place the answer in correct scientific notation, when appropriate and simplify the units. work with the units, cancel units when possible, and show simplified units in the final answer.
K=1/2(3.6*10^2 kg) (2.32*10^5 m/s)^2

- anonymous

- schrodinger

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- theEric

Alright, I'll start by making that equation "pretty"...\[K=\frac{1}{2}(3.6*10^2 [kg]) \ (2.32*10^5 [m/s])^2\]
Now, do you understand what this question wants you to do? All that is pretty much saying: calculate. Solve the problem and see what \(K\) is. And make sure you handle the units well.

- anonymous

ok so first we can convert the units

- theEric

We could. We could look at the units separately, to make it easier. I put units in \([\ ]\)'s to tell them apart by the way. So...\[[units\ of\ K]=[kg]\ \left( [m/s]\right)^2\]
Do you see what I did, there?

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## More answers

- anonymous

yea so wat would the next step be?

- theEric

Solving. So multiply units like you would variables or numbers. And that \(m/s\) is squared. So treat that like \(\Large\left(\frac{m}{s}\right)^2\) when you do your math.

- theEric

Understood? You can let me know what you get to see if we agree on what that is. If we have different answers, we can work through it together :)

- anonymous

alright thanks :)

- theEric

Let me know what you get, if I'm still on when you solve it!
After you work the units out, you try to find if they are actually one unit themselves!
And you can calculate the numbers separately - that's just putting it into a calculator correctly, though. Right?
Good luck!

- anonymous

thanks i will :)

- theEric

Alright. Have you started any work yet?

- anonymous

yea i tried and like to convert them but idk how u meant with ur method

- theEric

So, like, you have\[[kg]\ [m/s]^2\]\[=[kg]\ [m\div s]^2\]\[=[kg]\ [m]^2\div [s]^2\]So, in the end, you have\[[units\ of\ K]=\frac{[kg][m]^2}{[s]^2}\]

- theEric

Now it's memory work. I can only remember units from equations. Like, force is in newtons, and force is mass * acceleration which is kg * m/s^2. So newtons are "kg m / s^2"

- theEric

Do you know what "work" is in physics?
If not, I'd like you to look at the equation.

- anonymous

ok let me write that down

- anonymous

wait wat do u mean by m/s isnt it one unit?

- theEric

meters per second can be a unit, because it is a unit of velocity. However, meters and seconds are also units themselves, of distance and time respectively. A unit is just what you can measure something in.

- anonymous

ok so i put :
(3.6*10^2kg) (2.32*10^5s)^2
(3.6*10^2kg) (2.32*10^5)/ (2.32*10^5)

- anonymous

is that right?

- theEric

And you can treat units like variables. They can cancel each other out and multiply and stuff. You see it a lot when converting.

- theEric

I don't know if we're on the same page! I didn't get to the numbers yet, but you can do that separately.

- theEric

Sure. I'll do a similar probably, and you'll see how I do it. Then we'll go to your problem, and apply it. I'll think of a quick one.
Got it!

- anonymous

ok thanks :)

- theEric

I'll make the numbers have decimals, to make it more similar.
\[F=(65.3\ [kg])(9.81\ [m/s^2])\]So the units are different, and the numbers are different, but it's still physics.
We'll do this:
1. Solve to see what the units are, or I guess we'd say what the unit is.
2. Solve to see what the number is, making sure we have the correct significant figures.
3. Putting them together.

- anonymous

ok :)

- theEric

So\[F=(65.3\ [kg])(9.81\ [m/s^2])\]\[[units\ of\ F]=[kg][m/s^2]\]\[[units\ of\ F]=\frac{[kg]\ [m]}{[s]^2}\]
And kg m / s^2 is actually newtons, N.
Like I said, F=ma and so [N]=[kg][m/s^2].
Your equation is like \(K=\frac{1}{2}mv^2\), which is kinetic energy, which is an energy, and it's in joules. So remember that for yours.
Next, onto the numbers. Do you know how significant figures work with multiplying and dividing?

- anonymous

yes

- theEric

So now we look at the numbers, by which I mean\[F=(65.3)(9.81)\] Now that's simple, right? Its \(641.\). I had the same number of sig figs in each number. You always use the lesser amount.

- theEric

Now, put the numbers with the units and you have \(641\ [N]\).

- anonymous

so that would be the final answer?

- theEric

Yup! Now to yours! Your units are \[\frac{[kg]\ [m^2]}{[s^2]}=[J]\]joules.
And what is your number?

- anonymous

well for kg we have 3.6*10^2

- anonymous

and for m is 2.32*10^5

- theEric

You don't actually need to consider the units now. You can look at the numbers on there own! Like, this was my example:\[F=(65.3\ [kg])(9.81\ [m/s^2])\]\[\qquad\qquad\downarrow\]\[F=(65.3)(9.81)\]

- anonymous

so they r 3.6 and 2.32?

- theEric

You have
\[K=\frac{1}{2}(3.6*10^2 [kg]) \ (2.32*10^5 [m/s])^2\]
So just take the units away. Actually, here's the text for that equation:
K=\frac{1}{2}(3.6*10^2 [kg]) \ (2.32*10^5 [m/s])^2
Erase the "[kg]" and [m/s].
Then post what's left.

- theEric

"[m/s]"*

- anonymous

ok so it can be 3.6*10^2 and 2.32*10^5

- anonymous

is that right?

- theEric

Close....

- theEric

K=\frac{1}{2}(3.6*10^2 [kg]) \ (2.32*10^5 [m/s])^2
Erase the "[kg]" and [m/s].
Then post what's left.
You can surround what's left in `\[` and `\]` to make OpenStudy make it look cool!

- theEric

By doing so, you are no longer considering the units, but just the number amount.

- theEric

Is that something you'd rather not do?

- anonymous

no its fine. :)

- anonymous

im just trying to figure this out. im kida new at this

- theEric

Understood! That's when it is the hardest!
So here is everything:
K=(1/2) (3.6*10^2 [kg]) (2.32*10^5 [m/s] )^2
Erase all the units. What would it be without them?

- theEric

You can copy and paste that, or type it.

- theEric

Here's what I mean, but I'll show you with the math feature on here.\[K=\frac{1}{2}(3.6*10^2 \cancel{[kg]}) \ (2.32*10^5 \cancel{[m/s]})^2\]\[K=\frac{1}{2}(3.6*10^2) \ (2.32*10^5)^2\]

- theEric

\(\it{That's}\) what I meant

- theEric

And the \(\frac{1}{2}\) is exact, so it won't affect your significant figures.

- theEric

I have to go. Good luck, and take care! Just remember to use the right number of significant figures.

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