New answers tagged

1

According to the rules of qft there are virtual photons in the vacuüm. No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way, The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing ...


0

The space ship should undergo both linear motion and angular (rotational motion). Linear motion will take place in the direction in which the force is applied, and rotation about its centre of gravity. So the centre of gravity will act as an imaginary pivot.


2

You have the right starting point with energy basically, but I'm finding your homework hint more useful than where you go from energy of a differential unit. It says "The energy density should be easy to identify." The energy density is: $$ \frac{\text{energy}}{\text{volume}} = \frac{\text{mass}}{\text{volume}} \frac{\text{energy}}{\text{mass}} = \rho ...


0

Well if we neglect the hidden momentum the conservation law of momentum in electromagnetism is simple: The momentum can be stored in static fields ($D\times B$); the mechanical momentum ($mv$) + electromagnetic momentum ($D\times B$) $= constant$. The similar formula is valid for angular momentum (where it is not hidden momentum) See Feynman's Lectures ...


0

Your confusion can be met with a practical scenario. You know that it'e the electrical energy that is carried from the battery to the load by the electrons. A connection wire has very good conductivity. A resistor obstructs the flow of current. Inside the resistor, the electrons lose their energy in the form of heat due to collisions with the atoms and only ...


0

Resistors act to reduce current flow... I will give a counter-example to the claim that resistors "act to reduce current". Consider a $9V$ battery connected across a $100\Omega$ resistor; the battery current is $90mA$. Now, connect another $100\Omega$ resistor in parallel with the first. The battery current increases to $180mA$. Here's another ...


0

[Due to a silly rule I could not format links, I would appreciate if someone with reputation edited those for me. ] Consider the following analogy to water pipes: Wires are like pipes already filled with water; the water resembles the movable charge, in case of metal electrons. Voltage is a pressure difference between two points, e.g. because one is higher ...


0

What this means is that the resistor reduces the current compared to a circuit that didn't have the resistor in it. Say you have circuit with a bulb and a battery in which 0.5 A of current flows. If you then introduce a resistor in series with the bulb the current everywhere in the circuit will be less than 0.5 A. The current entering and leaving the ...


0

Yes, but where is the problem? :) If you have a resistor in your series circuit, then the current is reduced - in the whole circuit, of course, not just in the resistor. The current is the amount of charges per time. There are just less, in a certain time, but they still have to pass through all the other elements of the circuit too.


-1

Anna V gave the correct explanation. Only when weak bosons are created on mass shell, e.g. at collisions @ Ecm = M, can you apply total angular momentum conservation at a single vertex (production and decay). On the other hand, even for off mass shell bosons chirality (read helicity for ultra relativistic particles) imposses costraints at each vertex.


1

The concept of angular momentum only makes sense when we specify a rotation axis. So we will pick the axis passing through the initial center of rotation. When the string is cut, the point mass has a linear momentum $p = mv = mr\omega$. One can define the angular momentum of a particle about an axis as $\vec{L} = \vec{d} \times \vec{p}$ where $\vec{d}$ ...


4

Probably the best way to think about this is to say that $$p = mv\\ F=\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$ (Using the usual product rule for differentiation - thanks @ja72 for the suggestion). If velocity is constant the first term vanishes and your result follows.


0

Considering momentum: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$ Therefore, $$m_1u_1-m_1v_1=m_2v_2-m_2u_2$$ Factorising gives us: $$m_1(u_1-v_1)=m_2(v_2-u_2)$$ Allowing us to rearrange to: $$\frac{m_{1}}{m_{2}}=\frac{v_{2}-u_{2}}{u_{1}-v_{1}}$$ Using the fact that you say the velocities can be interchanged, we obtain a final answer of: ...


1

We consider friction to an impulsive force, in cases when normal force is impulsive. Here's how:We know that $f=\mu N$(only during slipping motion, for no slipping frictional force is equal to applied force RESISTING friction). Since friction is proportional to normal reaction, it will be impulsive only when normal force is impulsive.Thus, if in a situation ...


1

This is indeed confusing. The confusion comes from this very peculiar hypothesis: What if the person doesn’t apply a tangential friction force at his feet? It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay ...


0

Simple-Every physical or chemical process always aims at decreasing the energy and increasing the entropy(randomness of things).Lets take some examples When an object falls on the ground then it would break and result in a large number of pieces completely scattered(randomness) thereby increasing entropy.The pieces would never come together to make the ...


2

If you draw similar triangles, then you'll find that $r_A/r_B = v_y/v_x$, and so the product $r_A v_x$ is equal to $r_B v_y$. Try drawing a line from the tip of your lower $\vec{v}$ vector to the tip of your lower $v_y$ component to see this.


1

For linear momentum to be a constant, its value should not change. Being linear momentum a vector quantity, it's value is completely described by specifying both magnitude and direction. In a circular motion, even though the speed remains a constant, the direction of velocity (which is tangential to the point on a circle) is changing throughout the motion as ...


0

As force given by an impulse is given by $N \times$ (change in time) = $m(v-u)$ $N$ = $m(v-u)$/(change in time) Hence you need the recoil time of bat and mass of ball to calculate the force given to the ball by bat.


