<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR> Ok, If a 170 Lb man is standing up in an 8 foot canoe at the 'shore' end and the canoe is up against the shore, and the man walks to the other end of the canoe, how far out from the shore will the canoe move? <HR></BLOCKQUOTE>
If the canoe is hitting the shore, the canoe will not move, as it will not be able to move in a direction opposite the direction of the man walking. If the canoe does not have any contact with the shore, and the man is just walking in his canoe somewhere in the water, the canoe moves 4ft backwards, and the man goes 4ft forwards, both in relation to the water. Every action has opposite and EQUAL reaction....thus, each object moves 4ft.
The balloon in the bus should stay in the same position. This is only because the windows are closed (assuming the bus is now airtight). The inside of the bus is a seperate system of forces, and the air (since it has nowhere to go in an airtight bus) moves forward along with the bus, and thus the balloon moves as well. This won't happen in real life though, because a bus isn't airtight. But, if it was, the balloon wouldn't move. In more real situations, the balloon moves to the back.
I hope nobody is serious thinking you can't put different size tires on the front and back of a car.
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR> Cripers Shawalker, your explanations make you sound like the professor from Gilligan's Island!!!! But the heck with tire rotations. Tell us MORE about this bisexual girlfiend <HR></BLOCKQUOTE>
Thanks little buddy! Hehe...although this was probably accidental...I noticed you spelled girlfriend, girlFIEND. How ironic. Have a good one.
