User:RadicalOne/LitterBox

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This is my litterbox. Nothing - except this title - remains on this page for very long.
This page is not an article; it rarely contains meaningful content. My most frequent use for it is a text dump, and this is the main root of its name "Litterbox".

Time UBX
User:Tkgd2007/Userboxes/My time

 

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What happens when I play with the public litterbox
Needless to say, that last one never occurred.
 * (diff) (hist) . . Sandbox‎; 16:43 . . (+62,518) . . RadicalOne (talk | contribs) (I couldn't resist ;))
 * (diff) (hist) . . Sandbox‎; 16:46 . . (+759,560) . . RadicalOne (talk | contribs) (I couldn't resist ;) and I did learn something.)
 * (diff) (hist) . . Sandbox‎; 16:46 . . (+235,452,134,235,879,781,734,659,545,324,236,310) . . RadicalOne (talk | contribs) (Added all of Wikipedia)

Self Explanatory?
$$F_g = \frac{Gm_1m_2}{r^2}$$ and $$F_e = \frac{kq_1q_2}{r^2}$$

Coincidence?

Very Important
$$t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$

Very Notorious
$$i\hbar\frac{\partial}{\partial t} \Psi(\mathbf{r},\,t) = -\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf{r},\,t) + V(\mathbf{r})\Psi(\mathbf{r},\,t)$$

Very Convenient
$$t = \frac{M + C}{2500*(1 + S)*2^{N}*Sp}$$<BR><BR>$$t_{DY} = \frac{60000 + 50000}{2500*(1 + 13)*2^{4}*5}$$<BR><BR>$$ = \frac{1.1 * 10^5}{2.8 * 10^6}$$<BR><BR>$$ = 0.03928h$$<BR><BR>$$ = 2m 21s$$

Arrangement Testing
My Music Contact Me Wiki-editorial Chase My Tail User Notices Userboxes Email Contact My Litterbox

Idea
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1
A transparent 6.283m x 6.283m x 6.283m box with a 1-cm high-scooped rim with the shape of a one-cycle sine wave is filled to the middle of that rim with of a supersaturated solution of CuSO4. This box rests one meter in front of a curtain whose bottom is similarly scooped, with the shape of a one-cycle cosine wave. The lowest point on the curtain is level with the lowest point in the box's rim. A wide-beam helium-neon laser, emission wavelength 632.8 nanometers, is placed behind the curtain, aimed at the box, set such that the entire box will lie within the beam, and that the angle of incidence is 90°. If a 6.283-meter-diameter solar panel with a rating of 10V/m2 is placed in the path of and directly facing the beam, six meters behind the box, hooked up to a series circuit of an 10mA LED and a 200Ω resistor, what happens to the LED? Solution This question is essentially asking for the area between $$y = \sin{x}$$ and $$y = \cos{x}, 0 < \sin{x} < \cos{x}$$. This because the amount of deep blue Copper (II) Sulfate solution the red light from the helium laser has to pass through will absorb all of it, leaving only the upper half of the waves to consider. Additionally, when the curtain ($$\cos{x}$$) is below the box's rim ($$\sin{x}$$), no light will pass, which is not the case in the reverse.

Numerous red herring are introduced, such as the distances between the objects, since the light is a laser light and thus does not appreciably spread or lose intensity over a distance as short as this. Also notice that the diameter of the solar panel and the width of the box is very close to 2π meters, the period of both sine and cosine functions, making the areas on a 1:1 ratio with the mathematical functions.

To calculate this area, we must calculate the area under each function, and find the difference. Thus, where A is the area of illumination on the solar panel, $$A = \int_{0}^{\frac{\pi}{4}}{\cos{x} - \sin{x}},dx$$.

We could go through and solve it with all the Riemann sums - $$\lim_{n \to \infty}{\sum_{i=1}^n {\cos{x_i} - \sin{x_i}}}\Delta x$$ - but that would be unpleasant and tedious. A much simpler method would be to apply one of the fundamental theorems of Calculus, which states the following: Thus, we must find the antiderivatives of sinx and cosx. These are of course, -cosx and sinx respectively, since $${\operatorname{d}y\over\operatorname{d}x}(\sin{x}) = \cos{x}~\mathrm{a}\mathrm{n}\mathrm{d}~{\operatorname{d}y\over\operatorname{d}x}(-\cos{x}) = \sin{x}$$. Therefore, the area of the difference between the two functions on the given interval - also the area of illumination of the solar panel - can be written as
 * If $$f(a) = \lim_{x \to a}f(x)~\mathrm{o}\mathrm{n}~[a,b],~\int_{a}^{b}f(x) = g(b)-g(a),~g'(x)=f(x)$$.

