Function

Ah yes, well.. I forgot to put in the value of g.. My bad!

Hello everybody I have a question concerning the formulas of energy and its unit [math]E=\frac{m\cdot v^2}{2}[/math] [math]1 J=\frac{1 kg\cdot 1 m^2}{2\cdot 1 s^2}[/math] and [math]E=m\cdot g\cdot h[/math] [math]1 J=1 kg\cdot\frac{1 m}{1 s^2}\cdot 1 m=\frac{1 kg\cdot 1 m^2}{1 s^2}[/math] And because [math]E_k=E_p[/math] (e.g. a mass falling from a height its potential energy will be converted to kinetic energy when on its lowest point): [math]J=\frac{kg\cdot m^2}{s^2}=\frac{kg\cdot m^2}{2\cdot s^2}[/math] [math]\Leftrightarrow \frac{1}{2}=1[/math] What is wrong with my reasoning? Thanks! Function

Given: Vacid = 0,01 L VKOH = 0,02742 L cKOH = 0,168 M = 0,168 moles/L Asked: Macid = [math]\frac{m_{acid}}{n_{acid}}[/math] Conclusion: too few information given

Touché... Well, you know what I mean very simple in Dutch: no s at the end: Physics = fysica/natuurkunde; math(s) = wiskunde (or if you go a bit 'far'; mathematiek

Well, I just want to know if the last equation in my first post can be proven, in the form of, for example, |a| = |-a|, which is correct, and thus it is proven. A quick sketch of the tangent with tangent point P and intersections with asymptotes Q and R: To be proven: P is the middle of [QR] with Q and R given (in my first message). I wanted to prove this, other than by using the formula to get the distance |PQ| and |PR| or using the formula of calculating the mdidle of a segment ((x1+x2)/2;(y1+y2)/2). Therefore, I reasoned this: In order for P to be the middle of [QR], 2 requirements must be met: |xp-xr| = |xq-xp| and |yr+yp| = |yq-yp|. (I use the point symmetrical property: Q and R are point symmetrical over P) When I fill in the coördinates of R, Q and P, you get the last equation of my first message. Can that equation be proven, so I get something in the form of |a| = |-a| at the end, and thus I get something which proves that P is the middle of [QR]? Thanks. (Sorry for my lack of mathematical English, btw... I live in Belgium, and we don't really handle mathematical/scientifical English)

(Of course, I don't simply want to quadrate both sides)

Hello everyone In math class, we've proven something that had to do something with tha tangent of a hyperbola: the tangent point P is the middle point of [QR] with Q the intersection of the tangent and one assymptote, and R the intersaction of the tangent with the other assymptote. After we've calculated the values for: [math]Q\left( \frac{a\cdot\cos{t}}{1-\sin{t}};\frac{b\cdot\cos{t}}{1-\sin{t}} \right)[/math] [math]R\left( \frac{a\cdot\cos{t}}{1+\sin{t}};\frac{-b\cdot\cos{t}}{1+\sin{t}} \right)[/math] [math]P\left( \frac{a}{\cos{t}};b\cdot\tan{t} \right)[/math] we started expressing that P is middle of [QR], and getting an equality. However: I claimed this could've gone a bit more easy (my math teacher asked me therefore to present her a proof): P is the middle of [QR] [math]\Leftrightarrow Q [/math] and [math]R[/math] are point symmetrical over P. [math]\Leftrightarrow |x_q-x_p|=|x_p-x_r|[/math] [math]\Leftrightarrow \left|\frac{a\cdot\cos{t}}{1-\sin{t}}-\frac{a}{\cos{t}}\right|=\left|\frac{a}{\cos{t}}-\frac{a\cdot\cos{t}}{1+\sin{t}}\right|[/math]. But what now? I can't just scrap the "absolute value"-signs.. Is there a way to proof that equation? Thanks! -Function

let's all just speak Dutch can't be easier: Math(s?) = Wiskunde.. that's it. No s, no discussions on how to write it

Hello everyone! For a dissertation for biology, I've decided to perform a water quality test (apart from the usual biotic life; micro-invertebrats etc.). I'd like to know if there's a way to indicate (and the amount of; just indicating the presence won't give us information on the quality of the water...): Phosphate Ammonium Nitrate Nitrite Concentration of dissolved oxygen Thanks!

Small question in order to help you: what's the meaning of "0.168M"?

a) scale, grams b) scale, kilograms (use density of water, substract glass mass) c) Google Maps (don't know; what's "here"?), a car, perhaps, after resetting your kilometers counter?, kilometers d) scale, grams e) ok, now this is a bit more complicated: fill a glass (perfect cilinder, of which height and radius are known) with water. Calculate the volume of the water. Drop the pebble in the water, and recalculate the volume of the 'water'. Substract original volume of new volume, and you have the volume of the pebble. Weigh the pebble, using a scale, and use the mass and volume to calculate density. f) can't measure it with equipment, for the earth is not supported. You could base yourself on the formula for gravitational force (g): [math]g=G\cdot\frac{m}{r^2}[/math] with m = mass of the earth, r = radius of the earth and G = gravitational constant, approx. [math]6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}[/math] and g = approx. [math]9.81\frac{m}{s^2}[/math] and r = approx. 6.371 km. Just isolate m out of the formula, and you have the mass of the earth. Express result in kilograms or megaton, gigaton, whatever you prefer.