## RosieS12 Group Title 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 one year ago one year ago

1. theEric Group Title

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.

2. RosieS12 Group Title

ok so first we can convert the units

3. theEric Group Title

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?

4. RosieS12 Group Title

yea so wat would the next step be?

5. theEric Group Title

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.

6. theEric Group Title

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 :)

7. RosieS12 Group Title

alright thanks :)

8. theEric Group Title

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!

9. RosieS12 Group Title

thanks i will :)

10. theEric Group Title

Alright. Have you started any work yet?

11. RosieS12 Group Title

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

12. theEric Group Title

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}$

13. theEric Group Title

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"

14. theEric Group Title

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

15. RosieS12 Group Title

ok let me write that down

16. RosieS12 Group Title

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

17. theEric Group Title

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.

18. RosieS12 Group Title

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)

19. RosieS12 Group Title

is that right?

20. theEric Group Title

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

21. theEric Group Title

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.

22. theEric Group Title

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!

23. RosieS12 Group Title

ok thanks :)

24. theEric Group Title

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.

25. RosieS12 Group Title

ok :)

26. theEric Group Title

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?

27. RosieS12 Group Title

yes

28. theEric Group Title

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.

29. theEric Group Title

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

30. RosieS12 Group Title

so that would be the final answer?

31. theEric Group Title

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

32. RosieS12 Group Title

well for kg we have 3.6*10^2

33. RosieS12 Group Title

and for m is 2.32*10^5

34. theEric Group Title

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)$

35. RosieS12 Group Title

so they r 3.6 and 2.32?

36. theEric Group Title

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.

37. theEric Group Title

"[m/s]"*

38. RosieS12 Group Title

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

39. RosieS12 Group Title

is that right?

40. theEric Group Title

Close....

41. theEric Group Title

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!

42. theEric Group Title

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

43. theEric Group Title

Is that something you'd rather not do?

44. RosieS12 Group Title

no its fine. :)

45. RosieS12 Group Title

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

46. theEric Group Title

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?

47. theEric Group Title

You can copy and paste that, or type it.

48. theEric Group Title

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$

49. theEric Group Title

$$\it{That's}$$ what I meant

50. theEric Group Title

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

51. theEric Group Title

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