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3

Indeed your first suggestion is wrong :$ \Delta x = v_o t + gt $ Instead it should be $ \Delta x = v_o t + gt^{2} $(You can recheck it) Where you are wrong is here: According to your question v is the final velocity since $(v=v_{0}+gt)$ So $\Delta x\neq vt$ but instead it should be $\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ ...


3

The equations of motion for the position determine the accelerations: they are second-order differential equations in time: $$\vec F = m\vec a = m\ddot{\vec x }$$ So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the $F=ma$ formula ...


2

According to definition of 4-force: $$F^{\mu}=m_0 \frac{d V^{\mu}}{d \tau}$$ $$\to F_{\mu}V^{\mu}=\eta_{\nu \mu}F^{\nu}V^{\mu}=m_0 \eta_{\nu \mu} \frac{d V^{\nu}}{d \tau} V^{\mu}=m_0\frac{d}{d\tau}(V^{\mu}V_{\mu})-m_0 \eta_{\nu \mu} \frac{d V^{\mu}}{d \tau} V^{\nu} $$ But we know that: $$V^{\mu}V_{\mu}=c^2$$ and $\eta_{\nu \mu}$ is symmetric so: $$m_0 \eta_{\...


2

Recall that the four force $F$ on a particle of constant invariant mass $m$ is $$F = m\, \mathrm{d}_\tau V\tag{1}$$ where $V$ is the four-velocity. Moreover, recall that the Minkowski norm of a four-velocity is unity (or $c$, more generally, in nonnatural units). So therefore we have $$\mathrm{d}_\tau (V_\mu\,V^\mu) = 0 \tag{2}$$ as a universal identity....


2

It is like water in a hose. If the hose is full of water, water flows out the end immediately when you turn on the faucet. A drop of water at the faucet pushes a drop next to it, which pushes the next drop. Water doesn't flow that fast. If the hose is empty, it takes a while to reach the end.


2

A bit of 1, a bit of 3... The technical name is flow velocity, as correctly stated in the Wikipedia article about NS equations. But one could ask what "flow velocity" means. From the Wikipedia article: flow velocity [...] is a vector field which is used to mathematically describe the motion of a continuum. Although correct, this definition is ...


2

You are correct, it is the velocity of a small volume of fluid centered at the point, that is a macroscopic motion, but it is also the result of the average velocity of the particles in that volume.


2

While I don't know the terminal velocity of a cat (which may or may not be 100 kph). I know a lot of people say 100 kph, but I don't know if that is verified or just a calculated estimation someone came up with. I have experimented quite a lot with falling objects, including small animals and bugs. I have documented the falling speeds of many small ...


1

4-vectors have invariant length as defined by $$\vec{v}\cdot \vec{v} = g_{ab}v^a v^b.$$ Coordinate velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}t}$ does not have this property. Proper velocity $\frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau}$ does. Coordinate velocity is defined. It's not a 4-vector, so it's not that useful.


1

The thing about the relativity principle is just that: all non-accelerating frames are equal, in the sense that no inertial frame is more "real" or "accurate" than an other inertial frame. If two frames are moving in a Minkowskian manifold with a constant velocity relative to each other, it doesn't matter which one you choose. No matter what, all the ...


1

As the force $F$ is constant, and we know that the force of weight $mg$ is constant too; we can say the net force acting on the ball is constant whole the motion. But, from $t=0$ to $t=t_1$ the net force is $F-mg$ and after $t=t_1$ the net force is $-mg$. So, velocity graph will be linear but it has a breaking at $t=t_1$. For $0\le t\lt t_1$ the gradient of ...


1

Let's take the first equation of motion which is : \begin{equation} v=u+at \end{equation} Integrate this equation to get: \begin{equation} \int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt \end{equation} this gives: \begin{equation} x=ut+\frac{1}{2}at^2+x_0 \end{equation} The integration constant can be done away by putting the proper limits on $x$.(Assuming the ...


1

This is a simple variation on the so called "Twin Paradox", which is not a paradox in the logical sense (i.e. not a logical contradiction). Each cycle of the oscillator's motion is like the journey of the spacefaring twin. One possible cycle on a spacetime diagram is drawn below (source: Wikipedia "Twin Paradox" Article with my own additions). The ...


1

First of all, recall that one may vary the velocity $v$ independently of the position $q$ in the Lagrangian $L(q,v,t)$. In fact, the (Lagrangian) canonical momentum is defined as $$\tag{A} p(q,v,t)~:=~\frac{\partial L(q,v,t)}{\partial v}. $$ This is explained further in e.g. this, this, and this Phys.SE posts. Let us define for later convenience $$\tag{B} F(...


1

The position vector of a particle in polar coordinates is given by $$\mathbf r ~=~ r\mathbf e_\mathrm r$$ Velocity $\bf v$ is \begin{align}\mathbf v&=\frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\mathbf e_\mathrm r}{\mathrm dt}\\ &= \frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\theta}{\mathrm dt}~\mathbf ...


1

This is a simple question of conceptual understanding. The only force acting on the ball after it is released from the hand is that of the ball's weight due to gravity. Since it is the only force, the consequent acceleration is also the only acceleration. Gravity is being taken to be $10 m/s^2$ in a downward direction. For the answer to be correct, our ...



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