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anonymous
 4 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?
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u know the acceleration on earth and moon for a given distance say x calculate teh time of motion

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Find h with this: \[v^2=v_0^2+2ah\] Since v=0 at highest point, you can find then: \[0=v_0^2+2ah \\h=\sqrt{\frac{v_0^2}{2a}}\\\frac{h_{m}}{h_{e}}=\frac{\sqrt{\frac{v_0^2}{2a_m}}}{\sqrt{\frac{v_0^2}{2a_e}}} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 9.81 }{ 1.67 }12.0\] h=70.6, but how do u solve the second part?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[v=v_0gt\] again, \[v=0\] then you can find time traveled by wrench for each g's

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm not sure how to get Vo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can find v0, but don't really need to. \[0=v_0gt\\t=\frac{v_0}{g}\\\frac{t_e}{t_m}=\frac{v_0/g_e}{v_0/g_m} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but dont i need both Vo and t?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ Te }{ Tm } =\frac{ 1.67 }{ 9.8 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait, i'm not native eng speaker. What do they mean when they say how much longer? If they mean \[\Delta t=t_mt_e\] then we're wrong. What we found is \[t_m=0.170t_e\].
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