In the KE formula you’d round the velocity off to zero (because you really need m/s to calculate enough Joules produced, otherwise it’s negligible) and when you multiply out the equation, no matter how large or “small” the mountainous mass is, you get . . .ZERO.
FFS, it's like carnival of the idiots. I do not expect everyone to have knowledge of basic physics and mathematics. In fact, I only expect some small percentage of the population to have even the rudiments of physics and mathematics. But I do expect that when someone CLAIMS to have such knowledge, and attempts to use it in a public forum to demonstrate that they are correct, that they don't end up performing the scientific equivalent of a very public loss of bladder control.
Here's a hint - when you've learnt to count beyond five, come back and try again...
ABE: Should add something technical for everyone else:
KE has little to do with this. The plate is not in free motion. It is being constantly driven by the mantle and by gravity, and is continually losing energy to massive friction and deformation. Without the driving mechanisms, the plate would halt practically instantly. It is the rate of energy transfer, or power, that is relevant. The actual instantaneous KE of a plate is not large by any means, because of the v
2 term, and is measured in a mere several KJ. But if this is being transferred many times a second, then it is more than sufficient. A quick back-of-the-envelope calc shows me that the entire Himalayas gains on the order of 10KJ every second because of its uplift. It appears then that the Indian Plate's motion is about perfect to create this uplift. What a suprise...
Edited by cavediver, : No reason given.
Edited by cavediver, : No reason given.
Edited by cavediver, : Because AdminNosy asked nicely...