jeff einstein
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so sir i was right from the start i have tried to prove my point that different masses have different accelerations when dropped at same heightDaveC426913 said:Yes, you are correct.
You can convert F=ma to a=F/m.
For a given F: If m is small, then a is large. If m is large, then a is small.Looked at another way: Newton's first law says the F is equal and opposite, so
m1a1 = m2a2
And again, if m1 is smaller than m2, then a1 will have to be larger than a2
but if this is true....DaveC426913 said:Yes, you are correct.
You can convert F=ma to a=F/m.
For a given F: If m is small, then a is large. If m is large, then a is small.Looked at another way: Newton's first law says the F is equal and opposite, so
m1a1 = m2a2
And again, if m1 is smaller than m2, then a1 will have to be larger than a2
Sorry, we are now mixing up two things. I may be culpable for that.jeff einstein said:disprove this i guess...
jeff einstein said:but if this is true....
You have been instructed by multiple people over multiple pages carefully explaining the situation. Carefully and repeatedly explaining where your original statements were wrong and also where they were almost right. With multiple corrections from multiple people carefully explaining the physics and the source of the confusion.jeff einstein said:disprove this i guess...
PeterDonis said:The acceleration of both objects, however, is not the same, because it is the force divided by the mass of the object.