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darkwhale

Astronauts on our moon must function with an acceleration due to gravity of 0.170g . If an astronaut can throw a certain wrench 12.0m vertically upward on earth, how high could he throw it on our moon if he gives it the same starting speed in both places? How much longer would it be in motion (going up and coming down) on the moon than on earth?

  • one year ago
  • one year ago

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  1. salini
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    u know the acceleration on earth and moon for a given distance say x calculate teh time of motion

    • one year ago
  2. darkwhale
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    2. t=Vo/A

    • one year ago
  3. imron07
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    Find h with this: \[v^2=v_0^2+2ah\] Since v=0 at highest point, you can find then: \[0=v_0^2+2ah \\h=\sqrt{\frac{-v_0^2}{2a}}\\\frac{h_{m}}{h_{e}}=\frac{\sqrt{\frac{-v_0^2}{2a_m}}}{\sqrt{\frac{-v_0^2}{2a_e}}} \]

    • one year ago
  4. darkwhale
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    \[\frac{ 9.81 }{ 1.67 }12.0\] h=70.6, but how do u solve the second part?

    • one year ago
  5. imron07
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    \[v=v_0-gt\] again, \[v=0\] then you can find time traveled by wrench for each g's

    • one year ago
  6. darkwhale
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    i'm not sure how to get Vo

    • one year ago
  7. darkwhale
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    oh is it 12m/s?

    • one year ago
  8. imron07
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    You can find v0, but don't really need to. \[0=v_0-gt\\t=\frac{v_0}{g}\\\frac{t_e}{t_m}=\frac{v_0/g_e}{v_0/g_m} \]

    • one year ago
  9. darkwhale
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    but dont i need both Vo and t?

    • one year ago
  10. darkwhale
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    \[\frac{ Te }{ Tm } =\frac{ 1.67 }{ 9.8 }\]

    • one year ago
  11. imron07
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    Correct :)

    • one year ago
  12. darkwhale
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    so its 0.170?

    • one year ago
  13. imron07
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    Wait, i'm not native eng speaker. What do they mean when they say how much longer? If they mean \[\Delta t=t_m-t_e\] then we're wrong. What we found is \[t_m=0.170t_e\].

    • one year ago
  14. darkwhale
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    yeah Δt=tm−te

    • one year ago
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