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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

Physics
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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.
ok so first we can convert the units
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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Other answers:

yea so wat would the next step be?
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.
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 :)
alright thanks :)
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!
thanks i will :)
Alright. Have you started any work yet?
yea i tried and like to convert them but idk how u meant with ur method
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\[[units\ of\ K]=\frac{[kg][m]^2}{[s]^2}\]
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"
Do you know what "work" is in physics? If not, I'd like you to look at the equation.
ok let me write that down
wait wat do u mean by m/s isnt it one unit?
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.
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)
is that right?
And you can treat units like variables. They can cancel each other out and multiply and stuff. You see it a lot when converting.
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.
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!
ok thanks :)
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.
ok :)
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?
yes
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.
Now, put the numbers with the units and you have \(641\ [N]\).
so that would be the final answer?
Yup! Now to yours! Your units are \[\frac{[kg]\ [m^2]}{[s^2]}=[J]\]joules. And what is your number?
well for kg we have 3.6*10^2
and for m is 2.32*10^5
You don't actually need to consider the units now. You can look at the numbers on there own! Like, this was my example:\[F=(65.3\ [kg])(9.81\ [m/s^2])\]\[\qquad\qquad\downarrow\]\[F=(65.3)(9.81)\]
so they r 3.6 and 2.32?
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.
"[m/s]"*
ok so it can be 3.6*10^2 and 2.32*10^5
is that right?
Close....
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!
By doing so, you are no longer considering the units, but just the number amount.
Is that something you'd rather not do?
no its fine. :)
im just trying to figure this out. im kida new at this
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?
You can copy and paste that, or type it.
Here's what I mean, but I'll show you with the math feature on here.\[K=\frac{1}{2}(3.6*10^2 \cancel{[kg]}) \ (2.32*10^5 \cancel{[m/s]})^2\]\[K=\frac{1}{2}(3.6*10^2) \ (2.32*10^5)^2\]
\(\it{That's}\) what I meant
And the \(\frac{1}{2}\) is exact, so it won't affect your significant figures.
I have to go. Good luck, and take care! Just remember to use the right number of significant figures.

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