- #71
ChrisVer
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I editted it ( just to make sure you read what I meant).
very nice. good reasoning. But did the brick transform into an elephant while it was falling?! lolavito009 said:...
You need an understanding of Newton's first and second law and the concept of terminal velocity. According to Newton's laws, an object will accelerate if the forces acting upon it are unbalanced; and further, the amount of acceleration is directly proportional to the amount of net force (unbalanced force) acting upon it. Falling objects initially accelerate (gain speed) because there is no force big enough to balance the downward force of gravity. Yet as an object gains speed, it encounters an increasing amount of upward air resistance force. In fact, objects will continue to accelerate (gain speed) until the air resistance force increases to a large enough value to balance the downward force of gravity. Since the brick has more mass, it weighs more and experiences a greater downward force of gravity. The elephant will have to accelerate (gain speed) for a longer period of time before there is sufficient upward air resistance to balance the large downward force of gravity.
Once the upward force of air resistance upon an object is large enough to balance the downward force of gravity, the object is said to have reached a terminal velocity. The terminal velocity is the final velocity of the object; the object will continue to fall to the ground with this terminal velocity. When the air resistance force equals the weight of the object, the object stops accelerating and falls at a constant speed called the terminal velocity. In the case of the brick and the feather, the brick has a much greater terminal velocity than the feather. As mentioned above, the brick would have to accelerate for a longer period of time. The brick requires a greater speed to accumulate sufficient upward air resistance force to balance the downward force of gravity. In fact, the brick never does reach a terminal velocity; there is still an acceleration on the brick the moment before striking the ground.
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right, so the states you are talking about are 'almost discrete', but not quite, so they are not fully energy eigenstates, but have some spread around the most likely value. (Is that right?)ChrisVer said:I editted it ( just to make sure you read what I meant).