Tag Info

New answers tagged

0

You have that $$ E=2A_0\cos\left(\frac{\phi}{2}\right)\sin\left(\omega t + \frac{\phi}{2}\right) $$ which is correct. To get the intensity, you then square and time average this: \begin{align} I=\langle E^2\rangle&=4A_0^2\cos^2\left(\frac{\phi}{2}\right)\left\langle\sin^2\left(\omega t + \frac{\phi}{2}\right)\right\rangle\\ ...


0

Unless I have misunderstood your question, I do not believe that both masses would have the same $x$ value. Let me just run through the problem as I understand it. I am assuming that the weight is hanging from the spring as opposed to resting onto of it, so to find the equilibrium point of the spring use the equation $F=kx$. $F_g=kx$ $mg=kx$ ...


1

The angular acceleration $\alpha$ is just $\alpha = a/r = 1/200\textrm{rad/s}^2$ where $a$ is the tangential acceleration and $r$ the radius of the circle. For (b) you can use the angular constant acceleration equations. You have $\alpha$, $t$, and the starting angular velocity $\omega_0=0$. Using $$\omega_f = \omega_0+\alpha t$$ gives $$\omega_f = 0 + ...


1

Here's how. The ball gains a velocity $v$ due to gravity before hitting the ground. So each time it hits the ground its velocity is changed from $v$ to $-v$ (taking down as positive) during the collision, then returning again with $v$. The force $F_1= m\frac {dv}{dt}$ is experienced by the ball due to the collision, however this force is felt after the ...


0

I actually tried this, for an introductory physics class, punching holes in a gallon plastic container in which the area of top of the container, A, was about 23,000 mm^2 and the area of the small hole,A', was about 1 mm^2 the equation of continuity, i.e. Av=A'v' says that the velocity of the water at top surface,v ,is almost negligible compared to the ...


1

The experimental observation is that the plot of distance traveled to height of hole "appears" parabolic It's not "parabolic", it's the square root of a parabola (as you have shown below): $$ d=2\sqrt{Hh-h^2}\;. $$ This is just semantics, but it is good to get it right; a parabola only has powers 0,1, and 2, e.g., $ax^2+bx+c$, not square roots... ...


1

The first approach is incorrect because you identify the time a volume of water gets accelerated by the full force of gravity with the time it would take it to fall that distance. If that were the case, it would see less acceleration because the ram effect would decrease the pressure it sees from above to zero. The second approach gives you the correct ...


0

You have the right idea, but maths is a bit wrong because you have put the vector signs in the wrong places. $$ (|\vec{A}|+|\vec{B}|)cos (\alpha) - mg + |\vec{C}|sin (\alpha) = 0 \\ (|\vec{C}| cos (\alpha ) - (|\vec{A}|+|\vec{B}|)sin (\alpha) = 0 \\ \vec{B} \times \vec r_{OB} + m \vec g \times \vec r_{OG} = 0$$


1

The surface of the conductor has constant electrostatic potential $V$ and the electric field is proportional to the gradient of the potential: $\nabla V$. By definition the gradient of a scalar quantity is always perpendicular to the level curves (surfaces).


1

Your statement implies that the Electric field $\boldsymbol E$ will be parallel, during the whole motion, to the instant velocity $\boldsymbol u(t)$, i.e: $$\boldsymbol E = \boldsymbol E_\parallel + \boldsymbol E_\perp= \boldsymbol E_\parallel.$$ If you have a curved trajectory there must be a component of the force which is perpendicular to the velocity in ...


1

Here is an analogy. Let's consider dropping a ball on the earth's surface. We all know that Earth's gravitational field is a vector towards its centre. Only if the ball is dropped from rest, or was a given a downward shove, it's path will be straight(at least till it hits the ground). On the other hand, if you lob the ball horizontally, it's path is curved. ...


5

The reason is the samen as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) untill they dont feel a force anymore. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...


0

When the charge has gained momentum in a particular direction, it will experience a force and acceleration whose direction is that of the field lines, yes, but the velocity will not be in this direction. Compare with circular motion - the velocity is tangential to the circle, but, the acceleration is perpendicular to it.


1

Prove I have taken that line charge is placed Vertically and one test charge is placed. Now the electric field experienced by test charge dude to finite line positive charge. $$Ex = \int dx cos \alpha$$ $Ey$ will be cancel out as they will be opposite to each other. $$Ex = \int k \frac{dq}{x^2+y^2}cos\alpha$$ $$Ex = \int k \frac{\lambda ...


2

No the point of contact is not at rest. It moves with the block. You are probably confusing with rolling motion in which the point of contact is always at rest. There the point of contact is rest because the lowest point on the disk has two contributions, one due to forward motion of disk as a whole (v) and one in the backward direction due to rotation ...


1

According to Newtonian mechanics, it is true that the table exerts an equal but opposite force against gravity that results in $\Delta y = 0$, where $y$ is the up/down dimension. However, sliding block is clearly moving in the $x$ dimension (i.e. horizontally across the table). And it is also acted on by a force, namely friction. The block does not generate ...


