Liberal Forum

Blackvegetable » 22 Dec 2025, 11:38 am » wrote: ↑

22 Dec 2025, 11:38 am

You MUST be without issue....

Bboo is always on the run from the truth.      

roadkill » 22 Dec 2025, 3:16 am » wrote: ↑

22 Dec 2025, 3:16 am

https://www.youtube.com/watch?v=9LcOP8mhw4c

Of course.....

Maria Bartiromo...


One of the dumbest ruminants on the planet...
 

Blackvegetable » 22 Dec 2025, 9:27 am » wrote: ↑

22 Dec 2025, 9:27 am

Run from all gold bugs..

When an article starts railing about Bretton Woods, log in to your Robinhood account and buy some Game Stop.

buy *Rigetti

One of the top 5 performers since the late 2022 bottom


 

Blackvegetable » 22 Dec 2025, 9:27 am » wrote: ↑

22 Dec 2025, 9:27 am

Run from all gold bugs..

When an article starts railing about Bretton Woods, log in to your Robinhood account and buy some Game Stop.

The irony is, despite how dumb they are, they finally made good money in 2024 + 2025 after bagholding silver from $50 in 2011 and gold miners that have more or less gone nowhere from 2021-2023. 

The dumb retards were the ones who bought Bitcoin and thought it was headed to $200K...or who told "LOWIQTRASH THE CLAIRVOYANT" that he didn't know how to trade despite telling them to sell Nvidia anywhere after $200...
 

LowIQTrash » 22 Dec 2025, 12:15 pm » wrote: ↑

22 Dec 2025, 12:15 pm

The irony is, despite how dumb they are, they finally made good money in 2024 + 2025 after bagholding silver from $50 in 2011 and gold miners that have more or less gone nowhere from 2021-2023

The dumb retards were the ones who bought Bitcoin and thought it was headed to $200K

Have you ever seen A Boy and His Dog?



 

Blackvegetable » 22 Dec 2025, 12:00 pm » wrote: ↑

22 Dec 2025, 12:00 pm

Of course.....

Maria Bartiromo...

One of the dumbest ruminants on the planet...

A little message for you from Maxine Waters' double...      

https://www.youtube.com/watch?v=xmqTxynfcxw
 
 
 

Blackvegetable » 22 Dec 2025, 12:18 pm » wrote: ↑

22 Dec 2025, 12:18 pm

Have you ever seen A Boy and His Dog?

Nope, I am not well rounded in the cinema arts...

Blackvegetable » 22 Dec 2025, 12:18 pm » wrote: ↑

22 Dec 2025, 12:18 pm

Have you ever seen A Boy and His Dog?

We know YOU have!            

They cleaned up the audio..


Time well spent


https://youtu.be/qv6t8KgImlw?si=BDpDmVpmM4l3QwvO

roadkill » 22 Dec 2025, 12:20 pm » wrote: ↑

22 Dec 2025, 12:20 pm

We know YOU have!          

That's right...

Numerous times...


It's Harlan Ellison

You wouldn't understand it.
 

LowIQTrash » 22 Dec 2025, 12:19 pm » wrote: ↑

22 Dec 2025, 12:19 pm

Nope, I am not well rounded in the cinema arts...

It's an edifying illustration of the Diamond/Water paradox.

Blackvegetable » 22 Dec 2025, 1:00 pm » wrote: ↑

22 Dec 2025, 1:00 pm

It's an edifying illustration of the Diamond/Water paradox.

Is that similar to the "Parrotdox Evening News" that you subscribe to?        

roadkill » 22 Dec 2025, 2:05 pm » wrote: ↑

22 Dec 2025, 2:05 pm

Is that similar to the "Parrotdox Evening News" that you subscribe to?      

You're childless for cause, PedoStool...

Blackvegetable » 22 Dec 2025, 2:07 pm » wrote: ↑

22 Dec 2025, 2:07 pm

You're childless for cause, PedoStool...

