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RosieS12

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

  • 9 months ago
  • 9 months ago

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  1. theEric
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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.

    • 9 months ago
  2. RosieS12
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    ok so first we can convert the units

    • 9 months ago
  3. theEric
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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?

    • 9 months ago
  4. RosieS12
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    yea so wat would the next step be?

    • 9 months ago
  5. theEric
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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.

    • 9 months ago
  6. theEric
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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 :)

    • 9 months ago
  7. RosieS12
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    alright thanks :)

    • 9 months ago
  8. theEric
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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!

    • 9 months ago
  9. RosieS12
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    thanks i will :)

    • 9 months ago
  10. theEric
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    Alright. Have you started any work yet?

    • 9 months ago
  11. RosieS12
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    yea i tried and like to convert them but idk how u meant with ur method

    • 9 months ago
  12. theEric
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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\[[units\ of\ K]=\frac{[kg][m]^2}{[s]^2}\]

    • 9 months ago
  13. theEric
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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"

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

    • 9 months ago
  15. RosieS12
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    ok let me write that down

    • 9 months ago
  16. RosieS12
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    wait wat do u mean by m/s isnt it one unit?

    • 9 months ago
  17. theEric
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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.

    • 9 months ago
  18. RosieS12
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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)

    • 9 months ago
  19. RosieS12
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    is that right?

    • 9 months ago
  20. theEric
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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.

    • 9 months ago
  21. theEric
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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.

    • 9 months ago
  22. theEric
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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!

    • 9 months ago
  23. RosieS12
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    ok thanks :)

    • 9 months ago
  24. theEric
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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.

    • 9 months ago
  25. RosieS12
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    ok :)

    • 9 months ago
  26. theEric
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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?

    • 9 months ago
  27. RosieS12
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    yes

    • 9 months ago
  28. theEric
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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.

    • 9 months ago
  29. theEric
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    Now, put the numbers with the units and you have \(641\ [N]\).

    • 9 months ago
  30. RosieS12
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    so that would be the final answer?

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

    • 9 months ago
  32. RosieS12
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    well for kg we have 3.6*10^2

    • 9 months ago
  33. RosieS12
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    and for m is 2.32*10^5

    • 9 months ago
  34. theEric
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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:\[F=(65.3\ [kg])(9.81\ [m/s^2])\]\[\qquad\qquad\downarrow\]\[F=(65.3)(9.81)\]

    • 9 months ago
  35. RosieS12
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    so they r 3.6 and 2.32?

    • 9 months ago
  36. theEric
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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.

    • 9 months ago
  37. theEric
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    "[m/s]"*

    • 9 months ago
  38. RosieS12
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    ok so it can be 3.6*10^2 and 2.32*10^5

    • 9 months ago
  39. RosieS12
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    is that right?

    • 9 months ago
  40. theEric
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    Close....

    • 9 months ago
  41. theEric
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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!

    • 9 months ago
  42. theEric
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    By doing so, you are no longer considering the units, but just the number amount.

    • 9 months ago
  43. theEric
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    Is that something you'd rather not do?

    • 9 months ago
  44. RosieS12
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    no its fine. :)

    • 9 months ago
  45. RosieS12
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    im just trying to figure this out. im kida new at this

    • 9 months ago
  46. theEric
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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?

    • 9 months ago
  47. theEric
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    You can copy and paste that, or type it.

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

    • 9 months ago
  49. theEric
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    \(\it{That's}\) what I meant

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

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

    • 9 months ago
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