$$\begin{alignat}{2} A & = \int_{0}^{\frac{\pi}{4}}\cos{x}-\sin{x} \\ & = \int_{0}^{\frac{\pi}{4}}\cos{x}-\int_{0}^{\frac{\pi}{4}}\sin{x} \\ & = \sin{\frac{\pi}{4}}-\sin{0}-((-\cos{\frac{\pi}{4}})-(-\cos{0})) \\ & = \sin{\frac{\pi}{4}} + \cos{\frac{\pi}{4}}-1 \\ & = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}-1 \\ & = \frac{2}{\sqrt{2}}-1 \\ & = \sqrt{2}-1 \\ \end{alignat}$$ Therefore, (20.5 - 1)m² of solar panel are illuminated. Thus, the voltage produced is the rating per square meter times this, or 10(20.5) - 10 Volts, about 4.4142V. Now we need to calculate the resultant current in the circuit. As per Ohm's law, V = IR, or I = V÷R. Thus, I will be 10(20.5) - 10 divided by 100, about 0.0414 amperes, or 41.4 mA. This, being more than four times the current limit on the LED, will cause it to burn out.

<!--===2=== A radio telescope with a dish radius of nine meters and depth of 4.5 meters, stationed at 90° [N] latitude, is aimed at Polaris, and left overnight to scan for signals and/or stellar anomalies. The telescope is made entirely from a titanium alloy that costs $56 per kilogram; the telescope's materials cost $21 280 000. Unfortunately for the research, it rains overnight, from 2000 to 0400, at a rate such that a pool six meters in diameter forms in the bottom of the dish. Additionally, this rain contains a small amount of sulfuric acid due to acid rain, creating a pH of 3.1. The telescope is mounted on the end of a horizontal, uniform beam twelve meters long, with a cross-sectional area of 0.125m², made out of an alloy of density 8g/cm3, which is attached to a concrete wall by a joint with a rating of ±1 GN*m, and there is a strut directly under the telescope from the ground that can exert a maximum upward force of 80 meganewtons. What will the astronomers, returning the next morning, find? Solution This problem unifies elements from many fields, including calculus, statics, basic chemistry, and a bit of astronomy. We know the rating of the maximum torque the beam's joint can withstand. We need to figure out, if with the added weight of the water collected in the dish overnight, the total torque of the dish-beam-wall system now exceeds that value.

We start by calculating the torque that the joint is subjected to under normal conditions. Recall that $$\tau = F \times d = m*g*d*\sin{\theta}$$. Since θ = 90°, as the force is directly downwards, the sine can be eliminated. Thus, the torque from any one mass, in Newton-meters, is the product of the mass of the object, the gravitational constant, and the distance from the joint to that object's point of attachment. We need to also remember to sum all the torques acting on the rod. This includes that from the mass of the rod itself. We know the volume of the rod - the cross-sectional area times the length, or 12/16 cubic meters - and its density, so we can easily calculate its own mass: $$m = \rho * V = \frac{8g}{cm^3} * \frac{10^6cm^3}{1m^3} * \frac{12}{8}m^3 = 12 \times 10^6g = 12\mathrm{M}g$$. Recalling that the force of gravity is g * m, we know that the force of gravity on the rod is thus 117.6MN. We know that the point of action for this force is the center of mass, which, on a uniform beam like this, is halfway across, six meters from either end. Therefore the torque from the rod's mass is $$\tau _W = F * d = 117.6 * 6 = 705.6 \mathrm{M}N\cdot m$$.

We apply a similar method to calculate the torque from the radio dish; however, the mass is calculated differently. We know its materials cost $21280000, and that these materials cost $56 per kilogram. Therefore, there are 380 000 kg of material in the telescope. Since the telescope is on the end of the beam, its exerted torque is $$\tau _{tel} = F * d = m * g * d = 380000 * 9.8 * 12 = 44.69 \mathrm{M}N\cdot m$$.

Now we must calculate the torque, in the reverse direction, exerted by the vertical strut. Since it can exert a maximum force of 80 MN, its torque is thus 80*12 = 960 MN*m.

Therefore, under normal conditions, the joint is subjected to a total torque of $$\Sigma _{\tau} = \tau _W + \tau _{tel} - \tau_{strut} = 705.6 \times 10^6 + 44.69 \times 10^6 - 960 \times 10^6 = -209.71\mathrm{M}N\cdot m$$.

