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darkwhale

  • 2 years ago

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?

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

  2. darkwhale
    • 2 years ago
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    2. t=Vo/A

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

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

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

  6. darkwhale
    • 2 years ago
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    i'm not sure how to get Vo

  7. darkwhale
    • 2 years ago
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    oh is it 12m/s?

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

  9. darkwhale
    • 2 years ago
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    but dont i need both Vo and t?

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

  11. imron07
    • 2 years ago
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    Correct :)

  12. darkwhale
    • 2 years ago
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    so its 0.170?

  13. imron07
    • 2 years ago
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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 \[t_m=0.170t_e\].

  14. darkwhale
    • 2 years ago
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    yeah Δt=tm−te

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