Anyone else just hear that massive explosion?

Always be mindful, no matter how bad a day you are having, no matter how terrible your life might be, there is a rock out there on it's way to the 3rd Rock from the sun....

All the idiots down here arguing over religion and who owns what lands, turn your heads to the skies , ye ignorant fools! for you will always be dust.....

/Moose on a sober Sabbath podium.... :sneaky:
 

Mystery surrounds unidentified flying object​

Is it a bird? Is it a plane?

Well, depending on who you ask, it could be a meteorite.

Residents in Kariega, Humansdorp, Despatch, and even Plettenberg Bay reported seeing and hearing the unidentified flying object early on Sunday morning.

The situation saw people flocking to social media and WhatsApp groups to report the unusual sighting and sound that reportedly plunged into the ocean in the vicinity of Oyster Bay.

Paradise Beach resident Bennett Roberts said he was sitting on his stoep at about 9am when he heard a loud booming noise “like someone was dancing on the roof”.

“They happened intermittently in quick success and moved in a path from south to west right over my house.

More: https://www.heraldlive.co.za/news/2024-08-25-mystery-surrounds-unidentified-flying-object/
 
tetrasect said:
It's an estimate. With a wide range.

Don't think anyone can work it out scientifically without proper measurements.
Click to expand...
Even then,it chose probable words in sequence based on the question. It did not interpret the question of verify the answer
 
PsyWulf said:
Even then,it chose probable words in sequence based on the question. It did not interpret the question of verify the answer
Click to expand...

How can anyone verify the answer? What are you talking about?

It is an estimate. Based the on the fact that the explosion could be heard 200km away and by comparing it with other meteorites, along with the fact that energy dissipates with the square of the distance.

It also said:

Important considerations:
  1. This is an approximate estimate. Actual energy could vary based on specific conditions.
  2. The altitude of the explosion greatly affects how energy is distributed and perceived on the ground.
  3. Atmospheric conditions play a crucial role in how far the sound and shock wave travel.
  4. Terrain features can affect how the explosion is felt and heard at different distances.
  5. The composition and speed of the meteor also influence the nature of the explosion and its effects.
Click to expand...

It's simply not possible to do anything but make an educated guess, since we do not have any actual measurements.

And just because ChatGPT works the way it works doesn't mean it's wrong.
 
tetrasect said:
How can anyone verify the answer? What are you talking about?

It is an estimate.
Click to expand...
ChatGPT cannot estimate
It lacks logic to understand how to estimate

It is a probabilistic language model

And just because ChatGPT works the way it works doesn't mean it's wrong.
Click to expand...
Anybody guessing might guess right or wrong. But that's like saying a Streetsweeper is not necessarily wrong about nuclear physics. LLMs say what sounds logical,some of them have had some extra wiring for calculus and other things to help overcome this limitation,but the limitation is there
 
PsyWulf said:
ChatGPT cannot estimate
It lacks logic to understand how to estimate

It is a probabilistic language model


Anybody guessing might guess right or wrong. But that's like saying a Streetsweeper is not necessarily wrong about nuclear physics. LLMs say what sounds logical,some of them have had some extra wiring for calculus and other things to help overcome this limitation,but the limitation is there
Click to expand...

Doesn't mean it's wrong. In fact it can answer scientific questions very well.

For example:

Question:
If I added 3.57gr of malic acid to a 1 liter soda bottle with a magnesium rod in it and filled it up with 5 degree Celsius water, what would the pressure inside the bottle be and how much hydrogen would be dissolved in the water? Also, what other chemicals would be present and at what concentration?

