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Muhammad Arifur R.
Tutor of Science & Mathematics
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Geology
TutorMe
Question:

According to Bowen's Reaction Series, quartz is the mineral that forms at the very last stage of crystallization, provided that there is excess silica. But we clearly see an abundance of quartz around us. Look at the buildings: the main constituent is sand, which is mostly of quartz. Look at the beaches, most of them are covered with sand. Even dirt, which comes from mudstones, contains quartz as framework grain. Explain - despite always forming lately, why quartz is so abundant in nature.

Muhammad Arifur R.
Answer:

If we analyze Bowen's Reaction Series, we can see that at the very beginning of the crystallization, when the temperature is around 1200°C, olivine starts to crystallize first. As the temperature falls slowly, minerals of medium silication like albite, pyroxene, amphibole, biotite, anorthite crystallize. These minerals occupy most of the space in magma. At around 600°C, if there's an excess of silica $$(SiO_2)$$ in the magma, quartz begins to crystallize very slowly. This is why quartz is a mineral of high silication $$(> 60\%)$$. Now, as other minerals have occupied most of the space, there's little space for quartz to crystallize. So, due to $$\text{lack of free space}$$ and $$\text{extremely slow crystallization}$$, quartz grains become tightly bonded within the crystal lattice. That's why quartz has a very well-ordered atomic arrangement. This makes quartz one of the hardest $$(7 - \text{Mohs Hardness Scale})$$ minerals of the earth. Consequently all of these minerals undergo weathering & erosion by geomorphic agents (wind, water). Quartz being so well-crystallized, doesn't weather & erode as much as mica, feldspars do. So even when the other minerals have entirely been transformed to clay minerals like kaolinite, quartz sustains its composition and lattice structure. And quartz doesn't react with most of the chemical entities. So even after undergoing geologically extensive transportation, quartz is not much affected by weathering & erosion. And that's why there's an abundance of quartz around us. Quartz has a protection from weathering & erosion - a shield of well-crystallized crystal structure.

C Programming
TutorMe
Question:

$( S_n = 1 + 2 - 3 \times 4\div 5 + 6 - 7 \times 8 \div 9 +10- \cdots n $) Consider the above sum where the first $$n$$ positive integers are listed in ascending order, and the mathematical operations $$(+,-,\times,\div)$$ are repeated in that order. Find the value of $$S_{31417} $$. Give your answer to 3 decimal places. (Little More) Details: - BODMAS (Order of Operations) is obeyed. - A couple of explicit examples: $(\begin{align} S_4 &= 1 + 2 - 3\times4 = (1+2) - (3\times4) = 3 - 12=-9 \\ S_5 &= 1+2-3\times4\div5 = (1+2) - (3\times4\div5) = 3 - 12\div5 = 0.6. \end{align} $)

Muhammad Arifur R.
Answer:

Answer: $$-3.598$$ Explanation: You can identify the sequence of the operators. $(s=1+(2-3\times \frac{4}{5})+(6-7\times \frac{8}{9})+(10-11\times \frac{12}{13})+\dots \dots$) Now, get ready for the loop. As you see, that $$1$$ is kept out of the braces, you have to set the initial value of Sum, $$s=1;$$. Then you take four variables: a,b,c,d. You see, always, $(b=a+1$)$(c=b+1$)$(d=c+1$) And it occurs again and again. In the 1st loop, $$a=2$$. You can declare it inside the $$for$$ loop. To get the numbers of the 2nd loop, $$a+=4$$. Because then you get $$a=6$$ in the 2nd, and $$a=10$$ in the 3rd loop and blah blah blah! So, the for loop is $$for(a=2; ; a+=4)$$ . Look carefully, we didn't specify the condition for the loop, because using the $$break;$$ statement, we can do that job. Then, inside the loop, first produce the other numbers, $$b,c,d$$. Now comes the most important. If $$a=n$$ , $$s+=a;$$ then $$Break$$ the loop. But if $$b=n$$, then $$s+=a-b;$$ , and $$Break$$ ! Similar for $$c=n$$ and $$d=n$$ . Notice the corresponding lines of these statements in the sample code below. If none of them is equal to $$n$$, that means that they haven't reached the value of $$n$$ yet, which will happen from the beginning. For that, the statement is, $$s+=a-((b*c)/d);$$ . Remember, no $$break;$$ this time, as it needs to continue until any of the variables reach the $$n$$ . So, that's it. Here's the sample code in C++ after all this detailed explanation... Compile and run it, enter $$31417$$ and get the answer! #include<stdio.h> #include<math.h> int main() { double a, b, c, d, s, n; printf("What's the n: "); scanf("%lf", &n); s = 1; for (a = 2;; a += 4) { b = a + 1; c = b + 1; d = c + 1; if (a == n) { s += a; break; } else if (b == n) { s += (a - b); break; } else if (c == n) { s += a - (b * c); break; } else if (d == n) { s += a - ((b * c) / d); break; } else { s += a - (b * c / d); } } printf("The Summation, s= %lf", s); printf("\n"); return 0; }

Physics
TutorMe
Question:

An ice cube of side length $$a_0$$ is placed in a cylindrical container with a base area of $$A$$. After some time, ice starts to melt uniformly. Find the expression for the length of the ice cube's edge $$a$$, right at the moment when it leaves the contact with the base of the container. Given, Density of water = $$\rho_w$$ Density of ice = $$\rho_i$$

Muhammad Arifur R.
Answer:

Answer: $(\text{Edge Length, a}={\frac{ {a_0}^3}{A}}$) Its one of the many possible solutions. Explanation: Assume, Edge length (before)=$$a_0$$ Exactly when the cube starts floating, Edge length=$$a$$ Height of the submerged portion=$$h$$ Density of ice=$$\rho_i$$ Density of water=$$\rho_w$$ So, Mass of previous cube = Mass of current cube + Mass of water Water volume $( =\text{A}h-a^2h$) So, $(\rho_i{a_0}^3=\rho_ia^3+\rho_w h (\text{A}-a^2)$) $(\rightarrow {a_0}^3-a^3={\frac{\rho_w}{\rho_i}} h (\text{A}-a^2)$) Then, Mass of the current cube = Mass of $$a^2 h$$ Volume of water. $(\rho_i a^3=\rho_w a^2 h$) $(\rightarrow h={\frac{\rho_i}{\rho_w}} a$) Plug the value to previous equation: $({a_0}^3-a^3={\frac{\rho_w}{\rho_i}}h (\text{A}-a^2)$) $(\rightarrow {a_0}^3-a^3={\frac{\rho_w}{\rho_i}}{\frac{\rho_i}{\rho_w}} a (\text{A}-a^2)$) $(\rightarrow {a_0}^3-a^3=a(\text{A}-a^2)$) See, the density is gone from the expression! So let's do the final job: $({a_0}^3-a^3=\text{A}a-a^3$) $(\rightarrow a=\boxed{{\frac{ {a_0}^3}{A}}}$) The problem seems so complex at first glance. But its simple, and excellently interesting, isn't it?

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