4

Let's figure it out. First let's figure out the force you would need to slide the object. In order to slide it you'd need a force that overcomes friction. $$ F_{\text{slide}} > \mu m g $$ where $\mu$ is the coefficient of friction. In order to tip it, you'd need to cause a torque that will cause the can to rotate about its far bottom corner. If you ...


1

You can find the expression for the electric field of a finite line element at http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html - which gives for the Z component of the field of a line that extends from z=a to z=b $$E_z = \frac{k\lambda}{z}\left[\frac{b}{\sqrt{b^2+z^2}} + \frac{a}{\sqrt{a^2+z^2}}\right]$$ You can follow the approach in that ...


0

Does $$L_+ L_- Y_{lm} $$ ,where $Y_{lm}$ is a spherical harmonic function, equals to zero. If so, why? It may or may not equal zero depending on the value of $m$. If $m$ is equal to $-l$ then yes, otherwise no. If $m=-l$ then applying the lowering operator annihilates the state (i.e., gives zero) since there is no state with an $m$ lower than $-l$. ...


0

As you correctly point out, the solution follows by integrating the differential equation. The general idea is to expand the left hand side of your differential equation by realizing that $$\text{d}\ln n^c(T) = \frac{1}{n^c(T)} \text{d}n^c(T) = \frac{1}{n^c(T)} \frac{\partial n^c(T)}{\partial T}\text{d}T = \frac{T}{n^c(T)} \frac{\partial n^c(T)}{\partial ...


1

http://arxiv.org/abs/1001.3702 : "We give a rigorous computer-assisted proof that the triangular bi-pyramid is the unique configuration of 5 points on the 2-sphere that globally minimizes the Coulomb (1/r) potential. We also prove the same result for the (1/r^2) potential. The main mathematical contribution of the paper is a fairly efficient energy estimate ...


4

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


3

If the rope is "radially directed" it means every point has the same angular velocity $\omega$. Assume a length $2\ell$, then you can integrate the force on the rope from $r-\ell$ to $r+\ell$ - gravitational force must equal centripetal force. This gives you an equation for $\ell$ as a function of $\omega$ and $r$. See if that gets you going.


0

The mechanism of a speedometer determines the travel speed of the vehicle indirectly by measuring the rotation speed of the transmission output or wheels. One turn of the wheel causes the vehicle to cover a distance equal to the circumference of the wheel, so there is a direct relation between the wheel's angular speed and the vehicle's linear speed; ...


0

Yes. With low air pressure, the axle is closer to the road: the radius of rotation is reduced. Some cars have systems that warn you when your tire pressure is low. The way they work is by measuring the rotation rate of the tires. If the rotation rate exceeds a certain limit, the dashboard light glows. With under-pressured tires, your speedometer will ...


1

No. Automobile tires to not expand radially to any great extent - the steel belts will keep that from happening. So, the tire radius still determines how far the car travels per rotation. Now, if your tires are slipping on the road, or are slipping with respect to the rims, than yes you have speedometer problems, but you have lots of other problems as ...


0

I think you have the right approach for the non-grounded. For the grounded case, we can use the uniqueness theorem, which says that given a charge distribution and the voltage on the boundaries, there is only one solution for the voltage. The grounded sphere has a $V = 0$ surface at $R + \delta R$ and at infinity, and no charge outside. I can solve this by ...


1

How to prove it? EDIT: Consider a state $|\psi_n\rangle$, which is an eigenstate of the Hamiltonian $H$ having eigenvalue $E_n$. Then consider the action of $H$ on these two other states: $$ |\alpha\rangle \equiv AB|\psi_n\rangle $$ and $$ |\beta\rangle \equiv BA|\psi_n\rangle $$ EDIT: Ask yourself: Are these eigenstate of the Hamiltonian? If so, ...


1

We have, for a point charge $q$ at position $\vec r(t)$: $$\rho(\vec x, t) = q\delta^3(\vec x - \vec r(t))$$ $$\vec J(\vec x, t) = q \frac{d\vec r}{dt}\delta^3(\vec x - \vec r(t))$$ Let us for now work without worrying about what the derivative (more precisely, gradient) of the delta function actually is. We will also enforce the convention that $\vec\nabla$ ...


4

You need to learn about the Eikonal equation and the equivalent ray path equation, which I talk about in my answer to the question Physics SE question "Ray tracing in a inhomogeneous media", and, if you need to know how it comes as the *slowly varying envelope approximation" from Maxwell's Equations, I talk about this in my answer to the question , "Optics: ...


0

The sled ($m = 11.1\;\mathrm{kg}$) shown in the figure leaves the starting point with a velocity of $25.1\;\mathrm{m/s}$. Use the work-energy theorem to calculate the sled’s speed at the end of the track or the maximum height it reaches if it stops before reaching the end. The straight sections of the track (A, B, D, and E) have a coefficient of friction ...


