GingerAle

bmichelle9, are you a naturally rude idiot, or have you received special training?

Put the question back so others can learn from it.  

The layout of Heureka’s presentation is very clear. He enumerated the formula and its details to at least two prerequisite requirements.    

You’ve had two months to study this and said nothing until now; so, you should blôôdy well do it yourself and present it on here for evaluation, if you are unsure of your work. 

Use Boyle's Law

    For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. (The written form introduces ambiguities. Volume is measured in cubic units. It is not the cube of the volume.)

Boyle's Law expressed  mathematically:

\(P_1 V_1 = P_2 V_2\\ \text{ }\\ \text{1)}\\ V_2 = \dfrac {P_1 \cdot V_1}{P_2}\\ V_2 = \dfrac {125\; units \cdot 2\;units^3}{325\; units} = 0.76923\; units^3\\ \)

\(\text{2)}\\ P_2 = \dfrac {P_1 \cdot V_1}{V_2}\\ P_2 = \dfrac {125\; units \cdot 2\;units^3}{10\; units} = 25\; mass \;units\; per \;area \;units^2 \\ \)

No! 75mcg  is not 75mg

75mcg = 75 * 10^-6g

75mg = 75 *10^-3g

75mg is 1000 times more than 75mcg

Jeff, this is an example of messy:  Complex mathematics presented in ASCII text.

Mr. BB presentations as the Answer Man are usually just a plop of slop –a vast sea of ASCII soup coupled with descriptions of mathematical symbols pasted and plastered from Wolfram Alpha’s automated solutions. I usually need a double dose of Dramamine to prevent seasickness when I try to decipher them. This particular presentation is a little better than his usual slop.  This still requires wading, but you just need wading boot instead of the usual hip boots.  I didn’t need the Dramamine this time. 

Actually, I’m impressed that Mr. BB can query Wolfram proficiently enough to obtain this answer. Aside from Wolfram’s claim for natural language query, higher-level math usually requires precise wording or symbolic code to prevent absurd interpretations.

I was going to teach my cat how to do this, but my dog volunteered to teach him for me. (Good dog, Mr. Peabody!)

\(\text {Sir Alan's presentation in LaTeX}\\ \tiny \text {Reproduced verbatim (as much as practicable) with corrections from listed errata. }\\ \text { }\\ \small \text {definition: } \hspace{25 mm} years : = 315000000 {{\mkern 1mu\cdot\mkern 1mu}} s\\ \small \text {half-life: } \hspace{28 mm} \tau: =3{{\mkern 1mu\cdot\mkern 1mu}} years\\ \small \text {initial isotope power: } \hspace{6 mm} Q_0=100{{\mkern 1mu\cdot\mkern 1mu}}W\\ \text{ }\\ \small \text {isotope power as } \\ \small \text {a function of time: } \hspace{10 mm} Q(t): = Q_0{{\mkern 1mu\cdot\mkern 1mu}}e^{-ln(2)\cdot(\frac{t}{\tau})}\\ \small \text {plate area: } \hspace{22 mm} A: = 9{{\mkern 1mu\cdot\mkern 1mu}} mm^2\\ \small \text {plate distance: }\hspace{16 mm} x: = 5 {{\mkern 1mu\cdot\mkern 1mu}} mm\\ \)

\(\small \text {power received at plate } \\ \small \text {as a function of time: } \hspace{9 mm} P(t): = \frac{A \cdot Q(t)}{x^2}\\ \small \text {beginning of third year: } \hspace{9 mm} t_1: = 3 {{\mkern 1mu\cdot\mkern 1mu}} years\\ \small \text {middle of eighth year: } \hspace{12 mm} t_2: = 7.5\; years\\ \small \text { }\\ \small \text {total ENERGY received: } E:= \int_{t_1}^{t_2} P(t) \; dt\\ \text{ }\\ \hspace{55 mm} E=\frac{A}{x^2}\cdot Q_0 \cdot \frac{\tau}{ln(2)} \cdot (e^{\frac{ln(2)\cdot t_2}{\tau}} -e^{\frac{ln(2)\cdot t_2}{\tau}}) \text{ }\\ \hspace{55 mm} E = 2.224 \cdot 10^9 J\\\)

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By reproducing this, it seems I’ve demonstrated two proofs. Q.E.D.

1) If you prefer the format in LaTeX, you can do it yourself. It has the same messiness but with a different font. For an example of true messiness, see my troll post for the “Answer Man,” aka Mr. BB.

2) You are an idiot on multiple levels! You’ve risk alienating the only person (in regular attendance) who can answer this type of question.  You are well too far on the wrong side of the education gulf to be this arrogant. If you continue, you will only ever earn a dimwit degree in Stupid. 

Mr. BB has a dimwit degree in Stupid. He is about as educated as you are but he’s more arrogant. He uses his degree as a license to torture forum members with presentations so incompetent that it amuses my cat.  My cat likes him for some reason. I really don’t know why. Maybe because he makes him laugh or maybe because my cat is a (Monopoly) banker and embezzler, or maybe it’s because he likes rodents. It’s probably all three.     

Jeff, be sure to send Alan his $100.LOOOOL (Bitcoin should should work just fine).

This is an incredible presentation of incompetent SLOP: Wrong definition of the electrical resistance constant, wrong units for cross-sectional area, and wrong answer.

Classification: BatShìt-Stupìd.

------

R=0.000000012 ohms per meter per mm^2  <---- The symbol is ρ (rho) and the unit is the ohm-meter. (There is no “per” in this constant.)

L=0.5 meters <---  This should be 0.50. The trailing zero determines the precision of the measurement.   

A=20000mm^2 <--- This unit is wrong. This should remain in meters because the constant is defined in meters.

----------------

Formula: R_L=ρL/A

(0.50)(1.2E-8)/(0.02) = (3.0E-7) ohms @ 20°C <--- The constant is calculated at 20°C

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Mr. BB, stick to simple interest rate computations—my cat says you usually get those right.

I bought all the volumes. I’ve been counting the zeros to make sure I got what I paid for. I’ll let you know if they are all there—if any of you are still around, that is. 

After I’m finished, I’ll send the volumes to Sir CPhill, for a second verification. I know he’ll still be here: He can’t leave until the Roman zero in Sisyphus’ bolder is certified as authentic – and then he still has to write all the digits of PI in Roman numerals.  This will be a nice diversion for him, when he wants to take a break.