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The idea which helped me to differentiate between them is as follows: Image is a point of intersection of reflected/refracted rays. They might actually intersect (real image) or one might have to geometrically produce them to get them to intersect (virtual image). One might always find that the real image is formed if this point of intersection is on the ...


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This is a fun, high-quality qualifying exam question. The algebra is not hard; the physical insight takes some real thought; there are many ways to be partly right. Here's my take on it. Acceleration From Einstein's equation $E^2 = p^2 + m^2$ we have for each photon $E = p = hf_0$ (in the reference frame of the laser). We can use the power of the laser ...


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You have to be careful to use the same sign conventions for $x$ and for $v$. If you're careful enough you can talk about $v$ as a vector with only one component, and take the sign seriously. Let's choose the positive-$x$ direction to be "towards the right". If the bicycle's speed is to the right, $v > 0$, then when the friend catches the bus she's to ...


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I'm going to assume some things before I answer: This golf ball doesn't feel the effects of air resistance, or is in a place where that does not matter. You know some common equations for kinetic energy and gravitational potential energy. We ignore any effects of weather and assume the only things contributing to the ball's motion is the golfer and gravity ...


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Not quite - you want the cold sink to be as cold as possible. The cold sink is the cold air (or water or coolant) flowing over the heatsink. The heatsink is designed to efficiently connect the chip to the cold sink, so it needs to be as low resistance as possible. In thermodynamic calculations consider the chip==heatsink and the airflow to be the cold ...


2

The first equation is for a hollow cylinder that has a thickness, hence the $r_i$ and $r_o$ (inner and outer radii respectively). The second equation is for a thin hollow cylinder, for which the thickness is negligible, so $r_i$ can be assumed to equal $r_o$. As expected, setting $r = r_i = r_o$ and substituting into the first equation yields the second ...


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Energy goes as $E^2$. So when you double one and half the other, their ratio becomes 16. Answer (D) is correct. Looking at your equation, the relationship between $E$ and $D$ is $$D=\epsilon_0\epsilon_rE$$ So you expression can be rewritten as $$energy = \frac12\int \epsilon\cdot E\cdot E \cdot dV$$ or alternatively $$energy = \frac12\int \frac{D\cdot ...


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You must assume circular orbits, as stated in another answer. The ground speed will also be variable due to inclination, unless it's equatorial. An extreme example is an east-west orbit versus a west-east orbit. In one case, the ground velocity is added, in the other case, it's subtracted. But with the query limited to equatorial orbits, we can continue. ...


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Put 1 cell in each row. In other words, put all the cells parallel to each other. This way all the currents from individual cells add up and you have the least internal resistance of the cells, giving you the largest current through your external resistor.


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The current through the resistor R will be a maximum when the potential difference across that resistor is a maximum. For typical passive resistors, more potential difference means more current. You job is to find the arrangement of cells that would cause the greatest potential difference across the external resistor.


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Geostationary satellites travel in circular orbits about the equator. If you express the velocity as angular velocity, it will equal the angular velocity of Earth's rotation, about $7.3 \times 10^{-5} radians/second$. If you had another satellite in circular orbit about the equator, but not geostationary, you could subtract the angular velocity of the ...


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Your graph is not showing you that the photocurrent is higher for higher frequencies. It is showing the effect of frequency on the kinetic energy of the emitted electrons. The $x$ axis is the reverse potential i.e. making the metal positive with respect to some nearby electrode. Because the metal is positive the photoelectrons are attracted to it and this ...


3

That $C$ is the specific heat for the given cycle, i.e. $$dQ=nCdT$$ This is for $n$ moles of gas.(not the $n$ you stated in question) I will assume $$PV^z=\text{constant}$$ $$nCdT=dU+PdV$$ $$\int nCdT=\int nC_vdT+\int PdV$$ We will integrate it using Pranjal's method : $$nC\Delta T=nC_v \Delta T+\int \frac{PV^z}{V^z}dV$$ As numerator is a constant, ...


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We know that field due to point charge is zero at a distance infinity (and beyond) from it. Let us consider a square with equal charges at the corners. Let us assume that side of the square is beyond infinity, then some portion (not one or two points) inside the square will have zero net electric field due to the charges. The same holds good for other ...


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In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where $\mathbf{E}=\mathbf{0}$. Letting the charges be located at $(\cos(2\pi k/N),\sin(2\pi k/N))$ for $k\in\{1,2,...,N\}$, we can generate plots of $|\mathbf{E}|^{-1}$ for various $N$. The zeros of ...


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I think it is worth adding that increasing the intensity of incident light will increase the rate at which electrons are ejected from the surface (assuming incident light frequency is higher than the threshold frequency). I am NOT saying that more electrons will be ejected.