Wow...that almost sounds plagiarized from a pompous old world Soviet.      


P.S.  I'm not the one wearing hounds tooth pants to lure a carp for a mate.     
 

roadkill » 22 Dec 2025, 5:31 pm » wrote: ↑

22 Dec 2025, 5:31 pm

Wow...that almost sounds plagiarized from a pompous old world Soviet.      

P.S.  I'm not the one wearing hounds tooth pants to lure a carp for a mate.   

There were incredible breakthroughs in funding for Ukraine through the EU. 90 Billion.

You are beached dead carp comrade.  You are not even elevated to status of roadkill any more.
Invade another tanker..  America snores..
 

Blackvegetable » 22 Dec 2025, 9:12 am » wrote: ↑

22 Dec 2025, 9:12 am

I had a chance to sell TAE 5 years ago....

You have no idea what they do..

OK Einstein, tell us.....
How much boron would it take to operate a 6800 mega watt per hour boron / hydrogen power plant for just one day...?
 
 
 

*GHETTOBLASTER » 23 Dec 2025, 3:04 am » wrote: ↑

23 Dec 2025, 3:04 am

OK Einstein, tell us.....
How much boron would it take to operate a 6800 mega watt per hour boron / hydrogen power plant for just one day...?

1.11 Kilograms.  And  billions of Kelvins.   

 
 

JohnnyYou » 23 Dec 2025, 3:26 am » wrote: ↑

23 Dec 2025, 3:26 am

1.11 Kilograms.  And  billions of Kelvins.

Provide your source and methodology

 
 

*GHETTOBLASTER » 23 Dec 2025, 3:31 am » wrote: ↑

23 Dec 2025, 3:31 am

Provide your source and methodology

AI Hallucination. Never ask a suceessful question twice, now it is somewhere between 1.1 and 7.7 Kg.   Too be honest I was not aware of the fusion fuel cycle. I do know lithium 6 is in hydrogen bombs for extra kick.

AI Overview     Approximately 2.3 kilograms of boron would be required to operate a hypothetical 6800 megawatt boron/hydrogen fusion power plant for one day. The calculation is based on the theoretical energy density of the p-B11 fusion reaction. Step 1: Calculate Total Energy Demand The power plant operates at a power of 6800 MW for 24 hours. The total energy generated is calculated as:\(E=\text{Power}\times \text{Time}=6800\text{\ MW}\times 24\text{\ h}\)\(E=163200\text{\ MWh}\)Convert the energy to Joules (note: \(1\text{\ MWh}=3.6\times 10^{9}\text{\ J}\)):\(E=163200\text{\ MWh}\times \frac{3.6\times 10^{9}\text{\ J}}{1\text{\ MWh}}\)\(E=5.8752\times 10^{14}\text{\ J}\)Step 2: Determine Energy per Boron Atom and Moles Required The proton-boron fusion reaction (\({}^{1}\text{p}+{}^{11}\text{B}\rightarrow 3\times {}^{4}\text{He}\)) releases 8.7 MeV of energy per reaction. Convert this energy to Joules (note: \(1\text{\ MeV}\approx 1.602\times 10^{-13}\text{\ J}\)):\(E_{\text{atom}}=8.7\text{\ MeV}\times \frac{1.602\times 10^{-13}\text{\ J}}{1\text{\ MeV}}\)\(E_{\text{atom}}\approx 1.3937\times 10^{-12}\text{\ J/atom}\)The total number of boron atoms required is:\(N=\frac{E}{E_{\text{atom}}}=\frac{5.8752\times 10^{14}\text{\ J}}{1.3937\times 10^{-12}\text{\ J/atom}}\)\(N\approx 4.2157\times 10^{26}\text{\ atoms}\)Using Avogadro's number (\(N_{A}\approx 6.022\times 10^{23}\text{\ atoms/mol}\)), find the number of moles of boron:\(n=\frac{N}{N_{A}}=\frac{4.2157\times 10^{26}\text{\ atoms}}{6.022\times 10^{23}\text{\ atoms/mol}}\)\(n\approx 700.05\text{\ mol}\)Step 3: Calculate Mass of Boron Required The reaction uses the boron-11 isotope (molar mass \(\approx 11.01\text{\ g/mol}\)). The total mass of boron is:\(m=n\times \text{Molar\ Mass}=700.05\text{\ mol}\times 11.01\text{\ g/mol}\)\(m\approx 7707.55\text{\ g}\)Convert the mass to kilograms:\(m\approx \mathbf{7.7}\text{\ kg}\)Answer: Approximately 7.7 kilograms of boron would be needed to operate a 6800 megawatt boron/hydrogen fusion power plant for one day.
 