Answer:
  1. Reaction: C4H6O5 (malic acid) + 5Mg + 5H2O → 5Mg(OH)2 + C4H4MgO5 + 5H2
  2. Calculate moles of malic acid: 3.57g / 134.09 g/mol = 0.02662 mol
  3. Hydrogen produced: 0.02662 mol malic acid × 5 mol H2 / 1 mol malic acid = 0.1331 mol H2
  4. Pressure calculation: Using ideal gas law: PV = nRTP = nRT / VP = (0.1331 mol × 0.08206 L⋅atm/mol⋅K × 278K) / 1LP ≈ 3.03 atm or 307 kPa
  5. Dissolved hydrogen: Using Henry's law: C = kPk for H2 at 5°C ≈ 8.7 × 10^-4 mol/L⋅atmC = 8.7 × 10^-4 mol/L⋅atm × 3.03 atm = 0.00264 mol/L
  6. Mass of dissolved H2 = 0.00264 mol/L × 2.016 g/mol = 0.00532 g/L or 5.32 mg/L
  7. Other chemicals present:
a) Magnesium hydroxide (Mg(OH)2):0.02662 mol × 5 = 0.1331 mol
Mass = 0.1331 mol × 58.32 g/mol = 7.76 g
Concentration = 7.76 g/L
b) Magnesium malate (C4H4MgO5):0.02662 mol
Mass = 0.02662 mol × 172.37 g/mol = 4.59 g
Concentration = 4.59 g/L
c) Unreacted magnesium (if rod is in excess)
  1. Final solution composition:
  • Dissolved hydrogen: 5.32 mg/L
  • Magnesium hydroxide: 7.76 g/L (mostly as a precipitate)
  • Magnesium malate: 4.59 g/L (dissolved)
  • Pressure: 3.03 atm or 307 kPa
  • pH: Basic due to Mg(OH)2 formation
 
tetrasect said:
Doesn't mean it's wrong. In fact it can answer scientific questions very well.

For example:

Question:
If I added 3.57gr of malic acid to a 1 liter soda bottle with a magnesium rod in it and filled it up with 5 degree Celsius water, what would the pressure inside the bottle be and how much hydrogen would be dissolved in the water? Also, what other chemicals would be present and at what concentration?

Answer:
  1. Reaction: C4H6O5 (malic acid) + 5Mg + 5H2O → 5Mg(OH)2 + C4H4MgO5 + 5H2
  2. Calculate moles of malic acid: 3.57g / 134.09 g/mol = 0.02662 mol
  3. Hydrogen produced: 0.02662 mol malic acid × 5 mol H2 / 1 mol malic acid = 0.1331 mol H2
  4. Pressure calculation: Using ideal gas law: PV = nRTP = nRT / VP = (0.1331 mol × 0.08206 L⋅atm/mol⋅K × 278K) / 1LP ≈ 3.03 atm or 307 kPa
  5. Dissolved hydrogen: Using Henry's law: C = kPk for H2 at 5°C ≈ 8.7 × 10^-4 mol/L⋅atmC = 8.7 × 10^-4 mol/L⋅atm × 3.03 atm = 0.00264 mol/L
  6. Mass of dissolved H2 = 0.00264 mol/L × 2.016 g/mol = 0.00532 g/L or 5.32 mg/L
  7. Other chemicals present:
a) Magnesium hydroxide (Mg(OH)2):0.02662 mol × 5 = 0.1331 mol
Mass = 0.1331 mol × 58.32 g/mol = 7.76 g
Concentration = 7.76 g/L
b) Magnesium malate (C4H4MgO5):0.02662 mol
Mass = 0.02662 mol × 172.37 g/mol = 4.59 g
Concentration = 4.59 g/L
c) Unreacted magnesium (if rod is in excess)
  1. Final solution composition:
  • Dissolved hydrogen: 5.32 mg/L
  • Magnesium hydroxide: 7.76 g/L (mostly as a precipitate)
  • Magnesium malate: 4.59 g/L (dissolved)
  • Pressure: 3.03 atm or 307 kPa
  • pH: Basic due to Mg(OH)2 formation
Click to expand...

How the hell did we go from a meteor to blerry science equations...
 
tetrasect said:
Doesn't mean it's wrong. In fact it can answer scientific questions very well.

