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Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake.

In general, an earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is perfectly balanced and "stable" is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released. (It will take very little energy to release the potential energy stored in the system). A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has (basically) nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy (at least in any system where the activation energy is much less than the stored potential energy, like a natural tectonic shift).

This holds true for any place where the energy required to take a system out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
3 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake.

In general, an earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is perfectly balanced and "stable" is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released. (It will take very little energy to release the potential energy stored in the system). A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy (at least in any system where the activation energy is much less than the stored potential energy, like a natural tectonic shift).

This holds true for any place where the energy required to take a system out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake.

In general, an earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is perfectly balanced and "stable" is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released. (It will take very little energy to release the potential energy stored in the system). A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy (in a naturally occurring earthquake).

This holds true for any place where the energy required to take a system out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake.

In general, an earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is perfectly balanced and "stable" is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released. (It will take very little energy to release the potential energy stored in the system). A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy.

This holds true for any place where the energy required to take a system out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake.

In general, an earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is perfectly balanced and "stable" is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy.

This holds true for any place where the energy required to take a system out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake.

In general, an earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy.

This holds true for any place where the energy required to take a system out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy.

This holds true for any place where the energy required to take a system out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy.

This holds true for any place where there is stored potential energy where the energy required to take it out of equilibrium is less than the potential energy stored in the system, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy.

This holds true for any place where there is stored potential energy, not just for tectonic plates.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph, not the activation energy.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so; very little in the pencil case, more in the curved niche case. But this activation energy has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam with energy the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so. It has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam of the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so. It has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam of the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so. It has nothing to do with the potential energy stored in the system. It is the potential energy stored in the system when the earthquake happens that is what is read by a seismograph.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam of the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case the amount of energy required to take it away from it's stable position is only the amount of energy required to do so. It has nothing to do with the potential energy stored in the system.

1 year ago
1 score
Reason: None provided.

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam of the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved niche, sitting just as high as the pencil, but that perfectly fits that ball, will have just as much potential energy to be released as the pencil situation, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case however, the amount of energy required to take it away from it's stable position is only the amount of energy required to do so. It has nothing to do with the potential energy stored in the system.

1 year ago
1 score
Reason: Original

So, we're to believe that HAARP somehow automagically transmitted a focused beam with the energy equivalent of 600,000 megatons of TNT across the planet to Turkey? And then sent that same amount of energy there again, resulting in a 7.7 Richter scale aftershock,

I'm not sure about HAARP, but a directed energy beam of the size of an earthquake is not required to cause an earthquake. An earthquake is caused by tectonic plates moving against each other. You can think of the interface between two tectonic plates like an "accident waiting to happen." In other words, there is a tremendous amount of potential energy there, waiting to be released. How likely it is to be released is based on how stable the structure is between the two plates, not on how much energy will be released. Think of this like an "activation energy."

For example, a ball sitting on top of a pencil, even if it is in a "stable" state is pretty likely to fall off. Thus all the potential energy stored in the ball will very likely be released, even if it starts off in perfect equilibrium. A ball sitting on top of a curved plate sitting just as high as the pencil that perfectly fits that ball will have just as much potential energy to be released, but a much less likely chance for it to be released. It is in a much more stable position, even though the potential energy is identical.

In each case however, the amount of energy required to take it away from it's stable position is only the amount of energy required to do so. It has nothing to do with the potential energy stored in the system.

1 year ago
1 score