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In the idealized case, changing the frequency to even higher values once a current started running in the first place should make no difference to the total current: The electrons are either knocked out or they're not.


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One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose $V(\infty)=0$, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ...


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Some hints to get you started: First, consider a Galilean referential in which the problem is simpler and assume some fixed $v_P$, without taking care of its magnitude. Second, use what you know of the distance between a line and a point (projection). Third, minimize the distance between the line and point wrt $v_P$ under the contraint that ...


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As an idealization if plates are assumed to be infinite in extent there is no field between A and C since they are both positively charged and connected. So q1=q2=0, and q3=q4=3Q/2. Using your equations we get the same result. Grounding means q1+q2=0, so the same conclusion.


2

You start with a total of $3Q$ on the outer plates - and that charge has nowhere to go. So when you connect the two plates together, they will share the charge equally, and have a net charge of $1.5Q$ on each. Grounding the center plate means that charge can flow freely. So what will happen? First - charge will flow between plate B and ground. The boundary ...


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Coriolis force: $$F_c=m v \wedge \omega$$ We need the velocity component perpendicular to the axis of rotation, which is $$v_c = v sin(\theta)$$ Now we integrate the acceleration twice with respect to time in order to get displacement: $$x = \int\left(\int{v sin(\theta)\omega dt}\right)$$ Noting that $$\theta = \frac{v.t}{R}$$ where $R$ = radius of ...


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Try with the following steps: Write down the equation of motion $x(t)$ and $y(t)$ for a generic particle shooted with angle $\beta+\alpha$; and with an initial position such that it start from the two lines as you wrote above. Given this trajectory $x(t)$ and $y(t)$, compute the tangent vector of the position as the function of the time $t$ (which is the ...


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The angle of the first diffracted beam (assuming we are talking about double slit, and not worrying about the width of the single slit) is $$\alpha=\frac{\lambda}{d}$$ Then we find the distance from $$\frac{spacing}{distance}=\alpha$$ So $$distance = spacing \frac{d}{\lambda}=1.5cm\frac{30\mu}{500nm}=0.9m$$ Diagram:


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As you said, you want that your wave function is normalized. So, it is not correct that the solution outside the wall is $$\psi=Cexp(\mu x)+Dexp(-\mu x).$$ In fact this function diverges for $x\rightarrow\pm\infty.$ In order to keep your wave function normalized, you must use one solution for $x<-a$ and another one for $x>a.$ For $x<-a$, you ...


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My quick read of a few articles on Q, the reaction quotient, suggest that it's generally rather small. Thus $ln(Q)$ is negative, so $ - T * ln(Q) $ is positive and the potential difference increases with temperature.


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The mass $M$ in the equation for orbital velocity would be equal to the mass of the Earth (assuming that the mass of the satellite can be neglected) and thus constant. The gravitational constant, $G$, like the name suggests is constant as well. But the radius of the (circular) orbit, $r$, can vary. So by using different altitudes at which the satellite would ...


2

As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.


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Honestly, the argument you're making here is a mess - the question is based on bad premises. So let me show you how to do it properly, and hopefully that will resolve your confusion. You're right that in order for a wavefunction to be normalized, it must satisfy $$\int_\text{all space} P(x)\mathrm{d}x = \int_\text{all space} \psi^*(x)\psi(x)\mathrm{d}x = ...


2

The rate of formation is much higher in the presence of dust. There needs to be a mechanism for the energy of formation of the hydrogen molecule to be dissipated. Dissipating energy via a photon involves a forbidden transition. Instead, the energy can be transferred to the vibrational lattice of a dust particle. See The Interstellar Abundance of the ...


2

Yes, you're right ! The uncertainty principle tells us that the thickness of the energy state $\Delta E$ is linked to the typical decay time of this energy level. If $\Delta E$ is large then $\Delta t \sim \tau$ (decay constant of the energy level) is small and then this energy state is very unstable.


1

Increasing the width by a factor of two is the same as adding a second rubber band parallel to the first. For each, $\Delta F=-k\Delta x$. To stretch the combined system a distance $\Delta x$, you have to apply a force $F$ to the first, and $F$ to the second, doubling the needed force. Thus, for the combined system you have $\Delta F_\mathrm{combined} = ...


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Take a rubber band. Stretch it by a distance $x$ with your hands. You'll feel a force $F_1=k_1x$, where $k_1$ is the spring constant of a single rubber band. Now take two rubber bands, and hold them side by side. Stretch it by a distance $x$ with your hands. Since you're stretching two of them, you'll feel twice the force, so $$F_2=2F_1=2k_1x=k_2x$$ where ...


1

You should show your work, but my guess is that you have to notice the change of variables: $$\frac{d\chi}{dx}=\frac{d\chi}{d\xi}\frac{d\xi}{dx}$$ You need to do this a second time (using the derivate of a product. See if you can continue from there.


