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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

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  1. theEric Group Title
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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.

    • one year ago
  2. RosieS12 Group Title
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    ok so first we can convert the units

    • one year ago
  3. theEric Group Title
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    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?

    • one year ago
  4. RosieS12 Group Title
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    yea so wat would the next step be?

    • one year ago
  5. theEric Group Title
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    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.

    • one year ago
  6. theEric Group Title
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    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 :)

    • one year ago
  7. RosieS12 Group Title
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    alright thanks :)

    • one year ago
  8. theEric Group Title
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    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!

    • one year ago
  9. RosieS12 Group Title
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    thanks i will :)

    • one year ago
  10. theEric Group Title
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    Alright. Have you started any work yet?

    • one year ago
  11. RosieS12 Group Title
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    yea i tried and like to convert them but idk how u meant with ur method

    • one year ago
  12. theEric Group Title
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    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\[[units\ of\ K]=\frac{[kg][m]^2}{[s]^2}\]

    • one year ago
  13. theEric Group Title
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    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"

    • one year ago
  14. theEric Group Title
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    Do you know what "work" is in physics? If not, I'd like you to look at the equation.

    • one year ago
  15. RosieS12 Group Title
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    ok let me write that down

    • one year ago
  16. RosieS12 Group Title
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    wait wat do u mean by m/s isnt it one unit?

    • one year ago
  17. theEric Group Title
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    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.

    • one year ago
  18. RosieS12 Group Title
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    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)

    • one year ago
  19. RosieS12 Group Title
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    is that right?

    • one year ago
  20. theEric Group Title
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    And you can treat units like variables. They can cancel each other out and multiply and stuff. You see it a lot when converting.

    • one year ago
  21. theEric Group Title
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    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.

    • one year ago
  22. theEric Group Title
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    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!

    • one year ago
  23. RosieS12 Group Title
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    ok thanks :)

    • one year ago
  24. theEric Group Title
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    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.

    • one year ago
  25. RosieS12 Group Title
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    ok :)

    • one year ago
  26. theEric Group Title
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    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?

    • one year ago
  27. RosieS12 Group Title
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    yes

    • one year ago
  28. theEric Group Title
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    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.

    • one year ago
  29. theEric Group Title
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    Now, put the numbers with the units and you have \(641\ [N]\).

    • one year ago
  30. RosieS12 Group Title
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    so that would be the final answer?

    • one year ago
  31. theEric Group Title
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    Yup! Now to yours! Your units are \[\frac{[kg]\ [m^2]}{[s^2]}=[J]\]joules. And what is your number?

    • one year ago
  32. RosieS12 Group Title
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    well for kg we have 3.6*10^2

    • one year ago
  33. RosieS12 Group Title
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    and for m is 2.32*10^5

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

    • one year ago
  35. RosieS12 Group Title
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    so they r 3.6 and 2.32?

    • one year ago
  36. theEric Group Title
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    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.

    • one year ago
  37. theEric Group Title
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    "[m/s]"*

    • one year ago
  38. RosieS12 Group Title
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    ok so it can be 3.6*10^2 and 2.32*10^5

    • one year ago
  39. RosieS12 Group Title
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    is that right?

    • one year ago
  40. theEric Group Title
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    Close....

    • one year ago
  41. theEric Group Title
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    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!

    • one year ago
  42. theEric Group Title
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    By doing so, you are no longer considering the units, but just the number amount.

    • one year ago
  43. theEric Group Title
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    Is that something you'd rather not do?

    • one year ago
  44. RosieS12 Group Title
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    no its fine. :)

    • one year ago
  45. RosieS12 Group Title
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    im just trying to figure this out. im kida new at this

    • one year ago
  46. theEric Group Title
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    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?

    • one year ago
  47. theEric Group Title
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    You can copy and paste that, or type it.

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

    • one year ago
  49. theEric Group Title
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    \(\it{That's}\) what I meant

    • one year ago
  50. theEric Group Title
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    And the \(\frac{1}{2}\) is exact, so it won't affect your significant figures.

    • one year ago
  51. theEric Group Title
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    I have to go. Good luck, and take care! Just remember to use the right number of significant figures.

    • one year ago
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