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(answered) – 1) The average density of the Moon is 3.34 g/cm3 – less dense


(answered) – 1) The average density of the Moon is 3.34 g/cm3 – less denseDescriptionSolution downloadThe Questionsee attachmentsee attachmentsee attachmentsee attachment1) The average density of the Moon is 3.34 g/cm3 ? less dense than theterrestrial planets. The densities of water, rock, and metal are roughly 1g/cm3, 2.5 g/cm3, and 8 g/cm3 respectively. With this information, whatcan you infer about the Moon?s composition?2) The giant-impact theory suggests that the Moon formed out of debrisfrom a collision between young Earth and a Mars-sized body. How doesthe Moon?s density support this theory?3) Recently, images from New Horizons have revealed that Pluto?s largestmoon, Charon, has a belt of fractures and canyons near the equator. Theregions south of the canyon have fewer craters than regions in the north.What can you infer about the relative age of the terrain in the south?4) What assumptions did you use to answer part a)?5) Which element would you want to use to date:Hint: Think about the half-life of each radioactive element. The textbookand lecture slides have this information.?i) Ancient Greek ruins (~3000 years old)??ii) A Martian meteorite (~1 billion years old)? (1 pt)6) Below is a Strontium-Rubidium dating curve for an unknown sample ofa given age. On the plot, draw and label what the dating curve looked likefor this sample right after it formed, and what it will look like when it istwice as old as it is now. What was the rock?s original ratio of Strontium87/Strontium-86?7)The Nakhla meteorite is a meteorite from the surface of Mars thatimpacted Eartha)Using your mass spectrometer, you measure 10 units of Potassium-40 forevery unit of Argon-40. Calculate the age of this rock assuming there wasno initial Argon-40. Remember that when Potassium-40 decays, it decaysinto Calcium-40 about 90% of the time and Argon-40 about 10% of thetime.b) Why do we assume that there was no initial Argon-40? (1 pt)c) How could you check the calculated age? (1 pt)


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