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If you don't want to do the calculation yourself, you can use an online calculator like the one provided by NOAA. Alternatively, if you do want to do the calculation yourself you can use the haversine formula. This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is, the shortest distance over the earth’s ...


9

Isothermal, as you are likely aware means constant temperature. Your problem has the givens: The 0 - 1 km layer has a constant temperature of 0 C A surface parcel initially has a temperature of 0.5 C We can likely assume the parcel is not saturated and you are aware that that a parcel with a positive relative temperature perturbation is buoyant and will ...


8

45E is directly opposite 135W. It's therefore obvious (hopefully) that the shortest path is straight over the North pole (which is at 90 degrees North). The path takes us round 120 degrees of the Earth's circumference, or 1/3rd of it.


8

You are correct in thinking that oxygen comes from photosynthesis. In fact it is so much associated with photosynthesis, as opposed to any inorganic process, that the presence of oxygen in the atmospheric spectrum of other planets (solar or exoplanets) is reckoned to be one of the best indicators of life beyond Earth - not that such oxygen has been ...


8

To form a cloud, generally speaking, pressure does not play a significant factor (with the exception of the effect on temperature). If you want to prove that to yourself, go outside when it is foggy and measure the air pressure. Fog is a cloud, and pressure likely hasn't changed as much as it does in the vertical direction. As far a forming a cloud, it is ...


8

From this paper from 1958 (page 2), ... gaseous nitrogen of the atmosphere represents a vast store of potential fertility. It is not directly available to plants. Nitrogen-fixing bacteria, how- ever, absorb this gas from the soil solution and convert it to cell protein. When the cells die, other microbes attack the protein and convert the nitro- ...


7

It does make sense. As you said - glaciers have lower 18O/16O. This is because they have less 18O overall. Therefore, the oceans have more 18O. But this is today. At the Cretaceous there were no continental ice sheets, so what we said previously is not relevant. There are no ice sheets to take all of the 16O, so it's all in the ocean. This results in a ...


7

You say you're "not given the declination angle", but you also say "the horizontal direction of magnetism of these lavas is due west". That's your declination angle, right there! Since I assume (from the very round-numbered location) that this is a homework exercise or similar, I won't work through the rest of the calculation here. However, if you need more ...


7

The database with the negative longitude angles is storing longitude as eastings and westings, whereas the other database is storing everything as eastings. It appears you prefer your data as eastings and westings. Your methodology for changing eastings greater than 180 degrees to westings is correct: where the longitude angle is greater than 180 subtract ...


7

This is primarily an exercise in interpreting a minimalist diagram. Believe me, you will have to learn to read even sketchier diagrams. That said, the test had the correct answer to the question. Unambiguously, there are only three plates. The left and right plates each contain a piece of continental crust and a piece of oceanic crust connected by a ...


6

1. What is the youngest/oldest rock in the geological history of this map. Your approach is the right one but isn't the limestone underlain by something? Drawing the AB cross section would be of great help here, did you manage to do that? 2. What type of fold is shown on the map. What is the definition of a syncline and does your interpretation agree ...


6

Remember that you are concerned about stability fields. The lines on your stability diagram are the places where two minerals are in equilibrium. One one side one mineral will be more stable, on the other side the other one will be. Let's talk about $\ce{Ca}$ first. The reactions all have the same $\ce{Ca}$ so it's activity isn't a factor in their relative ...


6

Everything you did looks reasonable. There is one error in the 'net inflow into lake' line where you take the area of the lake to be $100\cdot10^3\ \text{m}^2$ instead of $100\cdot10^6\ \text{m}^2$, but that might just be a typo. In general, this type of problem is much easier to work (especially when you think you've made an error) by keeping everything ...


6

There is a quite relevant paper in Science with a fine analysis and interpretation on $\delta$ 18O evolution during the Phanerozoic Eon, which quite summarily display how this ratio depend on the sea level among other things. Full reference here : Miller, K. G.; Kominz, M. A.; Browning, J. V.; Wright, J. D.; Mountain, G. S.; Katz, M. E.; Sugarman, P. J.; ...


6

Zircons from Australia at 4.4Ga, and perhaps basalts from a Canadian island at 4.5Ga. It would interest you to know the quest of Boston University geochemist Matthew Jackson, who is searching for the oldest basalt and mantle rock that exists. His work isn't entirely agreed upon, as they are using all kinds of difficult isotopes which are a subject of ...


5

The first step I did was divide each wt% by their molecular weights to get a molecular proportion: Actually that's not what you did. What you did is how much moles of each oxide you have assuming 100 grams of mineral. But that's less important now. You are mostly correct in your steps. Eventually you have 0.972 Fe, 0.375 Ti, and 0.972 + 2×0.375 = 1.722 O. ...