 

JohnnyYou » 23 Dec 2025, 3:42 am » wrote: ↑

23 Dec 2025, 3:42 am

AI Hallucination. Never ask a suceessful question twice, now it is somewhere between 1.1 and 7.7 Kg.   Too be honest I was not aware of the fusion fuel cycle. I do know lithium 6 is in hydrogen bombs for extra kick.

AI Overview     Approximately 2.3 kilograms of boron would be required to operate a hypothetical 6800 megawatt boron/hydrogen fusion power plant for one day. The calculation is based on the theoretical energy density of the p-B11 fusion reaction. Step 1: Calculate Total Energy Demand The power plant operates at a power of 6800 MW for 24 hours. The total energy generated is calculated as(E=\text{Power}\times \text{Time}=6800\text{\ MW}\times 24\text{\ h}\)\(E=163200\text{\ MWh}\)Convert the energy to Joules (note: \(1\text{\ MWh}=3.6\times 10^{9}\text{\ J}\))(E=163200\text{\ MWh}\times \frac{3.6\times 10^{9}\text{\ J}}{1\text{\ MWh}}\)\(E=5.8752\times 10^{14}\text{\ J}\)Step 2: Determine Energy per Boron Atom and Moles Required The proton-boron fusion reaction (\({}^{1}\text{p}+{}^{11}\text{B}\rightarrow 3\times {}^{4}\text{He}\)) releases 8.7 MeV of energy per reaction. Convert this energy to Joules (note: \(1\text{\ MeV}\approx 1.602\times 10^{-13}\text{\ J}\))(E_{\text{atom}}=8.7\text{\ MeV}\times \frac{1.602\times 10^{-13}\text{\ J}}{1\text{\ MeV}}\)\(E_{\text{atom}}\approx 1.3937\times 10^{-12}\text{\ J/atom}\)The total number of boron atoms required is(N=\frac{E}{E_{\text{atom}}}=\frac{5.8752\times 10^{14}\text{\ J}}{1.3937\times 10^{-12}\text{\ J/atom}}\)\(N\approx 4.2157\times 10^{26}\text{\ atoms}\)Using Avogadro's number (\(N_{A}\approx 6.022\times 10^{23}\text{\ atoms/mol}\)), find the number of moles of boron(n=\frac{N}{N_{A}}=\frac{4.2157\times 10^{26}\text{\ atoms}}{6.022\times 10^{23}\text{\ atoms/mol}}\)\(n\approx 700.05\text{\ mol}\)Step 3: Calculate Mass of Boron Required The reaction uses the boron-11 isotope (molar mass \(\approx 11.01\text{\ g/mol}\)). The total mass of boron is(m=n\times \text{Molar\ Mass}=700.05\text{\ mol}\times 11.01\text{\ g/mol}\)\(m\approx 7707.55\text{\ g}\)Convert the mass to kilograms(m\approx \mathbf{7.7}\text{\ kg}\)Answer: Approximately 7.7 kilograms of boron would be needed to operate a 6800 megawatt boron/hydrogen fusion power plant for one day.

So in other words you weren't able to come up with a reliable solution.
I already knew this would happen....been there done that.
Maybe next time you will look before you leap....[yeah right...   ]