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BloomLocke367

  • one year ago

Can you check my work @Shalante? A cheetah is known to be the fastest mammal on Earth, at least for short runs. Cheetahs have been observed running a distance at \(5.00\times10^2\)m with an average speed of \(1.00\times10^2\)km/hr. How long would it take the cheetah to cover this distance at this speed? I first converted to m/s and I got the time to be 19.8s, is that right?

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  1. BloomLocke367
    • one year ago
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    I used |dw:1441319140411:dw|

  2. anonymous
    • one year ago
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    I got 18.0 sec

  3. BloomLocke367
    • one year ago
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    how?

  4. BloomLocke367
    • one year ago
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    I did 27.78m/s=550m/t. I then divided 550m by 27.78m/s and got 19.8s

  5. anonymous
    • one year ago
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    \[\frac{ 100km }{ 1 hour }\times \frac{ 1000m }{ 1km }\times \frac{ 1hour }{ 60minutes }\times \frac{ 1minute }{ 60seconds }\]= 27.7m/s Notices 100km/1hr is the same as 100km/hr. 27.7m/s is velocity To get time=distance over velocity t=500m/27.7m/s=18.0s

  6. anonymous
    • one year ago
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    It is 500m not 550m 5.00 x10^2m is 500m. Unless you posted the wrong number

  7. anonymous
    • one year ago
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    Otherwise you did everything correct.

  8. BloomLocke367
    • one year ago
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    Uh oh... I typed the problem wrong. it is 5.50*10^2, not 5.00

  9. BloomLocke367
    • one year ago
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    thanks for pointing that out and making me double check xp

  10. BloomLocke367
    • one year ago
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    so am I correct?

  11. anonymous
    • one year ago
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    Ah, I see. Yea you are correct then.

  12. anonymous
    • one year ago
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    Nice job solving the problem.

  13. BloomLocke367
    • one year ago
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    thanks :)

  14. anonymous
    • one year ago
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    Got anymore? I have 25 minutes before I go.

  15. anonymous
    • one year ago
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    any more*

  16. BloomLocke367
    • one year ago
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    yes, there's a part b to that problem, I'm working it out right now

  17. anonymous
    • one year ago
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    Feel free to post it, so I can solve it faster and also respond earlier.

  18. BloomLocke367
    • one year ago
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    it is "suppose the average speed of the cheetah were just 85.0km/hr. What distance would the cheetah cover during the same time interval calculated in a?"

  19. BloomLocke367
    • one year ago
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    I already converted it into m/s and got 23.61m/s

  20. BloomLocke367
    • one year ago
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    I got 467.5m... is that correct?

  21. anonymous
    • one year ago
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    23.61m/s*19.8s=476.5m You are correct 4.74 x10^2m tho since we are multiplying and dividing and least sf in original value is 3. Brb let me switch computers. (In school)

  22. BloomLocke367
    • one year ago
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    oki c:

  23. BloomLocke367
    • one year ago
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    woe... why 4.74?

  24. anonymous
    • one year ago
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    Oops I mean 4.76

  25. anonymous
    • one year ago
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    Actually, it is better to round it off 467.5m becomes 467m. I was rushing when I shut down the computer lol. 2 mistakes already.

  26. anonymous
    • one year ago
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    468m I mean.

  27. BloomLocke367
    • one year ago
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    hahaha it's alright.

  28. anonymous
    • one year ago
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    If answer is in 4 sf then shortest sf is 3 then you do scientific notation Ex: 5067m is answer. Least sf is 3 5.07x10^3m

  29. BloomLocke367
    • one year ago
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    okay thanks for the tip XD I'll be right back, I'll be afk.

  30. anonymous
    • one year ago
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    and shortest sf is 3, then you Typing so fast with this nice keyboard. Making lots of typos and such.

  31. anonymous
    • one year ago
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    I had to go right now. Remember? Probably be back in like 90 minutes the least. Feel free to tag me or pm me later.

  32. BloomLocke367
    • one year ago
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    oki, I'm back now tho

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