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Before telling you why an observer in free fall does not feel any force acting on him, there are a couple of results that should be introduced to you. Newton's second law is only valid in inertial frames of reference: To measure quantities like the position, velocity, and acceleration of an object, you need a coordinate system $(x,y,z,t)$. Now the ...


23

It is incorrect to link the feeling of being accelerated to being accelerated itself. You can be under constant velocity or be continuously accelerated, yet you need not feel anything at all. Let me explain. The reason you feel compressed or stretched when you are accelerated in a lift is because of the presence of the normal force from the ground on you. ...


5

Well, everything depends on what you mean by "to experience a force". I suspect that you are thinking of some psycho-physical idea. Indeed both floating in space and freely falling we perceive similar sensations. The reason is simply due to the fact that, in both situations, all particles of our body moves with the same speed (due to a spatially uniform ...


3

falling in a gravitational field is physically indistinguishable from floating in interstellar space Yes. Indeed, this is one of the founding principles of general relativity and is (one of the forms of) the equivalence principle. Your argument is that we can feel acceleration, and gravity makes you accelerate, so shouldn't you feel acceleration while ...


3

I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ...


3

Suppose your object is a sphere with a radius $r$ and mass $m$. The aerodynamic drag on a sphere is given by: $$ F_{drag} = \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 \tag{1} $$ where $\rho$ is the density of the air and $C_d$ is the drag coefficient. The drag coefficient varies with speed (the NASA article I linked shows how $C_d$ changes with speed) but over a ...


2

Yes, escape velocity should really be escape speed. The Wikipedia article on escape velocity states this explicitly. I doubt there is any logical reason for using the term escape velocity and I suspect it is an accident of history. You might want to ask on the History of Science SE how the term originated - a quick Google failed to retrieve any information ...


2

If the raindrop's vertical velocity is constant as the train is both stationary and moving, the time taken for the raindrop to travel down the window would be: $$t = \frac{1\ \text{m}}{5\ \text{m}/\text{s}} = 0.2\ \text{s}$$ Remember, the time $t$ would not depend on the speed of the train. The exercise also specifically states that the raindrop's vertical ...


1

This is one of those questions that can drive you crazy, since there is a great deal assumed and not stated. Let me try an alternate possibility. Conceivably, the problem wants you to assume that, when moving, the overall speed of the raindrop remains fixed at 5 m/sec, but it travels in a straight line at an angle due to horizontal wind forces, and this may ...


1

The graph shows a constant speed, which is positive, so we know that the velocity is not changing sign. Depending on the context of the graph (is it dealing with 1D motion or curvilinear motion) we could tell a lot. Technically, on a traditional velocity vs time graph, one is plotting a component of velocity, complete with signs. I don't think the plot is ...


1

The simple answer is that displacement, in this context, is distance. There are other uses, such as the weight of a ship, but that is not germane. Consider that the area under the line is x times y. In this case the x axis has the units of seconds, and the y axis has the units of meters per second. So when multiplying them out, $$sec\times\frac{meters}{sec} ...


1

Okay, so here's the basic physics: it comes down to differential equations. Big objects in air tend to see $v^2$ drag. Plugging this into $F = m a$, this means that something which is falling straight downward with speed $v$ will obey the nonlinear differential equation: $$\frac {dv}{dt} = \frac 1\lambda v^2 - g$$for some length $\lambda$. The terminal ...


1

Lets just say were solving for any specific speed. Not the terminal velocity. I'm still not sure what to do. It might help to start with the speed an object attains after falling a short distance, ignoring air resistance and assuming the effect of gravity is constant. If we fall from Mount Everest (about 8000 meters) to sea level, we can treat gravity ...


1

You need a coordinate system to decide a body’s position, velocity, acceleration, momentum or force on it. Assume the body is in free fall near the Earth. 1) First consider a coordinate frame (3 perpendicular rods and a clock) with its origin in free fall near the free falling body. By the equivalence principle we know the rods are falling in unison with ...


1

We don't need to appeal to relativity to explain why you don't feel any force in free fall. Plain old Newtonian mechanics predicts that too. What you actually feel when you feel a force being applied to you is that the external force applies only to a small part of your body (the soles of your feet if you're standing up and feel the normal force from the ...


1

There is another aspect somehow overlooked by the other answers. Consider a pile of iron filings accelerated towards a magnet. If you were to arrange so that they all have the same magnetic force per unit mass they would appear to experience no force relative to each other while being accelerated towards the magnet, and if you had weak bonds holding them ...


1

Depends on the Reynolds number. Stokes can be used for purely laminar flow. Complete answer to be found here.



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