1

The velocity after a time t1 of accelerating is the starting velocity of the deceleration phase. Thus $$a t_1 = b t_2$$ (not worrying about the sign here. I suppose you could) Further you have $$\frac12at_1^2+\frac12bt_2^2=s$$ Now you have two equations with two unknowns. Solving: rearrange first equation $$ \frac{a}{b} = \frac{t_2}{t_1}\\ t_2=\frac ab ...


0

Since plate B is grounded, charged conductors near it can induce equal and opposite charge on plate B. Infact a ground is necessary to charge an object. http://www.physicsclassroom.com/class/estatics/Lesson-2/Charging-by-Induction When a charged conductor is brought near a grounded conductor opposite charges are induced on the grounded conductor but overall ...


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The origin is assumed to be at rest wrt ground frame. So, your pendulum with lift moves up and the origin stays at its original place only. That explains that factor.


2

You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator $(b^2-c^2)^2$: $$ \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi ...


2

I'm assuming the proton at x=0 and the alpha particle at x=2 are held in place. I'll call these the "held" particles. There will be two forces which act on the "free" proton (at x=-1). I'm also assuming that this is just a classical mechanics problem, and will solve it that way. Before we go into the calculations, let's think about what is going to happen. ...


1

It is essentially two capacitors in parallel. It will therefore have twice the capacitance than if everything else were held constant and one of the +Q plates were removed.


1

I'll answer my own question. Since $\boldsymbol{\omega }=\omega _{z}\hat{\boldsymbol{k}}$ the angular momentum reduces to $ \boldsymbol{L}_{O}=-I_{xz}\omega _{z}\hat{\mathbf{i}}-I_{yz}\omega _{z}\boldsymbol{\hat{j}}+I_{zz}\omega_{z}\boldsymbol{\hat{k}}$. We can split the rod in three pieces, calculate moment(product of inertia for each body and sum up. ...


1

As an alternate answer using parallel axis theorem, note that the inertia tensor of a rod pointing in the $y$-direction rotating about its center is $$\mathbf{I}_\text{rod,y}=\left( \begin{array}{ccc} \frac{b^3 \rho }{12} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \frac{b^3 \rho }{12} \\ \end{array} \right)$$ and similar for the $x$ and ...


0

See What Really Happens in the Franck-Hertz Experiment with Mercury? The article explains how the curve depends upon the design of the tube.


1

I thought a cross-section would be $cm^2$, not $cm^{-2}$. The information you really lack is probably the macroscopic cross section. That is, $\Sigma = n \sigma$, where n is the number density of the target objects, in number per cubic centimeter. Thus, this gives a value of inverse length units. Using this, you can readily speak in terms of the ...


0

Also, you can use a sort of change of variables. $\int_a^b \dot{x}^2 dt = \int_a^b v^2 dt = v^2 \int_a^b dt = v^2({t_b}-{t_a})$ where the last part is using the fact v is constant for a line to pull it out of the integral. The result follows.


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The problem is in assuming the force is forwards at a contact point right underneath the axis of the wheel, together with the no-slip condition. This leads to zero friction force, since their is no friction when their is no slipping. A solution can only be found when deformation of the wheel is taken into account, allowing the force to act on a different ...


0

I don't know exactly where to start either but one could think of the black body radiation coming from a source far away that was emitted by a source at temperature $T_1$ where $g_{00}$ had some value say $g_{00}^1$. Now,if we make use of the fact that $g_{00}$ influences the photon frequency via gravitational Doppler effect, then imagining balance of ...


0

From a physics point of view, you have some kind of mass-diffusion problem. It is true that a mathematician will name it "heat equation" because the classical problem with the operator $\partial_t - \Delta$ is the heat equation. Homogeneous Neumann BC is indeed appropriate to model a "no flux" condition at the boundaries, be it a flux of heat or of mass. A ...


5

$\int^t_0 A x^2 dt = x_0 + A x^2 t$ is incorrect. You are assuming $x$ as a constant. $x$ is a function of time x(t). Try $\dfrac{dx}{dt}=Ax^2 \implies \dfrac{dx}{x^2}=Adt$. Now integrate both the sides in appropriate limits. $$\int_{x_0}^{x(t)}\dfrac{dx}{x^2}=\int_0^t Adt$$ $$\int_{x_0}^{x(t)}x^{-2}dx=\int_0^t Adt$$ ...


3

First problem: you say $v(t) = A x^2$, but that is a function of position, not time. Putting the definition right: $$ v = \frac{dx}{dt} = A x^2 $$ You can regroup terms on the same variable: $$ \frac{dx}{x^2} = A dt$$ And then do the integration: $$ \int \frac{dx}{x^2} = \int A dt$$ This is homework, so I will leave the integral limits and the ...



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