# Math Help - I need help converting.

1. ## I need help converting.

I'm not even sure this belongs in the algebra section, I know so little about math. It's a joke, really. I'm in a chem 101 class in University and I feel so screwed because I don't even know what to do, yet too embarrassed to ask for help because I SHOULD know this crap.

(Yes, I know, they won't care and would be happy to help; I understand that. Doesn't change my feelings, however).

Yet I didn't go to high school and it doomed me. I got my GED and I've been to a community college, but I transferred to a University. I also switched my major from psychology to chemistry because it interests me more.

And it does. I just need to learn the math. Which I will. On my own.

Sadly, I need to learn how to solve this by Tuesday and today is Sunday. I'm sure that won't be a problem if someone thoroughly explains it to me, and I'd be forever grateful to you if you did.

Here's the problem:

Convert 18.301 pm/ms2 to cm/min2.

How do I solve this?

I haven't taken a math class since 8th grade and I'm in my 20s. I know how to do basic algebra, and I WILL learn quickly because I am smart, I just need the explanations... thanks, guys.

2. ## Re: I need help converting.

You must know that when using equations with physical units the units of both sides of the equation must match up. For example
$\text{Pressure}\times \text{Volume} = \text{Work}$

Unit of pressure is $\frac{N}{m^2}$

Unit of volume is $m^3$
Unit of work is $Nm$

When you just put the units into the equation above you find that they are equal on both sides
$\frac{N}{m^2}\times m^3 = Nm$
$N\times \frac{m^3}{m^2} = Nm$
Which is true.

When you want to convert $\frac{pm}{(ms)^2} \text{ to } \frac{cm}{min^2}$ think about what you have to multiply by to get from 1 to the other. You should get the answer
$\frac{pm}{(ms)^2}\times \frac{cm}{pm} \times \frac{(ms)^2}{min^2}=\frac{cm}{min^2}$

$\frac{pm\cdot cm\cdot (ms)^2}{(ms)^2 \cdot pm \cdot min^2}=\frac{cm}{min^2}$

Just like fractions with algebraic expressions the units cancel when they are at the top and bottom of the fraction

So this is the equation you are working with

$\frac{pm}{(ms)^2}\times \frac{cm}{pm} \times \frac{(ms)^2}{min^2}=\frac{cm}{min^2}$

The input and output for the equation is when you are converting from pm/ms2 and to cm/min2

$[input]\times \frac{cm}{pm} \times \frac{(ms)^2}{min^2}=[output]$

The input you already know, find the values of the written terms.

First the value of $\frac{cm}{pm}$, it is easiest to think of everything relating to meters so rewriting this with meters $\frac{cm}{pm}=\frac{cm}{m}\cross{m}{pm}$

Now you know that there are 100 centimeters per meter so $\frac{cm}{m}=100$, you know that there are 10-12 meters per picometer so $\frac{m}{pm}=10^{-12}$

Putting those values into the equation $\frac{cm}{pm}=\frac{cm}{m}\times \frac{m}{pm}$ you get $\frac{cm}{pm}=100\times 10^{-12}=10^{-10}$ . i.e there are 10-10 centimeters per picometer

Now you have the value of $\frac{cm}{pm}$

Next, the time part

You need to find the value of $\frac{(ms)^2}{min^2}$. Time squared is confusing to think in so lets change it to $(\frac{ms}{min})^2$ Just as we looked at the pm and cm problem relating to meters it is easier to think of ms and min relating to seconds.

$\frac{ms}{min}= \frac{ms}{s} \times \frac{s}{min}$

There are 1000 milliseconds per second so $\frac{ms}{s}=1000$

There are 60 seconds per minute so $\frac{s}{min}=60$

Put those values into the equation $\frac{ms}{min}= \frac{ms}{s} \times \frac{s}{min}$ and you get $\frac{ms}{min}= 1000 \times 60$

$\frac{ms}{min}= 60000$

But you wanted to know $(\frac{ms}{min})^2$, that is just $60000^2$

The equation you were trying to find the conversion factors for was
$[input]\times \frac{cm}{pm} \times \frac{(ms)^2}{min^2}=[output]$

We can put these values in now.
$[input]\times 10^{-10} \times 60000^2=[output]$

And that is the formula you must use to convert from pm/ms2 to cm/min2

3. ## Re: I need help converting.

Thank you so much....... The depth you went into in answering my question is simply outstanding. Thank you, thank you, thank you.

I wish I wasn't so far behind in math. It's going to take me a long time to catch up, but I feel like I can if I really apply myself.

Just to add, I knew NOTHING about dimensional analysis before this. Thank you again. Now I understand these problems. Man, that was a lot simpler than I thought....