View Full Version : Unit conversions help!
lettertwelve
Sep3-08, 09:25 PM
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
none
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
correct??
lettertwelve
Sep3-08, 09:30 PM
correct
are you more than 100% sure :P
do you have time to check 2 more for me??
your reasoning is correct as well and im more than 100 percent certain
lettertwelve
Sep3-08, 09:37 PM
your reasoning is correct as well and im more than 100 percent certain
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..
LowlyPion
Sep3-08, 09:43 PM
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..
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.
lettertwelve
Sep3-08, 09:46 PM
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.
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
Astronuc
Sep3-08, 09:46 PM
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
none
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
correct?? 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.
LowlyPion
Sep3-08, 10:15 PM
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
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.
Cloudens
Sep11-08, 01:34 AM
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:
?hrs=1.00km(10^3m/1km)(1cm/10^-2m)(5.25cm/1min)(60min/1hr)
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.
stewartcs
Sep11-08, 08:21 AM
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
none
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
correct??
No.
CS
stewartcs
Sep11-08, 08:24 AM
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:
?hrs=1.00km(10^3m/1km)(1cm/10^-2m)(5.25cm/1min)(60min/1hr)
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.
Start off using the simple relationship d = rt, then solve for t.
t = d/r
Then it is just a matter of dimensional analysis.
CS
LowlyPion
Sep11-08, 08:30 AM
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:
?hrs=1.00km(10^3m/1km)(1cm/10^-2m)(5.25cm/1min)(60min/1hr)
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.
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.
lukas86
Sep11-08, 09:43 AM
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
LowlyPion
Sep11-08, 10:49 AM
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
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?)
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