# Tag Info

1

The short answer is that you are basically correct; you just need to be more careful with your notation and your minus signs. Here's the long answer. By the definition of a conductor, the sphere is at some constant potential. Additionally, the potential of the system goes to zero as $r\rightarrow \infty$. There are a lot of different ways to think about ...

2

It doesn't matter. If you have a scalar $\alpha$ then $$\alpha (\vec{B} \times \vec{C}) = (\alpha \vec{B}) \times \vec{C} = \vec{B} \times (\alpha \vec{C}).$$ You can prove this simply by writing out the components of each of the three expressions and showing that no matter which order you do it in you will get the same results.

1

You want to use energy conservation still. The total energy of the system is still $\frac{1}{2}kd^2$. The difference here is that you will have an extra term in your kinetic energy due to the rotation.

2

You should realize that the first equation you write gives you the value of $a$. In these kinds of problems, you are always given some force, and you are expected to apply Newton's laws to the problem. Therefore $F_{\mu} = -mg\mu \underbrace{=}_{\text{$2^{nd}$Law}} ma \quad \quad \to \quad \quad a = -g\mu$ Now you know the acceleration, you can find how ...

1

Remember the law $F=ma$; you already know the force from friction $F_{\mu}= -mg\mu$. Hence you can get $a=-g \mu$, and one of your unknowns is gone.

1

Based on how you phrased the question, I think the answer comes from intuition and definition-chasing. Static friction: This is a contact force which acts against forces trying to slide two surfaces against each other. It is limited in maximum magnitude because we don't expect two surfaces to be inexoriably fused, at least in the scales of problems in ...

-1

It is possible to arrive at the expression for Ohm's law by using a simple classical model. The simplest treatment I've seen of this happened to be on an optics book: Pedrotti and Pedrotti's "Introduction to Optics", Chapter "Optical Properties of Matter", paragraph "Conduction current in a Metal". Basically free electrons in metals can be thought to obey ...

0

The determinant is fairly easy to calculate. You know already, essentially, the eigenvalues of the stiffness matrix; more accurately, you know the eigenvalues of the matrix $\mathbf{m}^{-1}\mathbf{k}$, because the $\omega_i$ are zeros of the equation $$0=\det(\mathbf{m}^{-1}\mathbf{k}-\omega^2).$$ (The more aesthetically minded would replace ...

0

Static friction is always less than or equal to some maximum total relative force, that if exceeded will result in the motion of the object. The static friction will change both direction and amplitude to keep the object still with respect to whatever is providing the friction. So long as the force applied to the object, from another means, is less than or ...

0

Ohms law is not a "law" as much as it is an excellent approximation of an unavoidable material property. Imagine a charge in vacuum with an contain electric field along z. This charge will continue accelerating, therefore the current is unbound. In this example the system has an inductance (current changes with time at constant applied potential), but no ...

2

The centrifugal force on the ring is the pseudo force when in the ring's reference frame, which causes it to move outwards, given by $$\vec{F} = m\frac{v^2}{r} = mr\omega^2$$ Where m is the mass of the object, v is the tangential velocity of the object, and omega is the angular velocity To find the time required for the ring to fall off, you need ...

2

You are right that in this theoretical problem the breaking force to achieve constant velocity is the same for all velocities. You either stipulate that the bike has the desired speed as a initial condition, or that it coasts with no breaking applied until the desired velocity is reached. In the real world, there will always be some friction or drag forces ...

1

Draw a force diagram and determine the braking force required to counteract the (portion of) gravitational force pulling the bike down the slope. If the net force is zero, delta-v will be zero. I think the problem you were given assumes the bike starts out with velocity v . If it doesn't then you'll need to derive a braking force whose magnitude varies ...

0

No, that isn't right. Think about the gravitational field of a sphere. Equi-potential surfaces form concentric spheres about the original sphere. An object in the gravitational field of a sphere will follow the field lines (or lines of action of force) - which in this case are radially inward. Now if you imagine a field with two spheres, they will be ...

0

Because then why would they tell us the distance between the centers? Your mistake is that you have considered $U_f$ as the gravitational potential energy (up to sign, which you may want to be careful with) of the object on moon's surface due to the moon, and $U_i$ as the gravitational potential energy on Earth's surface due to the Earth. But what ...

1

There is no way to derive Ohm's "law" from simple definitions of electric field and current. You have to understand the dissipative mechanisms at work in a system to validate the notion of Ohm's law. Sam29's answer introduces these dissipations through the concept of the resistivity $\rho$ or its reciprocal $\sigma$ of a material. The effectiveness of ...

0

Let's start with $E=Fq$, like you have up top. Rearranging to solve for q, we have $q=E/F$. We know that a change in charge creates a current, $dq/dt=I$, so substituting $\frac{d}{dt}\frac{E}{F}$ for $\frac{dq}{dt}$, we now have, $\frac{d}{dt}\frac{E}{F} = I = V/R$ You could also use the relationship $E=-\bigtriangledown{V}$ to obtain a relation.