5

There are no dips given, so you have to work by pure geometric deduction. The sediment subcrops on the east side all seem to be roughly parallel to the contours, which suggests near horizontal stratification. On the western side the diorite-shale subcrop is higher, suggesting about an 80 metre downthrow to the east. The aplite boundary shows no curvature in ...


5

Ok, based on the comments above, this is where I am at... earth's radius is $6371 \text{ km}$ earth is a sphere There is approximately $3 \cdot 10^{15}$ kg of CO2 in earth's atmosphere The density of CO2 at 1atm is $\dfrac{1.977 \text{ kg}}{\text{ m}^3}$ So, first, with assumption (3 and 4) we can compute the volume of CO2 in the atmosphere to be $\approx ...


5

This is a really bad question, because all three can lack fossils, or all three can have fossils. The definition of "minerals" is also a bit unclear, because fossils are also made out of minerals. I'll assume that by minerals they mean non-biogenic minerals. Let's go through each one of them: a clastic sedimentary rock This rock will have minerals, ...


5

What does 'nitrogen' mean? Nitrogen might mean gaseous $N_2$. It might also mean compounds, which contain nitrogen, such as nitrate, ammonia and ammonium. $N_2$ $N_2$ cannot be metabolized by most plants because it is to inert. You need bacteria or plants (e.g. Legume family plants) to perform nitrogen fixation. During this processes the $N_2$ is ...


5

There are three plates, because there are two plate boundaries (marked in blue): a convergent margin to the left, and a divergent margin to the right. Lithospheric plate #2 contains only oceanic crust. Lithospheric plates #1 and #3 contain both continental and oceanic crusts. Note the difference between the terms plate and crust. The place where ...


4

It does seem like it's impossible to know unless you have additional information. However, I think there is a hint in there. See this white halo around the intrusion? My guess (and I could be wrong here) is that it's not there for artistic reasons but rather it's there to provide a very strong hint. My feeling that this is some kind of metamorphosed contact ...


4

Inclination is given by $$I=\arctan(2\cot\theta)$$ where theta is equal to 90° minus the observer's latitude (in the Northern Hemisphere), therefore $$I=\arctan(2\cot(90-40))=59,21^\circ$$ To get declination the following figure helps: Declination is the angle between geographic north and magnetic north. (I suppose 80°N, 72°W are the coordinates of the ...


4

From the dips it is obvious that the stratigraphic structure is a syncline, in which case you correctly deduce that the youngest layered rock is the mudstone. But look carefully. The dips on either side of the fold axis are more than at right angles to each other, indicating an open fold of about a 110 degree deformation. Now consider the basalt dykes: ...


4

I am assuming you are asking for the case of sea level being 200m higher and in isostatic equilibrium. In that case we can make use of Airy's isostasy model: https://en.wikipedia.org/wiki/Isostasy#Airy Applied to a water column over the mantle, you have to replace $\rho_c$ with your $\rho_w$. The total increase in ocean depth is $x = b_1 + h_1$, where $h_1$ ...


4

The magnitude of these seasonal variations differ from location to location. The graph below portrays variations in CO2 levels at Point Barrow Alaska (PTB), La Jolla California (LJO), Mauna Loa Observatory (MLO), Christmas Island (CHR), Samoa (SAM), and the South Pole (SPO) over the last 60 years. Source: Scripps $\text{CO}_2$ Program, Global Stations $\...


4

The respiration of plants is not evenly distributed through seasons since the northern and southern hemisphere have a different amount of land and plants. The southern hemisphere has far less land. In the northern winter, the carbon released increases and then by ~May the northern hemisphere has enough carbon uptake (due to increased photosynthesis) to ...


4

Although this is a somewhat poorly researched question, I think there is an opportunity to make a nice point here. First of all, let's look at the Mohs scale: Talc Gypsum Calcite Fluorite (fluor)Apatite K-Feldspar Quartz Topaz Corundum Diamond Their (simplified) chemical formulae are: $\ce{Mg3Si4O10(OH)2}$ $\ce{CaSO4\cdot2H2O}$ $\ce{CaCO3}$ $\ce{CaF2}$ $\...


4

As long as we deal with a spherically symmetric planet, the mass of the shell above you does not influence the observed gravity; the little mass close by exactly cancels the large amounts of mass far away (you would float inside a planet if it looked like a pingpong ball and had all its mass centered on the surface!), this is called Newton's Shell Theorem. ...


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