Therefore, the joint is normally under no torque, as the strut will simply only exert enough force as needed to counteract the masses of the rod and the telescope under normal conditions.

Now we need to calculate the mass of the rain. That will involve knowing its volume and its density. Recall that the density of water is defined as 1g/mL, and the solutes in rain are not in high enough concentrations to realistically offset this. Therefore, the density of the rain in the dish is 0.001kg/mL. Now we need to calculate the volume. Recall that Polaris is the North star, and that a radio telescope is parabolic; its cross-sectional shape is describable by a quadratic function. We know the dish to have a radius of nine meters. If we set the origin of the coordinate system to the center of the base of the dish, its edges now extend to x = -9m and x = 9m.

Normally, a pure quadratic function would have a value of 81m at that point, but since the depth of this dish is given as 4.5 meters, it must therefore be 1/18 as steep, and thus modeled by the function y = (1/18)x2. Now we calculate the area under the graph of that function in the interval [0,3], and revolve that around the y-axis for the difference between the volume of a cylinder of radius 9m and height 0.5 meters and the amount of water-acid solution we have. To calculate the area, we must use integral calculus:

$$\begin{alignat}{2} A_{2D} & = \int_{0}^{3}\frac{x^2}{18}\, dx \\ & = \left (\frac{1}{18}\right )\int_{0}^{3}x^2\, dx \\ & = \left (\frac{1}{18}\right )\frac{3^3}{3}-\frac{0^3}{3} \\ & = \left (\frac{1}{18}\right )\frac{27}{3}-\frac{0}{3} \\ & = \left (\frac{1}{18}\right )9-0 \\ & = \left (\frac{1}{18}\right )9 \\ & = \frac{1}{2} \\ \end{alignat}$$

Now, a rectangle of dimensions 3m by 0.5m would normally have an area of 1.5 square meters. But we know that the area under the parabola is occupying one third of that area in its two-dimensional positive cross-section, as 0.5 is one third of 1.5. One can then realize that the area above, where the water would be, is two-thirds of that area. Using similar curves, one can then deduce that the parabola's revolved volume will be two thirds of the cylinder created if the aforementioned rectangle was revolved around the y-axis. This means that we need only to calculate the volume of a cylinder with radius 3 meters and height 0.5 meters, then multiply that by 2/3 for the volume of the water.

$$\begin{alignat}{2} V_{cyl} & = 4\pi r^2h \\ & = 4\pi 9 0.5 \\ & = 18\pi \\ V_W & = \frac{2}{3} \cdot 18\pi \\ & = 12\pi \\ & \approx 37.5m^3 \end{alignat}$$

Now we need to calculate the mass of the fluid. It has a pH of 3.1, meaning it is acidic. Recalling that $$\mathrm{p}\mathrm{H} = -\log{[\mathrm{H_3O}^+]}$$, we can then determine that $$[\mathrm{H_3O}^+]=10^{-\mathrm{p}\mathrm{H}}$$. This of course yields a hydronium concentration of 7.94 * 10- mol/L. Recalling that H2SO4 is a diprotic acid, that means that the concentration of sulfuric acid is half that, or 3.97 * 10- mol/L. Now we can calculate the mass, since we know how much of each component we have. Since water has a defined density of 1.00, 12.566 cubic meters of it will thus weigh 12.566 metric tonnes, or 12.566Mg. The amount of sulfuric acid is not large enough to change the volume significantly, but it does alter the mass. Again recalling the density of water, we know we have 12.566kL of it, and when we take into account the [H2SO4], 4.99 moles of it. Recalling the molar mass of sulfuric acid to be 98.086 g/mol, we then know we have 489.52 grams of it. Now simply add the mass of the acid and the water, multiply by the gravitational constant, and we know the extra weight exerted on the telescope by the rain. This is of course (0.48952 + 37500) * 9.8, which equals 0.3675 meganewtons. We can now calculate the exerted torque, again recalling F*d, which comes to 4.41MN*m.

For the total torque that the joint is subjected to, we simply add this to the torque under normal conditions, assuming that the vertical strut is now exerting its maximal force of 80MN. This sum is $$\Sigma _{\tau} = \tau _W + \tau _{tel} - \tau_{strut} + \tau_{H_2O} = 705.6 \times 10^6 + 44.69 \times 10^6 - 960 \times 10^6 + 1.478 \times 10^9 = 1.268\mathrm{G}N\cdot m$$. That is of course far more than the joint is rated for, and thus the returning astronomers find their entire telescope assembly collapsed to the ground. -->