x ((6.022x10^23)/55.85) = 2.713E+23
If you submit that answer my class you would still be wrong. Notice that most of the given numbers have only 3 siginificant digits. That is the most that you can have in your answer. The above number simply shows you used a calculator, it should read 25.2, given the numbers used to generate it.

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