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43

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$ m g = k v^2 $$ And thus $$ v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...


19

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...


11

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...


6

The period of an elliptical orbit is given by: $$ T = 2\pi\sqrt{\frac{a^3}{GM}} $$ where $a$ is the semi-major axis. For a circular orbit of radius $r$ we have $a = r$. The two orbits you show do not have the same semi-major axis, so they do not have the same period. However if the elliptical orbit had $a^3 = 4r^3$ then the period of the elliptical orbit ...


4

You seem to be saying that friction couldn't speed it up, because nothing else is moving that fast. Well, how fast is it moving? We can imagine the gyroscope axis parallel to the z axis, and the casing to be aligned such that the x axis goes through it. If the casing is tipped slightly, the gyroscope resists that turning and one side of the shaft has firm ...


4

The wind is certainly doing work, because it applies a force and the point where the force is applied is displaced. However it isn't doing any work on the boat, it's doing the work on the water. The key point is that the net force on the boat is zero. We know the net force on the boat is zero because the boat is moving at constant velocity - if the net ...


3

Well, the car would have to experience some acceleration from rest to that constant velocity. This part of the graph will have the shape of a parabola if the acceleration is constant as the velocity is constantly changing. The graph would only be a straight line if this acceleration happened during time instant t=0, which is far from what happens in reality. ...


3

Actually, in your rotating torus your are mimicking gravity, which is point outward. I.e. the outside of the torus acts as a floor. The centrifugal acceleration would be $g\approx\omega^2 R$. This $g$ plays the same role as the gravitational acceleration on liquid or gas pressures under normal gravity conditions, so you can say $\Delta p = \rho g h$, or in ...


3

There are three possible forms of motion for a small object in a $\propto \frac{1}{|r|}$ potential: hyperbolic motion parabolic motion elliptic motion The first and second possibilities are unbound, so the object comes from infinity and will go back to infinity afterwards. The third one is bound, which means that the object cannot escape the ...


2

Objects in orbit come pretty close. If you don't mind venting the cabin or taking a walk outside, even air drag can be very nearly eliminated. All you have left are very small forces due to being in a non-inertial reference frame, and drag from the very, very thin atmosphere. Neither would be noticed without some very precise equipment. To reduce these ...


2

Impact on water is a very complex topic. Your simple calculation just figures out the velocity of a free-falling body after a 50 m drop. That just tells you the initial relative velocity of body and water surface. It doesn't tell you much about the force at impact, or whether the person survived. There are two things that might kill on impact: high local ...


2

You can see it by making tangent to the curve. Tangent is nothing but the instantaneous velocity in the distance time graph. The tangent at $t=0$ must be zero because the velocity is zero at $t=0$ since the car started from rest. You can easily see from the two graphs that the second graph shows it correctly.


1

Heuristically I'd say yes, it always must be curved if you're starting from rest. Think of it this way; suppose the graph isn't curved, that is, it's a straight line, then either the straight line is horizontal, or not horizontal. If the graph (straight line) is horizontal, then the distance isn't changing and you remain at rest. If the graph is a ...


1

"Velocity always perpendicular to position vector" means that the distance from the particle of interest to the origin never changes. The question is whether such motion always traces out a circle. Counterexample: consider a plane pendulum with its pivot at the origin. The velocity always obeys $\vec v \cdot \vec r = 0$, which is a good definition of ...


1

Not its not always a circle. For example, in 3 dimension, a particle can move on a spherical surface. The particle doesn't have the component of velocity in the radial direction. So this can be any path on the surface of the sphere. If the magnitude of the velocity changes, it will depend according to what function it changes. The magnitude of velocity ...


1

A sailboat is moving at a constant velocity. Is work being done by a net external force acting on the boat? This is a bit of a trick question it tricks you into thinking there is a net external force. The boat is moving at a constant velocity; that's a given. That means that the net external force on the boat must be zero. But if I use Work = Force ...


1

I am not satisfied with the way @Bernhard answered, since it just shows the maximum velocity, thus only answering partially the question. The air resistance can be written as : $$ R = \frac{1}{2}\,C_x\, \rho\, S\, v^2 $$ Note : The mass of the object is not in this equation. This is very important. Applying Newton's law to one of the object gives at any ...


1

Newtonian mechanics should be accurate enough. There are general procedures that can help answer the question "is it safe to ignore this effect." See, for example, dmckee's comment. This answer is more of a question-specific approach. The force on the "falling" object by Earth is equal in magnitude to the force on the Earth by the falling object. This can ...


1

Say you have a weight tied to each side a a rope which is strung over a pulley with friction. Here's a really easy way to see why the tensions on each side of the rope can't be equal. Imagine a really stiff pulley - in other words, ${\bf F}_{friction}$ is high. If that's the case, it'll be possible to balance unequal loads on this pulley system - i.e. a ...


1

Have you heard of superfluidity? It happens when you cool liquid helium below about 2 Kelvin. The helium then will flow freely and without any friction. If you induce a current vortex in liquid helium, it will remain flowing until the end of time (however, you cannot draw energy from it, because the liquid is frictionless) or until it warms up again. So, in ...



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