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What is the difference between the color and streak of a mineral?



Rock A has a piece of rock B included inside which is older and how do you know?



How do plate tectonics explain the formation of metamorphic rocks?



A layer of sandstone is found on top of a layer of limestone in a geologic cross section. what does this tell you about how the sea level changed in that area?



explain how a metamorphic rock can change into another type of metamorphic rock?



Answer those and I will give you 10 points!!!

2007-11-11 01:50:25 · 2 answers · asked by King Violation. 5 in Science & Mathematics Earth Sciences & Geology

2 answers

All in your lab manual but I feel generous today.

1) Streak is the same for a mineral regardless of its hand sample colour. Hematite can be a variety of colours in sample but it will always streak red.

2) Law of Inclusions. The inclusion must be older. It had to exist in order to be included. Rock B is older.

3) Metamorphic rocks form under high pressure and temperature. These conditions can be achieved at some plate boundaries, particularly convergent ones.

4) Assuming the limestone is older, it got shallower. Limestone is deep marine. Sandstone is tidal.

5) Add more heat and/or pressure.

2007-11-11 02:51:47 · answer #1 · answered by Lady Geologist 7 · 1 0

1) Colour is, well, the colour of the rock. It's not the most reliable tool, however, because it can change due to weathering. Streak is the powdered form of a mineral, and usually a better property to use when you're identifying rocks.

2) Rock B is older, because rock A probably formed around it. There would be no way for rock B to just insert itself into rock A!

3) Metamorphic rocks make up most of the Earth's crust. They are formed by continental collisions, which cause pressure, friction, and heat.

4) The sea level went down?

5) A Metamorphic rock could be formed, and then metamorphed later again into another type of rock. (Example: Slate --> Schist ---> Gneiss)

2007-11-11 16:40:06 · answer #2 · answered by Anonymous · 0 0

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