0

Try to keep this tidy. It is a straight forward calculation, but there are many terms, so tidiness is the key. Start with the Energy work relation: $$E_A - E_E = W$$ where $E_A$ is the energy at the beginning, $E_E$ the energy at the end and $W$ the energy loss due to friction. We have to split $W$ further into $$W = W_A + W_B + W_C + W_D + W_E$$ where ...


0

I think the $\Omega_N$ in your question is the number of states inside an energy sphere $E$; Now we need to consider micro-canonical ensemble, which means we need to get the number of states in an energy shell between $E$ and $E+\Delta$: $\Omega'=\frac{1}{N!h^{3N}} V^N \frac{\pi^{3N/2}}{(3N/2)!}\{[2m(E+\Delta)]^{3N/2}-(2mE)^{3N/2}\}$ Which is equal to: ...


0

This probably isn't going to be in time for your lab tomorrow, but I'll attempt to shed some light on this: In this sort of fluid-structure interaction with an oscillating/vibrating body, there are generally three fluid effects that need to be considered: added mass, added stiffness and added damping. Added mass is basically the fluid's mass adding to that ...


0

No, I don't think you have the correct entropy. I think the temperature is incorrect in the numerators, it should be (328 - 313) and (313 - 283).


1

At the moment the string becomes slack, you have a projectile like you say. Then, you want to solve for the time when $x^2+y^2=a^2$, when the distance of the particle from the origin is equal to the length of the string (i.e. when the string becomes taut again). Clearly t=0 satisfies this equation, but there is also another time that corresponds to the ...


1

Hint: conservation of energy gives $\frac{1}{2}m (\dot{x}^2+\dot{y}^2)+mgy(x)=mgy(x_0)$. But $\frac{dy}{dt}=\frac{dx}{dt}\frac{dy}{dx}$, and $dx=\frac{dx}{dt}dt$. You can use these relations to solve for $\dot{x}$ in terms of $x$, and then your integral of $\frac{dx}{\dot{x}}$ must be your integral of $dt$.


1

$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$ Thanks to Prof. Y. F. Chen I was able to figure it out. While in the integral the first term on the RHS can be converted into a surface integral as below: $$\int\nabla'\times(\frac{\vec{J}}{R})d^3x^{'}=\oint(\vec{n}\times\frac{\vec{J}}{R})d^2x^{'}$$ ...


0

I am not so sure about your first step. One sees a gain in speed due to conversion of potential energy into kinetic energy. That would give ${{m}\over{2}} v^2_\mathrm{f}-{{m}\over{2}} v^2_\mathrm{i}= m g h$ no need for angular functions as the hight is already shown in sketch, but you must remove losses due to friction as well. Looking at the standard ...


1

Using your diagram as a reference, the rays from the source are parallel to the (vertical) median of the prism (C). The surface(s) they are incident on are inclined by angles $a$ and $a'$ to this median. So the incident ray makes an angle of $a$ and $a'$ with the respective surfaces (note: not with the normal). From the law of reflection (equality of angles ...


5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...


1

Looks like textbook hybridization problem, did you check the usual suspects, or e.g. this one?


2

The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged


0

One way to think about this is the following. In general, the partition function (which is the integrand of the vacuum amplitude and not the vacuum amplitude itself) will be of the form $Z_\psi^{\pm}\propto\sum_{a,b}C[^a_b]Z^a_b(\tau)$ where $a$ and $b$ sum over the different sectors as given in the text and the $C$s are some phases. Many of these phases ...


1

Why not try to draw the Feynman diagrams first and characterize them by different topologies; then count the possible contractions for each cases?


0

In answer the specific questions that you pose: Your aim is to determine the drag force on the sphere (via dimensional analysis). Have you set up a free-body diagram of the problem? This might help you think about what parameters are actually important to the problem. Hint: the list of parameters you give are sufficient for a complete description of the ...


1

Gauss law is useful only in the cases of high symmetry systems like sphere, infinitely long (or very long and thin) cylinder, or infinite plane. You can't even apply it to curved sides of "real", short cylinder. You need to show somehow that gravitational field is the same everywhere on you Gauss surface, so that integral in the Gauss law turns out to be a ...


1

In my opinion, your solution is correct. The book's solution is correct as well, but you would need to draw a different diagram for it to make sense. Their assumption is to take by default, r as lying on the positive x axis. In this way, the distance between the $Q_2$ and the required point is the mod of $r-2$. Since this expression is being squared, it ...


1

Pay close attention to how you defined $x$: it's not the coordinate of the desired point on the $x$ axis, it's how far left of the positive charge that point is. So when your equation tells you $x = 4.83$, that means the desired point is $4.83$ units left of the positive charge. But, presumably the question you've been asked wants the coordinate of the ...


0

Ignoring the Coriolis effect, which is velocity dependent. I get for an x-y Cartesian coordinate system, with the y component toward the North Pole and x component pointing out from the equator at the position you are measuring the effective g=G , the following (in ordered pair vector representation); G= ...



Top 50 recent answers are included