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1

This happends because of the friction generated between the paper and the coin. Remember, friction, at low speed, is not affected by the speed. So when you pull the paper, at any speed, the friction will have always the same value. The distance covered by the coin is always $x=\frac{1}{2}\frac{F}{m}t^2$, where, $F$ is the force of the friction (constant), $...


2

The forces are the same (assuming the the coin is moving relative to the paper in both cases), is the friction force between the coin and the paper, and is in a first approximation at least, independent of the relative speed between paper and coin. The difference between the two cases is that is that the force acts during a shorter time in the fast motion ...


1

There is a limit on how much frictional force the penny can apply to the paper in order to remain attached to it. It is limited by its mass based on $F_f = \mu _s F_N = \mu _s * m_{penny}*g$. If you accelerate the paper, you are applying a force $F = ma $ and the penny's frictional force will try to oppose to keep the penny on the paper (inertia). For a ...


-4

This is a wonderful question and filled with interesting albeit incomplete or inaccurate answers. Launching an "orbital body" concerns two issues and two issues ONLY namely mass and intertia. So before we "launch" anything we must first understand THE EARTH IS MASSIVE...which works AGAINST the "inertial reality" of a conical shaped item which albeit is ...


2

This is the third time you've asked this question. The moment of inertia of a system depends on the choice of axis. If one is talking about the moment of inertia of an object with no other qualification, it is assumed that the axis goes through the center of mass, but the orientation of the axes in the body needs to be specified. In the case of a sphere, ...


3

You are correct that it will move, but also remember that conservation of momentum still holds, so, if you start out at rest, and push a large massed object, you have: $$\begin{align}\Sigma p_{0} &= \Sigma p_{f}\\ 0 &= mv + MV \\ v &= - (M/m)V \end{align}$$ So, if your mass is small relative to the house, you will recoil from the house at a ...



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