# Tag Info

0

It is a hypothetical condition as inertial will never let this condition happen. For the sake of argument I am using impulse. faster you stop an object more will be the force. Example using gloves to stop a fast ball in sports. $$F_{impact}*t=mv-mu$$ $$F_{impact}=\frac{mv-mu}{t}$$ $$F_{impact}=\lim_{t \to 0}\frac{mv-mu}{t}$$ According the equation the force ...

3

I am only going to leave a brief answer, seeing that the comments are very accurate. The paradox can simply be resolved by considering the elastic nature of all the objects. How so ever instantaneous might the $dt$ or the time of collision seem to the human eye, actually it occurs over a small duration, based on the elasticity of both the objects involved in ...

1

Let me discuss a simpler version of your rocket-question: one where there is no gravity, so that we don't have to worry about gravitational potential energy. Consider a rocket in free space (vacuum), and consider that the rocket is at rest. Now the rocket fires it's engine for a short time. The engine accelerates the rocket. The rocket now has kinetic ...

1

For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ...

3

There is a mutual attraction from gravity, and we generally only consider the smaller object here on earth because the earth is so massive, the acceleration of the earth is negligible. This is because $a = F/m$, and with equal $F$ between the two objects, the acceleration will scale as $a\propto 1/m$. For the earth, this leaves $a$ ridiculously small, but ...

2

Does the bigger mass EVER move towards the smaller mass? Yes. $F = KMm/r^2$ $M*a_{M}=F$ $m*a_{m}=F$ As you see the smaller the mass the higher the acceleration and in consequence the higher the traveled distance in a given time t. If the above is true, can we technically move the Earth by us(human population) jumping indefinitely? No. Each ...

1

yes, the earth will accelerate towards you , however the Earth's acceleration will be so small for all practical purposes that you usually do not consider it. Earth's acceleration is small because the mutual forces between you and the earth are the same, but the masses are different, so this results in different accelerations (remember: $F=ma$). Now if you ...

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In all cases, the two objects move towards one another. In fact they experience exactly the same gravitational force. However, because acceleration equals force over mass $$\mathbf{a} = \frac{\mathbf{F}}{m}$$ that equal forces causes the heavier object to accelerate much less than the lighter one. But technically, the Earth does move towards you very ...

0

No, conservation of Baryon number prevents a neutron decaying into gravitons. A neutron has Baryon number $B=1$. A graviton (or any gauge boson or lepton) has $B=0$. And famously, $1\neq0$.

2

This is the relative strength of interactions of elementary particles: strong 1 electromagnetic 1/137 weak 10^-6 gravity 6x10^-39 A free neutron decays through the weak interaction with a lifetime of 14.7 minutes. The gravitational interaction is 10^-33 times weaker than the weak. In the lifetime computations this would be squared .Even if baryon ...

3

You can do so. The energy put in is the integral of $VI$, the product of the voltage and current. It may be hard to calculate $I$ from $V$ because of the back emf and changing circuit resistance. The energy absorbed is the increasing magnetic field energy caused by the expansion of the current loop and the increasing kinetic energy of the wire and ...

1

Imaging the balls on a string. You are launching N balls per second, at a velocity $u$. This means the distance between the balls is $u/N$. And $N$ balls per second will pass a certain point in space. Now if the car is moving at a velocity $v$ (same direction as $u$), fewer balls per second can hit it - because subsequent balls on the string have further to ...

2

A physical system in GR is never isolated, in general, as it interacts with the curved metric, i.e., the gravitational background. (However a notion of isolated system can be given in the particular case of an asymptotically flat spacetime as discussed in auxsvr's answer.) Apparently this fact prevents the existence of conserved quantities because the ...

0

If the metric is asymptotically flat, it is straightforward to assign meaning to a quantity, such that it resembles the energy we know from special relativity. In particular, for the Kerr metric we may regard $(\partial_t)^a$ as a vector representing the stationary observer at infinity, where space-time is Minkowski, and the rest appears to said observer as ...

0

The ball only feels an impulse along the normal direction and not the tangential direction. Hence there is only a change in momentum in the normal direction and not the tangential. It is probably worth noting that although the overall momentum is conserved when a ball strikes a very large wall the momentum of the ball does change (and so will that of the ...

1

Being "near by" means that there is (unavoidably) an interaction between them. The problem is no longer $$\gamma \to e^+ + e^- \,,$$ but $$\gamma + A \to e^+ + e^- + A\,,$$ (here $A$ represents the spectator nucleus) and there is now a way to share out the energy and momentum. The interaction is generally thought of as mediated by the electromagnetic ...

3

Within the Newtonian framework of mechanics conservation laws are tricky to develop and are not obvious at first glance. Lagrangian mechanics generalises the concept of conservation laws by exploiting "symmetries". The connection between symmetries and conservation laws is made by Noether's theorem. An object has a symmetry if it is invariant under a ...

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From the definition of lagrangian mechanics, Noether's theorem shows that conservation of momentum and energy comes from invariance vs time and space. Yes, that's what we can read on websites like this. But note that we define our time using the motion of light, and our space too. Is the reverse true? Are Lagrangian mechanics completely ...

