# Tag Info

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The answer to this is "it's complicated." Indeed NASA has a 4 page trail devoted to the many oversimplifications, such as the assumption that the air flowing over the top must move faster to "catch up" to the air moving across the lower side of the wing. There are certainly two major aspects of this lift. The first is that air flowing ...

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The direction of instantaneous velocity at any time gives the direction of motion of a particle at that point in time. The magnitude of instantaneous velocity equals the instantaneous speed. This happens because, for an infinitesimally small time interval, the motion of a particle can be approximated to be uniform.

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Just like any other object moving in a fluid, an oil droplet is subject to drag from the fluid. The drag generally increases with velocity (the precise manner in which the drag increases with velocity depends on the Reynolds number), and so eventually the buoyant force is equal to the drag force and the oil droplet reaches a terminal velocity. There is one ...

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I am not entirely sure, but I think you made two wrong assumptions in your calculations. First of all, dv/dt is not equal to g. In general, for an object in free fall, the derivative of the velocity vector is equal to the gravitational acceleration vector, which points downwards and has the magnitude of g. But an object in the brachistochrone is not in free ...

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The Prandtl–Glauert effect vapour-cone will be there for all sufficiently fast (transonic) velocities. After a compression-shock bow wave, the moist air expands suddenly to below ambient pressure and cools below the dewpoint. The aircraft speed does not have to be supersonic. The cloud appears a bit below supersonic and, as does the compression shock, ...

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The easiest way to compute time derivatives in QM is in the Heisenberg picture. We know that for any operator $\hat{A}$ the Heisenberg equation of motion holds: $$\frac{d}{dt} \hat{A} = \frac{1}{i\hbar} [\hat{A},\hat{H}] + \frac{\partial}{\partial t} \hat{A}$$ More specifically for the position operator, the partial derivative will be zero, giving us \frac{... 1 For the same reason that a ball thrown upward slows down: when gravitational potential energy increases, kinetic energy must decrease, in order to keep the total energy constant. 1 “As” means “proportional to”. The sequences \frac12, \frac32, \frac52, \dots and 1, 3, 5, \dots are proportional. 1 Because we need the minimum speed by which the circle gets completed i.e. it stops when the circle is just completed. If the speed given to m is greater than \displaystyle \sqrt{\frac{4gl}{3}} then it completes the circle as well as have some velocity at the lowest point. The torque of mass 2m is greater than that of m about the center of rod and ... 1 Under what assumptions is the relationship \nabla\times\vec v=2\vec\omega applicable? In polar coordinates, \begin{align*} \vec v&=\dot{\vec r}=\frac{d(r\hat r)}{dt}=\dot r\hat r+r\frac{d\hat r}{dt}=\dot r\hat r+r\dot\theta\hat\theta=\dot r\hat r+r\dot\theta(\hat z\times\hat r)=\dot r\hat r+(\dot\theta\hat z\times r\hat r)=\dot r\hat r+\vec\omega\... 1 I believe the dependence on fluid velocity squared is only true for well developed turbulent flow. Consider the Darcy-Weisbach equation for flow through pipes:\Delta h_f=f \frac{L}{D}\frac{V^2}{2g}\tag{1}$$For laminar flow through circular pipes:$$f=\frac{64}{\text{Re}}$$With \text{Re} the dimensionless Reynolds number:$$\text{Re}=\frac{VD}{\nu}$$... 1 Radial acceleration is v^2/r and this is proportional to the total mass enclosed within the orbit. Adopting a rough Keplerian approximation$$ \frac{v^2}{r} = \frac{G}{r^2}\sum_i M_i,$$where M_i are the masses of the various components (baryonic, dark). If I understand correctly, your "theoretical" rotation curve arises from considering only ... 1 The 4-velocity is defined to be normalized, i.e.: g_{\mu\nu}u^{\mu}u^{\nu}=-1 \; , so if you choose comoving observers, for which u_i=0, then u_{0} = \frac{1}{\sqrt{|g_{00}|}}. From the above formula you can read the difference between the time felt by the comoving observer and the proper time, indeed \frac{dx^0}{d\tau} = \frac{1}{\sqrt{g_{00}}} \... 1 I disagree with the accepted answer. The forces acting on the ball besides gravity isn't a spring like force k x as stated. It is an air resistance term. It has the form of \beta v^2 The air stream is at constant velocity v, and the speed of the ball upwards is \dot x then the equation of motion is$$ m \ddot{x} = - m g + \beta\, (v-\dot x)\, {\rm ...

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It would accelerate. From the moment you detach it, there would be two forces acting on the paper ball: gravity and the ventilator. Lets assume that the force increases approximately linearly when going downwards. Then you could describe the motion by: $$m \frac{d^²x}{dt²} = -kx(t) -mg$$ This is the equation of a harmonic driven ...

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I think that one has to assume that there is no torque on the ball about the point it's rotating. So, it is like a pirouette problem \begin{align*} \mathbf{\tau}=\frac{d\mathbf{L}}{dt}=0&\Rightarrow \mathbf{L}\text{ is constant.}\\ \Rightarrow mv_1r_1&=mv_2r_2 \end{align*}

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Suppose you are moving at a high speed relative to someone else. The other person sees you passing by at a high speed and notes that communication between your neurons is happening more slowly. But from your perspective, you are stationary and the other person is moving past you at a high speed, so you would observe that communication between the other ...

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It's pretty much straight-forward conclusion of vector and calculus algebra. Position vector is defined as : $$\mathbf r = x\, \mathbf{\hat i} + y\,\mathbf{\hat j} + z\,\mathbf{\hat k}$$ If you value picture, then this is position vector in 3D space : Speed is position vector derivative against time by definition, so :  \begin{align} \mathbf v &= \...

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Velocity tells you how fast the position changes. Intuitively the velocity vector should have the following properties The direction of the vector should tell you the direction of movement The magnitude should tell you how fast the position is changing. Or to be more precise, the magnitude should tell you the speed. When you have two position vectors $\vec ... 1 It would be possible to answer this question in different ways. Let me try to use an argument which unifies the 1D and multidimensional cases. Among the possible ways to define the velocity at time$t$in 1D is that number that, when multiplied by a small time$\Delta t$, will provide the best (linear) approximation of the position at time$t + \Delta t$. ... 1 To start understanding vectors intuitively, just think of them as mathematical objects that combine dimensions together. If you do understand why the velocity is the change of position with respect to time in one dimension, think of two separate velocities, one in the x-axis, and one in the y-axis. In time t, the given object moves a distance$v_x \times t\$ ...

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