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shortcake1970

  • 2 years ago

if you were to travel to a star 50 light years from earth at a speed of 2.0 X 10 to 8th power m/s, what would you measure this distance to be?

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  1. calculusfunctions
    • 2 years ago
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    @shortcake1970 do you know the definition of a light year?

  2. calculusfunctions
    • 2 years ago
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    A light year is the distance that light travels in one Earth year.

  3. VeritasVosLiberabit
    • 2 years ago
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    light travels at 3*10^8 m/s

  4. calculusfunctions
    • 2 years ago
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    Yes that's approximately the speed of light, as @VeritasVosLiberabit said.

  5. VeritasVosLiberabit
    • 2 years ago
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    so it seems the 2*10^8 is meant to throw you off because it is irrelevant to finding the distance

  6. shortcake1970
    • 2 years ago
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    and the distance light travels in a year is 9.47 X 10^15 i have that much

  7. calculusfunctions
    • 2 years ago
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    Now find the number of seconds in one Earth year. Yes @VeritasVosLiberabit , that is extraneous information. Not needed.

  8. VeritasVosLiberabit
    • 2 years ago
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    it might be better to keep your calculations in MKS units unless told otherwise.

  9. calculusfunctions
    • 2 years ago
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    Alright @shortcake1970 now simply multiply that answer by 50.

  10. shortcake1970
    • 2 years ago
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    so do I just multiply?

  11. calculusfunctions
    • 2 years ago
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    Haha that's what I said.

  12. shortcake1970
    • 2 years ago
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    oops yup you just answered that

  13. shortcake1970
    • 2 years ago
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    Thank you!

  14. calculusfunctions
    • 2 years ago
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    Welcome!

  15. calculusfunctions
    • 2 years ago
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    Need help with anything else or are you okay?

  16. shortcake1970
    • 2 years ago
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    well I do have another questions I might need help with let me write it...

  17. calculusfunctions
    • 2 years ago
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    Sure

  18. calculusfunctions
    • 2 years ago
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    Are these questions you've tried yourself? It's important you try them yourself first.

  19. shortcake1970
    • 2 years ago
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    a beam of a certain type of elementary particle travels at a speed of 2.8 X 10^8 m/s. at this speed the average lifetime is measured to be 2.5 X 10^-8 s. what is the particle's lifetime at rest. let me tell you what formula i was thinking of using

  20. shortcake1970
    • 2 years ago
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    once i know which formula i can work it out myself :)

  21. calculusfunctions
    • 2 years ago
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    Sorry I was away for a minute. Just give me second. I'm just in the middle of something.

  22. shortcake1970
    • 2 years ago
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    ok i'm having a hard time typing the formula

  23. shortcake1970
    • 2 years ago
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    delta t=delta t o /square root of 1-v^2/c^2

  24. calculusfunctions
    • 2 years ago
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    OK I'm back. Yes\[\Delta t =\frac{ \Delta t _{0} }{ \sqrt{1-\frac{ v ^{2} }{ c ^{2} }} }\]

  25. calculusfunctions
    • 2 years ago
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    Do you understand? \[\Delta t = 2.5\times10^{-8}s\]\[v =2.8\times10^{8}m/s\]\[c =3.0\times10^{8}m/s\]\[\Delta t _{0}=?\]Rearrange the formula first to get\[\Delta t _{0}=\Delta t \sqrt{1-\frac{ v ^{2} }{ c ^{2} }}\]

  26. shortcake1970
    • 2 years ago
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    .90 x 10^8 s

  27. calculusfunctions
    • 2 years ago
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    Give me a second to calculate it.

  28. shortcake1970
    • 2 years ago
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    is the answer for the previous question 4.73 x 10^17?

  29. calculusfunctions
    • 2 years ago
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    It should be\[9.0\times10^{-9}\]

  30. calculusfunctions
    • 2 years ago
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    That exponent is negative 9

  31. shortcake1970
    • 2 years ago
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    oh i see

  32. calculusfunctions
    • 2 years ago
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    Do you? Did you spot your error?

  33. shortcake1970
    • 2 years ago
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    yes i spotted the error

  34. calculusfunctions
    • 2 years ago
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    Great. I have to go but is there anything else before I do?

  35. calculusfunctions
    • 2 years ago
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    I guess not. I might be back a little later. Bye.

  36. shortcake1970
    • 2 years ago
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    thank you so much.

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