0

For a system of point particles, the definition $$\vec{L}_i=\vec{r}_i\times\vec{p}_i$$ is always true; it's just a definition. I see no reason why that won't work here. The only choice you have to make is where to measure the position vectors $\vec{r}_i$ from. A particularly convenient position from which to measure $\vec{r}_i$ is the rotation axis. One ...

0

There are a number of ways you can examine the law in a microscopic view. One of them is this: An applied voltage creates an electric field, which superimposes a small drift velocity on the free electrons in a metal conductor. This drift velocity is way smaller than the speed of transmission in a conductor. Now, the basic relations are: $$I=\frac VR\\ ... 1 Equal and opposite forces do not imply equal energies. Conservation of momentum is what is important here. You'll notice that the two children end up with equal and opposite momenta when all is said and done. But energy is not being conserved because this is an inelastic collision (i.e. work is being done). Moreover, energy is a scalar, so don't get ... 4 In both of your solutions, you attempted to use Newton's 3rd law:$$\vec{F}_{1\rightarrow2}=-\vec{F}_{2\rightarrow1}.\tag{Newton's 3rd law}$$You did this correctly in your first method ("Newton's law method") but incorrectly in your second method ("Kinetic energy method"). In your first method, you explicitly set the magnitude of the forces equal to each ... 2 This vector potential can be written in every point on the plane except the origin as:$$ A = d\phi$$where \phi is the polar angle (\phi = \mathrm{tan}^{-1}\frac{y}{x}). This does not mean that A is exact, because \phi is singular at the origin. But this means that the magnetic field is zero at every point except the origin. At the origin ... 2 It is not immediately obvious, but the block has calculable angular momentum at the point just before impact. the block has velocity v tangential to the disk's center of rotation which is a distance r away., and so has angular velocity \omega=v/r. the block also has calculable moment of inertia around that center, I=mr^2. Then, it is simply ... 0 I suppose you know the mass and extent of the disk. Let's just ignore the mass of the stick. (We don't have to do that, but it makes everything simpler). We can then just use conservation of energy: You can calculate the kinetic energy of the block. Once it stops, this energy will be transferred completely to the rotational energy of the disk. Using the ... 1 If you're familiar with differential forms, then akhmeteli's answer is great, especially if you want to generalize to curved geometries. Let's try to be notationally and mathematically precise without using forms and be as explicit as possible. Let a vector potential A = (A^\mu) = (A^0, \mathbf A) be given. Consider a parametrized path x(\lambda) = ... 0 I guess it's because one assumes the medium in question is isotropic. 0 You can definitely represent a 3d QHO wavefunction as a composition of radial components and angular components (spherical harmonics). 4 This is just the potential of a standard harmonic oscillator. The presence of the linear term is just due to the fact that you're using a coordinate system where the minimum of the potential is not at x=0. You can re-write it as$$V(x)=\frac{1}{2}mx'^2+c$$where x'=x+\frac{\lambda}{m} is the coordinate centered at the minimum of the potential and ... 4 So I assume you actually need to prove Poincare invariance of \int d\tau A_\mu\dot{x}^\mu for a particle trajectory, rather than invariance of A_\mu\dot{x}^\mu, but the former expression is equal to \int_a^b dx^\mu A_\mu, where a and b are the initial and the final points of the trajectory, and Poincare invariance is indeed almost obvious for this ... 1 I think you should have a lowered index on the RHS:$$\begin{align} \frac{dh_{ab}}{dt} &= \frac{d}{dt}\left(h_{ma}h_{nb}h^{mn}\right)\\ &=\frac{dh_{ma}}{dt}\delta_{b}{}^{m} + \frac{dh_{nb}}{dt}\delta_{a}{}^{n} + h_{ma}h_{nb}\frac{dh^{mn}}{dt}\\ \frac{dh_{ab}}{dt}&= 2 \frac{dh_{ab}}{dt} + h_{ma}h_{nb}\frac{dh^{mn}}{dt}\\ \frac{dh_{ab}}{dt} &= ...

2

A space traveler on an artificial satellite will be in freefall around the planet it is orbiting. So the ink will not experience any acceleration relative to the pen due to the planets gravity. On earth the ink gets sucked up by the pen due to capillary action, but is counteracted by gravity. While in orbit the full "force" of capillary action can be used ...

