Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Unit conversions help!

  1. Sep 3, 2008 #1
    1. The problem statement, all variables and given/known data

    A snail travels on average 5.25 cm/minute. How many hours would it take to travel a distance of 1.00km?

    2. Relevant equations


    3. The attempt at a solution

    5.25cm/min X 60min/hr X 1m/100cm X 1km/1000m

    cm, min, and m cancel out and the answer ends up to be .00315

  2. jcsd
  3. Sep 3, 2008 #2
  4. Sep 3, 2008 #3
    are you more than 100% sure :P

    do you have time to check 2 more for me??
  5. Sep 3, 2008 #4
    your reasoning is correct as well and im more than 100 percent certain
  6. Sep 3, 2008 #5
    what about this one:

    solid copper has a density of 8.92 g/mL. How many cm^3 does 0.75 lbs occupy?

    i did: 8.92g/mL X 1000mL/cm^3 X 1lb/453.59237g

    i'm missing something i believe, but i don't know what, and how accurate is my lb/g part, as well as where do i put the 0.75lb? if you could turn me in the right direction..
  7. Sep 3, 2008 #6


    User Avatar
    Homework Helper

    After you cancel out your units what are you left with? That would suggest what you need to do to get an answer in the desired form.
  8. Sep 3, 2008 #7
    i believe i should only be left with lbs, but then i can't begin to figure out how i'd get rid of the cm^3
  9. Sep 3, 2008 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    Hold on here. What are the units of 0.00315? If it's hrs, then that is one fast snail.

    In the question, one is given speed and distance, and there is a relevant equation that gives time (duration) in terms of distance and speed (assume constant speed).

    However in the solution given, one has multiplied speed by ratios of time and distance, and so the units would be for speed, which cannot be correct.

    Please write an expression for time in terms of distance and speed, and then apply the appropriate ratios.
  10. Sep 3, 2008 #9


    User Avatar
    Homework Helper

    You were developing the units of a constant for converting. What is the statement of that constant after you do all that and cancel out your units?

    The question is asking you how many cm3 is in .75 lb.

    Use your new constant to convert it then at that point.
  11. Sep 11, 2008 #10
    Wouldn't you start the equation with the 1.00km instead of the 5.25cm/min and use that as a conversion factor.

    So it would be something like this:


    Which would equal 31500000 and since you have to use sig figs it would be 3.15 x 10^7 hr

    Correct me if I'm wrong.
  12. Sep 11, 2008 #11


    User Avatar
    Science Advisor


  13. Sep 11, 2008 #12


    User Avatar
    Science Advisor

    Start off using the simple relationship d = rt, then solve for t.

    t = d/r

    Then it is just a matter of dimensional analysis.

  14. Sep 11, 2008 #13


    User Avatar
    Homework Helper

    It won't live long enough.

    Think about what the numbers mean.

    5.25 cm/min is what in 1 hr? 60 times that? making it 315 cm in an hour?

    That's what 3.15 m/hr?

    Now if I have to go 1000 of those how fast can Slugo do it?

    Won't that be 1000m/3.15 m/hr? Doesn't that yield hours?

    2 weeks? He might make it back home in time to see the wife in her new shell.
  15. Sep 11, 2008 #14
    317 hours, so 13.23 days.

    Like LowlyPion wrote... 5.25cm/min >> 315 cm/hour >> 3.15m/hour >> 0.31746 hr/m *(1000m) >> 317.46 hrs
  16. Sep 11, 2008 #15


    User Avatar
    Homework Helper

    I think my point was to look at the conversions in steps to insure that each step meets a reasonableness test. If you are taking reasonable steps along the way, then maybe you will reach a reasonable result?
    107 hours was simply not a reasonable answer.
    That should have triggered your suspicions.
    (How many years is 107 hours you think? We should all live so long?)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook