# Tag Info

## New answers tagged velocity

0

For a spacetime interval: $\Delta \tau = \sqrt{\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}$ $$\frac{d\tau}{dt} = \frac{\Delta t - v_x\Delta x - v_y\Delta y - v_z\Delta z}{\sqrt{\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}}$$ Because the spatial coordinates are function of t. Dividing by $\Delta t$ numerator and denominator and letting deltas go ...

2

why doesn't the Lorentz factor of a component of 3-space momentum depend on only the corresponding component of velocity? Because that definition of $\mathbf{p}$ wouldn’t transform as a 3D vector should under rotations. Since the velocity $\mathbf{v}$ is a vector under rotations, the Lorentz factor $\gamma=1/\sqrt{1-v^2/c^2}$ needs to be a rotational scalar,...

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REFERENCE : My answer here How to add together non-parallel rapidities?. $=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$ From above reference consider the velocity vectors \begin{align} \mathbf{u}_1 \boldsymbol{=} \left(u_{1x},u_{1y},u_{1z}\right) & \boldsymbol{=}\left(u_1 ...

1

The following may be a useful approach. A simple traveling wave can be written as: $$y=\sin\left(kx-\omega t \right)$$ We want to follow the position of the wave at a constant phase, $\phi$. Let that phase=0 which leads to: $$\phi=kx - \omega t = 0$$ $$kx=\omega t$$ $$x=\frac{\omega}{k} t$$ Then the velocity will be $$v=\dot{x}=\frac{\omega}{k}$$ ...

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Here's a perspective from a former chemistry student. I'll try to explain using molecular motion. I once taught myself compute programming and made a particle-geometry collision simulator to test the following, and it seemed to work, so I'm drawing on the intuition that I learned from that project. Background: Pressure is one measure of kinetic energy ...

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It depends on the type of valve. Butterfly valves for example have a approximately linear behaviour only at a few percent of opening. Gate valves are more linear. But for small openings as 10% to 20%, an almost linear behaviour is expected. That means the flow doubles, provided the volume of the reservoir is big enough to have its level practically constant.

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I just can't wrap my head around why pressure decreases as velocity increases. That's somewhat backwards. That makes it sound like a decrease in pressure is caused by an increase in velocity, when it's more that an increase in velocity is caused by a decrease in pressure. If there is a pressure differential, that means there is a net force on the fluid, ...

1

I want to replace the masses with their relativistic mass where relativistic mass is the following equation: Yes, you can. But remember that you had better write the relativistic law of momentum conservation. When you do, you will find that it is as if you replace the masses in your classical equation with the corresponding relativistic masses. However, ...

10

i just can't wrap my head around why pressure decreases as velocity increases.. When velocity increases then you obviously have acceleration. Now what is causing this acceleration? As always (according to Newton's second law, $\vec{F}=m\vec{a}$) acceleration is caused by a force. In this case the force acting on a piece of fluid comes from the pressure ...

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I just can't wrap my head around why pressure decreases as velocity increases This is a classic misunderstanding of Bernoulli's equation. What Bernoulli's equation actually says is that the velocity will increase in the direction of decreasing pressure: $P_2-P_1=-\frac12\rho(v_2^2-v_1^2)$. This makes sense: if the pressure is higher on the left than on the ...

3

You have this equation, $p = mv$. If you increase the velocity and keep $p$ constant, yes you must have had an increase in mass. The question is, what does it mean for $p$ to be constant? It is defined by the product of mass and velocity. So you have done nothing to make predictions, you have just stated what you would call mass if you increased v and kept p ...

0

Note, for a small change in $p$ (and $v$, as their correlation is 1, or $m$ depending on normalization): $$\delta m = \frac d {dp}({\frac p v})\delta p+ \frac d {dv}({\frac p v})\delta v$$ $$\delta m = \frac{\delta p} v - \frac{p\delta v}{v^2}$$ Since: $$\delta p = m \delta v$$ we get: $$\delta m = \frac{\delta p} v - \frac{p\delta p/m}{v^2}$$ $$\delta ... 1 You are conflating work with energy, but there is a slight difference. In fact: Work=change in Energy, i.e. U=ΔK. Work is the accumulated force magnitude over distance along a path, so that an element of work dU=<F, ds>=F.ds.cos(θ) , where ds=elemental displacement along trajectory curve S , F=applied Force along S. We integrate this and then call the ... 5 So you aren't doing the same thing as your teacher here. Notice how in what your teacher did they have \Vert d \mathbf r \Vert =ds, so they did not say that$$\frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {dr}{dt}\right)^2 $$this would not be true in say circular motion, where the position vector is changing but its magnitude is not. However, ... 0 I think that in the first case, a step is missing:$$\frac{d\mathbf r}{dt}.\frac{d\mathbf r}{dt} = \frac{d\mathbf r}{ds}\frac{ds}{dt}.\frac{d\mathbf r}{ds}\frac{ds}{dt}$$\frac{d\mathbf r}{ds} is an unit vector in the direction of \frac{d\mathbf r}{dt}. So the dot product: \frac{d\mathbf r}{ds}.\frac{d\mathbf r}{ds} = 1 We can't do the same for \frac{... 3 Yes, \vec v=\frac{ds}{dt}\hat{t} So,$$ d\vec{v}\cdot d\vec {v}=(d^2s\hat t +ds\,d\hat t)\cdot(d^2s\hat t+ds\,d\hat t)=(d^2s)^2+(ds)^2$$But {dv}^2=(d^2s)^2 1 Since \vec v_{rel} is a scalar multiple of \vec r we have \hat v_{rel} = \hat r, so \displaystyle \frac {d \hat v_{rel}}{dt} = \frac {d \hat r}{dt} = \vec \omega \times \hat r \\ \displaystyle \Rightarrow |\vec v_{rel}| \frac {d \hat v_{rel}}{dt} = |\vec v_{rel}| ( \vec \omega \times \hat r ) = \vec \omega \times (|\vec v_{rel}| \hat r) = \vec \omega \... 0 If a rotating (constant) vector is decomposed into magnitude and direction \vec{\rm vec} = v\, \hat{e}  and the derivative of a unit vector is \dot{\hat{e}} = \vec{\omega} \times \hat{e}, then multiply both sides with v$$ \frac{\rm d}{{\rm d}t} \vec{\rm vec} = v \frac{\rm d}{{\rm d}t} \hat{e} = v \left( \vec{\omega} \times \hat{e} \right) = \vec{\...

1

Since $\hat{v}_{rel}$ is a unit vector $\dot{\hat{v}}_{rel} = \vec{\omega} \times \hat{v}_{rel}$. Multiplying by $|\vec{v}_{rel}|$ on both sides gives you that equation.

2

Figure the volume of 3 meters of 13 mm pipe, this is the flow rate per second in the pipe. Then divide the volume by 69 and you will have the average flow rate per second for each hole. The precise flow rate for each individual hole would depend on multiple variables, but the total of the 69 holes will equal the flow rate of the main pipe as flow rate in ...

2

If the ship programs the engines to have an increment of 0.1c every 60s, the interval between the announcements (until 0.9 c) is by definition 60s in the ship's time . The way the ship knows its velocity can be the average velocity of stars (blue shifting ahead and red shifting behind). Of course the crew feels an increasingly strong fictitious force to ...

9

what the crew would "see" on Earth (if they could) is life speeding up until it's well beyond a blur. What the people of Earth would see if they could "see" into the ship are people slowing down until they stop moving. No: if the ship is approaching Earth, then people on Earth will see clocks on the ship running fast, and people on the ...

2

The reference frame of the ship is non-inertial, so the positioning of the navigator and the captain with respect to the acceleration direction matters. By the equivalence principle, the ship's reference frame will look the same as being on the surface of a planet with a very strong surface gravity (51 thousand $g$s with the acceleration you have given!). So,...

0

Mass cannot travel at c (the speed of light in a vacuum), but it could travel near c. To an outside observer considered to be at rest relative to the ship, the time passing on the ship would become more dilated (passing slower) the faster the ship went. Any one in the ships frame of reference would observe time as passing normally for them. So the people on ...

2

As the navigator and captain are in the same reference frame, there will be no relativistic time dilation between the two. However, to accelerate an object to the speed of light would require an infinite amount of energy, and the energy required to keep accelerating increases with speed, so a constant acceleration of $0.1c$/min is not sustainable.

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In many discussions of Special Relativity, there are three velocities: the velocity of an object in some reference frame, the velocity of the same object in some other reference frame (one that is moving relative to the first frame), and the relative velocity between the two frames. These are often denoted as $\mathbf u$, $\mathbf u’$, and $\mathbf v$, or $\... 1 Calculate their relative velocity. Determine how long it will take for both vehicles to meet over that distance. From there it will be trivial to calculate the distance the SUV travels. 1 The key to showing why they behave the same is calculus. You are correct that a particle cannot simultaneously have a force on it and no forces on it, just due to sheer logic. However, what we can say is that the particle can have a force on it at some time point$t$(such as when the two objects are at the same velocity), and no force on it at some time$t ...

0

in the moment, both particles have the same velocity it is of cause the same, for example if both were cars of the same mass they would cause the same damage in a collision. What is different is not the velocity, but the acceleration. So if you compare at the same moment and in in this moment the acceleration stops both continue with 10mps.But for many ...

0

I’m referring to an instant in time where the velocities of two different particles are identical. One particle experiences a force and the other does not. I think their states of motion are different even if the velocities are identical. Am I correct? The state of motion of an object is defined by its velocity. See the following: https://www....

1

Pressure and density are proportional in gas. Sound speed depends on pressure AND density in opposite way, so the effect cancels out an the speed is only dependent of temperature and kind of gas. This is not related to sound speed in fluids, whir very high densities compared to gas

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The only galaxies that we observe to be blue-shifted are some galaxies in the Local Group such as Andromeda. These galaxies are relatively close neighbours to the Milky Way, so although they are still subject to the Hubble flow they also have an individual radial speed towards us that is greater than the expansion of space. Galaxies which are further away ...

0

The $\frac{dv}{dx}=0$ suggests that you are either moving with constant velocity, the object has become stationary or you're going to turn. Eg: 1.You are moving with constant velocity along a line. Since the velocity is constant with respect to distance, the derivative is zero. 2.Becoming stationery: A body starting with velocity $v$ on a surface being ...

1

In an inertial frame of reference an object at rest will remain at rest unless acted on by a force and an object in motion will remain in motion at a constant velocity unless acted on by a force. This is according to Newton's first law of motion. So an object considered to be moving with no friction or other forces acting on it will continue moving with a ...

