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RosieS12

  • 2 years ago

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

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  1. theEric
    • 2 years ago
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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.

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

  3. theEric
    • 2 years ago
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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?

  4. RosieS12
    • 2 years ago
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    yea so wat would the next step be?

  5. theEric
    • 2 years ago
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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.

  6. theEric
    • 2 years ago
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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 :)

  7. RosieS12
    • 2 years ago
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    alright thanks :)

  8. theEric
    • 2 years ago
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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. RosieS12
    • 2 years ago
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    thanks i will :)

  10. theEric
    • 2 years ago
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    Alright. Have you started any work yet?

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

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

  13. theEric
    • 2 years ago
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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"

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

  15. RosieS12
    • 2 years ago
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    ok let me write that down

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

  17. theEric
    • 2 years ago
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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.

  18. RosieS12
    • 2 years ago
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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)

  19. RosieS12
    • 2 years ago
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    is that right?

  20. theEric
    • 2 years ago
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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.

  21. theEric
    • 2 years ago
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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.

  22. theEric
    • 2 years ago
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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!

  23. RosieS12
    • 2 years ago
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    ok thanks :)

  24. theEric
    • 2 years ago
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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.

  25. RosieS12
    • 2 years ago
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    ok :)

  26. theEric
    • 2 years ago
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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?

  27. RosieS12
    • 2 years ago
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    yes

  28. theEric
    • 2 years ago
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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.

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

  30. RosieS12
    • 2 years ago
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    so that would be the final answer?

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

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

  33. RosieS12
    • 2 years ago
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    and for m is 2.32*10^5

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

  35. RosieS12
    • 2 years ago
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    so they r 3.6 and 2.32?

  36. theEric
    • 2 years ago
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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.

  37. theEric
    • 2 years ago
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    "[m/s]"*

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

  39. RosieS12
    • 2 years ago
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    is that right?

  40. theEric
    • 2 years ago
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    Close....

  41. theEric
    • 2 years ago
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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!

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

  43. theEric
    • 2 years ago
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    Is that something you'd rather not do?

  44. RosieS12
    • 2 years ago
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    no its fine. :)

  45. RosieS12
    • 2 years ago
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    im just trying to figure this out. im kida new at this

  46. theEric
    • 2 years ago
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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?

  47. theEric
    • 2 years ago
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    You can copy and paste that, or type it.

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

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

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

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

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