2

Yes. It is much easier to think of this in terms of conservation of momentum: Because light (and electromagnetic radiation in general) has momentum, you will have to gain momentum in the opposite direction to conserve total momentum --- just like if you were to throw the flashlight. It is difficult to think of this in terms of forces because we tend to ...

0

The continuity equation is $\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{\rho \mathbf{v}} = 0$, Now you can substitute directly for $\nabla \cdot \mathbf{\rho \mathbf{v}}$ with the expression for divergence in spherical co-ordinates $\nabla \cdot \mathbf{A} = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over ... 0 Your first equation is wrong, the relative speed of first stage to second stage in your equation is v+35, not 35. 3$\frac{dM}{dt} = \frac{\partial{M}}{\partial{t}}+\frac{\partial{M}}{\partial{x}}\frac{d{x}}{d{t}} = \frac{\partial{M}}{\partial{t}}+v\cdot\nabla{M}$(with no assumption on what is M) . So if$v\cdot\nabla{M} \neq0$you can have one of$\frac{dM}{dt}$and$\frac{\partial{M}}{\partial{t}}$that is zero when the other is not. ... 1 If you sit on the chair and push with same force, it is not an external force. Newton's law of motion: if external force is not acted on a body it will remain same in state. So the chair will not move. Your acted force is an internal force. 2 Imagine the person pushing you is wearing socks and standing on a slippery floor while they push you. Their feet might slip out from under them and they would get pushed backwards at the same time you are getting pushed forwards. This is Newton's third law - if you push forward on something, you are actually pushing yourself backwards at the same time. We ... 7 Because when you push on the chair, you're also pulling on the chair in the opposite direction without realizing it. For example, I just tried pushing the back of the chair I'm sitting in away, but to do so, I had to hold on to the seat of the chair. And if I went flying off of the chair, it would move - but I wouldn't be on it anymore. As others have ... 0 Actually, if you don't take into account the friction between you and your friend pushing the chair and the ground, the chair could have any velocity, but it would stay constant. The thing is that the force you exert on the chair is equal and opposite to the force applyied by your friend so the total force (resultant force) is zero. Knowing then that force ... 0 If you have no contact with floor and walls outside the chair other than through the wheels of the chair, you will not be able to fulfill the requirement of Newton's Third Law with respect to the floor and walls other than through the wheels of the chair. You can push on any part of the chair you like, but an equal and opposite force to propel you across ... 2 As dmckee says in a comment, the proof is ridiculously simple. Suppose we work in the centre of momentum frame so the total momentum is zero. The particle comes in with some momentum$p$and the antiparticle comes in with the opposite momentum$-p$, and the two annihilate. Suppose the annihilation produced a single photon. The momentum of a photon is: $$p ... 1 Suppose a\overline{a}\rightarrow\gamma is possible for a particle a with a definite nonzero mass, p_a^2=m^2>0 ("mostly-minus" metric, c=1). Conservation of momentum implies p_\gamma=p_a+p_{\overline{a}}\implies p_\gamma^2=m_\gamma^2=0=p_a^2+p_{\overline{a}}^2+2p_a p_{\overline{a}}=2m^2+2 p_a \cdot p_\overline{a} However, the scalar product on ... 1 You need to use vectors. Since L \neq r \times p, you need to use \vec L = \vec r \times \vec p instead, where the \times is the vector cross product of vectors, not the scalar multiplication of scalars. So you have$$\vec L= \left[(v_o t \cos \theta) \hat x+ (v_o t \sin \theta - \frac{1}{2}gt^2)\hat y\right] \times m\left[(v_o \cos \theta)\hat x+ ... 0 The most important thing that is constant is momentum in an inelastic collision. The force is constant for a specific wall and a moving bullet(considered to be appoint mass) irrespective of time. But, since in this case you take the wall to be immovable then its only the force which is constant. As you may want to say that the momentum that Earth gets due ... 1 Newton developed a formula for penetration depth of projectiles traveling at high speed. $$D\approx l_\text{bullet}\frac{\rho_\text{bullet}}{\rho_\text{wall}}$$ To a good approximation, the depth of penetration is constant. 0$m \frac{v^2}{r} = Forcem \omega^2 r = ForceForce = ma\omega = \sqrt{\frac{k}{m}}2 \pi f = \omegaf= \frac{1}{2 \pi} \omega\$ From here you can make substitutions and arrive at the result. Let me know how it goes.

0

Arturo's answer above is great, I just want to add mine because after working on understanding it, I think I have a simple explanation that might help someone. In the first equation listed in the original post, force and radius are inversely related if velocity and mass are held constant - this means that the frequency must change since velocity depends on ...

0

How do the Planets and Sun get their initial rotation? A simple answer is conservation of angular momentum. The correct answer is that simple answer is a bit too simple. The Sun Stars form from a gas cloud undergoing gravitational collapse. The cloud collapses into a protoplanetary disk as the protostar begins to grow. Conservation of angular momentum ...

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