2

I) The restricted$^1$ transformation (RT) $$\tag{1} (q,p)~\longrightarrow~ (Q,P) ~:=~(q, \sqrt{p} - \sqrt{q})$$ of OP's professor with inverse RT $$\tag{2} (Q,P)~\longrightarrow~ (q,p) ~:=~(Q, (P+ \sqrt{Q})^2) ,$$ and with Hamiltonian $H=\frac{p^2}{2}$ and Kamiltonian $K=\frac{p^3}{3}$ is indeed interesting. Apparently we should assume that $p,q,Q\geq ... 0 The contact point of the disk with the plane has null instantaneous velocity This implies that there is no slippage, and as such there are no non-conservative forces doing work on the disk. Assuming the disk is perfectly rigid and is not being subjected to any linear or angular accelerations, the disk will continue to roll forever, and will not come to ... 3 The question is asking what is the force exerted by the water on either of the two faces of the plate. The net force will be zero as force on either side sides cancel, so your intuition made sense. The force on a side comes from water pressure across the triangular surface. The pressure at any point on the triangle depends on the depth of that point. ... 0 I will assume the author means that each party has one qubit of the entangled state $$\rho=\frac {|00\rangle \langle 00|+|11\rangle\langle11|}{2}$$ Which is interesting because$\rho$isn't a pure state. Anyway this state is a "key"; in the sense that without it parties should not make any sense of the data being transmitted. I am not going any ... 0 A rolling disk will come to a stop eventually because any incidental friction will decellerate the center of mass. Ideally with a flat surface, and constant motion there should be no change as there will be not friction required to keep the disk rolling. In real life though, for sure a rolling disk will stop eventually. 1 You're almost right. But... 1138 kilowatts power output will give you 1138 kilowatt-hours in, well... one hour, not 1 second. Just leave out the$\times 3600$It's better to avoid weird units(like kilowatt-hours as much as possible, so another longer way is this:$1138 \text{ kilowatts}$mean$1,138,000\text{ joules/sec}$So in one year you'll get ... 0 First of all, your measurement unit is wrong. Energy is measured in, say, kWh, not kW/h. Second, you should not multiply by 3600, as the final result is in kilowatt-hours. 1 Your solution looks fine to me. Yes: the angular momentum is preserved in the horizontal plane (the weight is vertical and the reaction of the sphere surface is a central force) so your first relation is fine, just remember that$\theta$is not the vertical angle, but lies on the plane tangent to the sphere at point B. There are two kinds of rotational ... 0 First, you are right: in this problem angular momentum around the vertical axis is conserved. This is because all forces acting on the particle have no azimuthal component. Note, that even if we assume that the particle is rolling on the bowl surface without slippage, the angular momentum of its own rotation would be of negligible compared with the angular ... 0 I do not have a solution, just some steps to get there. I have parametrized the problem with spherical coordinates,$\varphi$is the azimuthal angle (around the hoop),$\psi$is the nutation angle (drop from horizontal plane) for a position vector $$\vec{r} = \begin{pmatrix} r \cos \varphi \cos \psi \\ -r \sin \psi \\ -r \sin \varphi \cos \psi ... 0 So a "pure" paramagnetic system has a positive magnetic susceptibility and for which the dipoles don't interact. Usually the effect of the "thermal bath" outweighs an applied magnetic field so the net magnetization is 0. I would say that in general the answer is yes seeing as it takes a Squid magnetometer to detect a paramagnetic system, but it is certainly ... 2 Those equations are rather tautological. The gist is that it doesn't matter whether you have some acceleration or gravitational attraction, they're really indistinguishable. So you basically get a g' \equiv g + a as new "compound down acceleration". Then you write down the exact same equations as for the earth-rest-frame case, but with this modified g' ... 1 If a net force is towards the positive x direction, then the x component of velocity of the object is increasing; period. Now, knowing this, are any of the statements (a) through (d) true? a) It can be moving in the negative x direction Sure, it could be moving in the negative x direction. All that is required is that the velocity is ... 1 The only thing that a net force in the positive x direction means is that the net acceleration is in that direction. It tells you nothing about its position or velocity. a) Imagine placing the x-axis vertically and letting the positive x direction be downward. Now imagine that the constant force is gravity and the particle is a ball. If you throw the ... 1 a), b), c), d) All yes. Imagine a ball that is moving at 10 meters per second to the left (negative x direction). Now, a force in the positive x-direction will slow it down initially, and finally make it go to the right, constantly accelerating to the right. This is all independent of the y-direction, in which it might also be moving. A force in the ... 1 You need to be careful about what exactly the inverse sine function is doing. If arcsin is given input x, it returns the angle, y, that sin(y) would have produced. If you consider \sin(x): You'll see that$$ \sin(0.523) \approx 0.5 \\ \sin(2.62) \approx 0.5 \\ \sin(6.81) \approx 0.5 \\... $$The inverse sine function doesn't just return a single ... 1 Here is how to solve these problems in general. Make a sketch for the balance of forces. and using trigonometry write down the x and y components of the vectors$$F_{BC}\cos(\varphi)+F_{BD}\cos(\theta) = F \\ F_{BC}\sin(\varphi)-F_{BD}\sin(\theta) = 0$$Now solve for$F_{BD}$and$F_{BC}$. 3 Cool question! Thanks to user lionelbrits for his answer that prompted me to pull out my mechanics books and check the definitions of "canonical transformation" given by different authors. If you look in Goldstein's classical mechanics texts in the section on canonical transformations, then you'll find that canonical transformations are essentially defined ... 1 The original coordinates satisfy the equations of motion when the integral of$p\, \dot{q} - H(p,q)$is minimized, and the new coordinates satisfy the equations of motion when the integral of$P\, \dot{Q} - K(P,Q)$is minimized. There is no requirement that$H$and$K\$ be numerically equal. The transformation is canonical if the Poisson bracket remains ...

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