# Tag Info

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Marshmallow is traditionally made by stirring a hot supersaturated solution of sugar and gelatine or agar into whipped egg whites. On cooling a material with (at least) four phases present is formed. The phases are: the protein in the egg and gelatine (or polysaccharide in the agar) form an elastic solid held together by crosslinks between the protein ...

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A rough calculation can be done using this frame : On the far left we have Faora and on the far right the soldier she is attacking in the next scene. If we assume the height of a soldier to 1m80 then they are at a distance of 10m It takes her 4 frames to cover that distance in the next scene and the video is 24 fps therefore the speed is $$V = ... 4 Yes, because acceleration also includes change in direction. For example, a race car on the track goes in a circle. If its speed is 150 mph for the entire race, it is still accelerating because it is not going in a straight line. 4 This is just a particle-in-a-box. You only have \ell, m quantum numbers in a more complicated system like an atom--in particular, there needs to be a rotational symmetry and this just isn't that kind of system. 3 In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics. 2 The gravitational force on a small mass m some distance R from the center of a large spherical mass M is given by$$ |F| = \frac{GMm}{R^2}. $$If your distance from the center is some altitude r above the radius of the Earth's surface R_\oplus, the force is$$ |F| = \frac{GMm}{(R_\oplus + r)^2} = \frac{GMm}{R_\oplus^2} \left( 1 + \frac{r}{R_\oplus} ...

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For some function $f$ of $x$, the logarithmic derivative is simply $$\frac{\mathrm{d}\log f}{\mathrm{d}\log x} = \frac{x}{f} \frac{\mathrm{d}f}{\mathrm{d}x}.$$ You can check that this follows from the chain rule applied to $\mathrm{d}g/\mathrm{d}y$, where $g = \log f$ and $y = \log x$. This is a common thing to see in astrophysics, since if we have a power ...

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The ping pong ball would lose a tiny amount of kinetic energy to the truck. The truck ends up with a momentum of just under twice what the ping pong ball had. However, energy is 1/2 m*v^2 = 1/2(m*v)^2/m. Since the truck is much more massive than the ping pong ball, it carries much less energy for a given momentum. The end result is that the small amount of ...

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Balance torques around the corner of the step, so r x cos(38.7) x mg= F x 0.3. F = 114 N

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What I believe to be true: The bubble changes shape with spherical symmetry, hence the flow is irrotational, and there is a velocity potential. We solve this problem using conservation of mass at the boundary $r=R>a$, and at $r=a$. By conservation of mass: $4\pi r^2*u(r)=4\pi a^2 *\dot{a}$ $\implies u(r)=\frac{a^2 \dot{a}}{r^2}$ Then the velocity ...

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Definition of potential difference is the amount of work per unit charge to move a charged particle from one place to the another place. The potential difference between point $a$ and point $b$ is as below, $$V_a - V_b = - \int_{\mathbf{r}_b}^{\mathbf{r}_a} \mathbf{E}\cdot \mathrm{d}\mathbf{r}.$$ What we call as potential with $V=\frac{kQ}{r}$ is the amount ...

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You are right in stating that potential and hence potential differences are dependent on field. The relation in fact is $\mathbf{E} = -\nabla V$ Hence, as we can see, if $E$ = 0, then $\nabla V$ is in fact constant, not $V$. Now, to compute the potential, we can rely on coloumb's formula, taking $V$ at infinity t be zero, for a differential ...

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Initial kinetic energy is $K_1=\frac{1}{2} m (v_x^2+v_y^2)$ with potential energy $U_1=0$. At the apogee, the potential energy is $U_2=m g h$ and the kenetic energy is $K_2=\frac{1}{2}m v_x^2$. Equate the two sums to get your answer. $$U_1+P_1 = U_2 + P_2$$ $$0+\frac{1}{2} m (v_x^2 + v_y^2) = m g h + \frac{1}{2} m v_x^2$$  \frac{1}{2} m v_y^2 = m g ...

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Can you tell from the image below if Q1 and Q2 are attracted or repelled? No, you do not have enough information. Will Q2 only be attracted to the sphere if Q2 is enough bigger than Q1? For any nonzero values of Q1 and Q2 you can compute the distance at which there is no net force. Will the positive charge inside the shell attract electrons interior to ...

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Assume plane waves. The tangential boundary conditions show that the transverse electric and magnetic field vectors must stay in the same direction on transmission or reflexion from the interfaces, assumed aligned with the wavefronts. Since we know the direction of the waves, let's say the $\vec{E}$ fields are all in the $\hat{X}$ direction, the magnetic ...

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Integration is finding the area under a curve that isn't necessarily straight. If you have a velocity time graph and find the area under it, this gives you the distance travailed. If you have a acceleration-time graph the area under it is the change in velocity. There are several techniques to integration, which I will not go into here. As mentioned in the ...

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