0

I am not sure if this answers OP's question, but momentum can mean one of two things. It is either a "generic momentum" $(q,p)\in T^\ast M$, which is just a covector on $M$ defined at some point, or what I am calling the canonical momentum, which is actually a map $\xi:TM\rightarrow T^\ast M$. Given a Lagrangian $L:TM\rightarrow\mathbb R,\ (q,\dot ... 0 I am not exactly sure about my answer, however, I think that the relativistic form for$R$is: $$R = \frac{1}{\gamma_a m_a + \gamma_b m_b}\left(\gamma_a m_a(0) + \gamma_b m_b (vt)\right)\space,$$ where, according to your data,$\gamma_a=1$and$\gamma_b=\gamma. Therefore, we have: \begin{align*} u = \frac{R}{t} = \frac{\gamma}{m_a + \gamma m_b}m_bv = \frac{... 1 I believe this may a useful approach.x_{1}$refers to the vertical position of ball 1 (starting at the top of the building) and$x_{2}$is ball 2.$h$is the height of the building. $$x_{1}=h-\frac{1}{2}gt^{2} \quad \left(1\right)$$ $$x_{2}=v_{0}t-\frac{1}{2}gt^{2} \quad \left(2\right)$$ Solve for the time when ball 1 reaches 60 m above the ground (... 2 This law holds true for any inertial frame. Thus the velocity referred here, can be with respect to any inertial frame. The momentum conservation law states: $$\sum_{i=1}^N m_i u_i = \sum_{i=1}^N m_i v_i$$ where, the subscript$i$stands for$i$-th particle.$N$is the total number of particles.$m_i$is the mass of the$i$-th particle.$u_i$and$v_i$are ... 0 As you said the Lagrangian is defined on the tangent bundle, whose elements, loosely speaking, are pairs of a coordinate and a derivative, e.g. $$(q, \dot{q}) = \left((q_i)_i, \; \dot{q}_j\frac{\partial}{\partial{q_j}}\right)$$ The Hamiltonian on the other hand is defined on the cotangent bundle, whose elements are pairs of a coordinate and a 1-form, e.g. $$... 1 Gravity will pull on both balls the same. If there was nothing else in the way, it would cause both balls to accelerate at the same rate. However, there is something else in the system. There is the track, and it provides a "normal" force which pushes perpendicular to the track. This normal force is partially opposing gravity. Near the bottom, ... 2 Short answer : Depends on the magnitude of force acting since two bodies of different mass can have same acceleration and thus same velocity after sometime if both were initially at rest or moving with same velocity. Explanation : Case 1 : Take two bodies , say of mass m_1 and m_2 (m_1 < m_2) and suppose both of them experience the same amount of ... 3 I am assuming one-directional motion for simplicity. Suppose you apply a Force F on objects with different masses. Then using Newton's Second Law:$$a=F/m$$Meaning that if mass becomes greater, acceleration becomes smaller. Further a=dv/dt. Therefore$$dv = a dt = \frac Fmdt \int{dv}=\Delta v= \int\frac Fm dt$$Clearly for the same force F, for ... 3 If I understand your question, you are asking about applications of Newton's Second Law: F=ma. In the case of acceleration, you have$$ a=\frac{F}{m} \qquad(1) $$which shows a greater mass, m, results in a smaller acceleration, a, for a given constant force, F. In the case of velocity, the expression is:$$ v=at $$Using a from (1) above, this ... 0 why the two balls hit the bottom at the same time Not going into a lot of details, just notice inclination of normal force vectors : So,$$ \frac {\alpha_2}{\alpha_1} \propto \frac {L_1}{L_2} $$Or in words - ball at higher altitude is pushed down by a greater force, but it has to overcome a greater distance too. While bottom ball has comparatively small ... 1 Without going into the math behind pendulums, we can still make it plausible that their periods do not depend on their amplitude. A basic feature of all harmonic pendulums is that they accelerate the swinging object with an acceleration proportional to its displacement. Meaning that if we increase initial displacement, we also get a higher initial ... 0 why the two pendulums released from different angles have the same time period ? The time period for a simple pendulum is independent of its initial conditions.$$T = 2\pi\sqrt{\frac{l}{g}}$$where$l$is the length of the string of the pendulum and$g$is the acceleration due to gravity As you can see its time period doesn't even depend on the mass or the ... 2 The velocity vector describes the change in the displacement vector, and hence, it can be in any direction relative to what direction the displacement vector points in, since there are many ways the position of the particle can change. A few counterexamples: Circular motion: Velocity vector is perpendicular to the position vector which denotes the particle ... 1 Displacement is the shortest possible distance between the initial point and final point of line of motion. Do velocity and displacement always have the same direction? Need not to be. Imagine a ball being projected vertically upward with some speed$v$. At some time t (less than time of ascent i.e.$t<\frac{v}{g}$), the direction of its velocity is ... 0 And answers own question: After z is measured, the distinction between redshift and blueshift is simply a matter of whether z is positive or negative. For example, Doppler effect blueshifts (z < 0) are associated with objects approaching (moving closer to) the observer with the light shifting to greater energies. Conversely, Doppler effect redshifts (z &... 2 In air the acceleration depends on the mass, so the objects will not reach the ground at the same time. The sum of the forces with air resistance.$m a = m g-\frac{1}{2} A \text{$\rho$c} v^2$or$a=g-\frac{A \text{$\rho$c} v^2}{2 m}$So that the acceleration$a$indeed depends on the mass. Without air resistance$a=g\$ and the acceleration is independent ...

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