For example:

Question:
If I added 3.57gr of malic acid to a 1 liter soda bottle with a magnesium rod in it and filled it up with 5 degree Celsius water, what would the pressure inside the bottle be and how much hydrogen would be dissolved in the water? Also, what other chemicals would be present and at what concentration?

Answer:
  1. Reaction: C4H6O5 (malic acid) + 5Mg + 5H2O → 5Mg(OH)2 + C4H4MgO5 + 5H2
  2. Calculate moles of malic acid: 3.57g / 134.09 g/mol = 0.02662 mol
  3. Hydrogen produced: 0.02662 mol malic acid × 5 mol H2 / 1 mol malic acid = 0.1331 mol H2
  4. Pressure calculation: Using ideal gas law: PV = nRTP = nRT / VP = (0.1331 mol × 0.08206 L⋅atm/mol⋅K × 278K) / 1LP ≈ 3.03 atm or 307 kPa
  5. Dissolved hydrogen: Using Henry's law: C = kPk for H2 at 5°C ≈ 8.7 × 10^-4 mol/L⋅atmC = 8.7 × 10^-4 mol/L⋅atm × 3.03 atm = 0.00264 mol/L
  6. Mass of dissolved H2 = 0.00264 mol/L × 2.016 g/mol = 0.00532 g/L or 5.32 mg/L
  7. Other chemicals present:
a) Magnesium hydroxide (Mg(OH)2):0.02662 mol × 5 = 0.1331 mol
Mass = 0.1331 mol × 58.32 g/mol = 7.76 g
Concentration = 7.76 g/L
b) Magnesium malate (C4H4MgO5):0.02662 mol
Mass = 0.02662 mol × 172.37 g/mol = 4.59 g
Concentration = 4.59 g/L
c) Unreacted magnesium (if rod is in excess)
  1. Final solution composition:
  • Dissolved hydrogen: 5.32 mg/L
  • Magnesium hydroxide: 7.76 g/L (mostly as a precipitate)
  • Magnesium malate: 4.59 g/L (dissolved)
  • Pressure: 3.03 atm or 307 kPa
  • pH: Basic due to Mg(OH)2 formation
Click to expand...
 
tetrasect said:
Hmm...

https://twitter.com/x/status/1827651577311867134
Click to expand...

tetrasect said:
It's an estimate. With a wide range.

Don't think anyone can work it out scientifically without proper measurements.
Click to expand...

Finding pieces of the meteorite helps to considerably narrow down the size estimate. I 'guessed' in my earlier comment on the video that the fragments would be fist-sized, which seems to be confirmed by that photo. The Chelyabinsk meteor was estimated to be 9000 tons (Wikipedia). I don't see how those fragments are consistent with such a large object.

P.S. There is a whole science around estimating things without being able to measure them exactly.
 
Gordon_R said:
Finding pieces of the meteorite helps to considerably narrow down the size estimate. I 'guessed' in my earlier comment on the video that the fragments would be fist-sized, which seems to be confirmed by that photo. The Chelyabinsk meteor was estimated to be 9000 tons (Wikipedia). I don't see how those fragments are consistent with such a large object.

P.S. There is a whole science around estimating things without being able to measure them exactly.
Click to expand...
The Chelyabinsk meteor was 18 meters in diameter and released around 30 hiroshimas when it exploded 29km above Chelyabinsk. There were thousands of injuries.

This local explosion was obviously not as big.

However, since energy dissipates at the square of distance, if the Chelyabinsk meteor had exploded 60km above Chelyabinsk (like you estimated this local meteors altitude), it would have had to have been the size of 120 Hiroshima bombs to create the same "damage"on the ground.

So altitude makes a MASSIVE difference.

So, if this local meteor exploded at a low altitude it might have only been a few hiroshimas, of it exploded at high altitude it could have been many.
 
Can you guys please do the meteor again, I wasn't listening the first time.
 
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