0

In my view the motion of the person, seen in lab frame, would be linear motion, because, at the beginning, the person has a tangential velocity $ωr$ and radial velocity $v$, and he will keep these two forever. But then does it makes sense to talk about conservation of angular momentum? I mean it will surely be conserved in the lab frame but the ...


0

After reading your question again I gave a try proving why and how angular momentum is conserved. In the example given by the question one needs to understand how the angular momentum of an object moving in a rotating, non-inertial frame is conserved. I will present in brief, my effort of proving how and why the angular momentum is an integral of motion. The ...


1

Defining precisely what are all the quantum numbers is a difficult question because it depends highly on the model under consideration, even for the standard model. In particular any U(1) symmetry leads to a quantum number, and similarly some U(1) subgroup of non-abelian groups that commute with all other interactions can also be associated to quantum ...


1

Yes and no. depends on the direction considered. Momentum is a vector quantity. Since there is no force in the x direction, momentum is conserved in that direction but not in the vertical direction because gravity (an external force for the system of Wedge and block) is acting. Yes, for your 2nd comment above. You can use COLM (conservation of linear ...


0

I've looked at this problem, and the preceding pages in Spiegel's book, and it would seem to me that any particle travelling in this orbit would have an infinite velocity at O, at least if the velocity was non-zero anywhere else in the orbit since the potential is infinitely negative when r=0 so the kinetic energy would have to be infinite. Wouldn't it be ...


1

For a 2D planar simulation with zero friction do the following Definitions Each body has 3 degrees of freedom. These are $(x_1,y_1,\theta_1)$ and $(x_2,y_2,\theta_2)$ defined at the center of mass. Each body has mass and mass moment of inertia. These are $m_1$, $m_2$ and $Iz_1$, $Iz_2$. The contact is at point A with coordinates $(x_A,y_A)$ and with ...


0

No. Momentum is still conserved. In particular, the component of momentum parallel to the ground is conserved. So if the ball is going to the right before hitting the ground, it will continue going to the right after. The formula you refer to is for one-dimensional collisions. That applies only if the elements are arranged so that there actually is ...


0

I don't understand why all answerers immediately jump to the topic of momentum, carrying by the field. Nowadays nobody doubt field can carry momentum. The questions like this and especially mine one: How to observe Newton's third law violation? have different aspect on how this is possible, that third law is violated. Third law is the main workhorse of ...


-1

An electric field is the force that fills the space around every electric charge or group of charges. Electric fields are caused by electrical forces. Electrical forces are similar to gravitational forces in that they act between things that are not in contact with each other. Electric fields are also analogous to magnetic fields resulting from forces acting ...


0

Your assertion that if we place a charged object (such as a sphere) in the "donut hole" region of the inductor, this object will be accelerated back and forth through the "donut hole" by the changing electric field created by induction. is completely incorrect. There is never any magnetic field outside the solenoid, which means that there is never any ...


2

First draw a circular path. In order for an object to follow that path at a constant speed, there must be a force acting on it towards the centre of the circle. At any instant, that force has no component in the direction of motion, and so, if that's the only force acting on the object, its speed will stay constant. Now draw a spiral path (i.e. one in which ...


0

Think of pushing a wagon full of sand down the road with a constant applied force. $F=ma$ handles that just fine. But now open a hole in the wagon so the sand slowly drains out as you push. If you still push with a constant force the mass of the wagon and remaining sand in it decreases with time and of course you'll get it going faster in the end. You ...


0

I have tried to illustrate the error which has been made about the second term. Assume that a mass at position $A$ ($\vec r_1$) is under the influence of an attractive central force which originates from point $C$. The velocity of the mass is $\vec v_1$ with radial and tangential components of velocity $\vec v_{1r}$ and $\vec v_{1\theta}$. A little ...


1

I do not get why system such as the rocket in space are defined as "variable mass" since the mass of the system is not varying. This depends entirely on where one draws the system boundaries. One possible boundary is the rocket plus all of the exhaust gases it has released. The center of mass of the rocket + exhaust gas cloud system moves per the ...


0

I do not get why system such as the rocket in space are defined as "variable mass" since the mass of the system is not varying. It is because the "system" refers to particles that are still in the rocket, no to the whole system including the expelled exhaust gas. This is often done when the point of the analysis is in describing and understanding ...


0

$F=ma$ is true only when the mass of the system is fixed. However, in your case where the spaceship is moving across space, the system is losing mass due to the ejection of fuel out of itself. For such cases, where the total mass of system is varying, we cannot use this version of Newton's 2nd Law. That version of the law that is written down in your ...


2

The two equations are the same to first order, which is all that is important. If I were writing down the equation for the total momentum P(t+dt) myself, I would probably jot down the first equation (that of Morin) since I would be thinking of the instantaneous velocity of the rocket at time t rather than at time t+dt. But, again, the distinction is not ...


0

Let I be the impulse transferred to the string due to the fall of the ball on the plank. Now as gravity is a non stop impulsive force so for calculation purpose it could be neglected. And the impulse would be transferred on the other side of the string as well. Let P' be the final momentum and p be the initial momentum. So,I=P'-P I=2mV-mu. -1 And, ...



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