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As a very rude guess, fresh snow (see page vi) can have a density of $0.3 g/cm^3$ and be compressed all the way to about the density of ice, $0.9 g/cm^3$. Under perfect conditions you could see a 13 feet uniform deceleration when landing in 20 feet of snow, or about 4 meters. Going from $30 m/s$ to $0m/s$ (as @Sean suggested in comments), you'd have ...

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This is another chance to use one of my favorite approximations ever! I first offered it as an answer to a question about how deep a platform diver will go into the water. Now is the chance to use it again! Issac Newton developed an expression for the ballistic impact depth of a body into a material. The original idea was expressed for materials of ...

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@Señor O gives a very good answer, but he assumes an ideal deceleration. Based on a viewing of the scene, Anna sinks a little under a meter, while Kristoff doesn't sink more than half a meter. Since they fell about 200 feet (about 60 m), my initial estimate for their impact velocity is (assuming no air resistance): $v = \sqrt{2gh} = \sqrt{2*60*9.8} ... 11 The fog you are seeing is condensation of atmospheric water, not sublimed$CO_2$. The water fog is made very near the boiling surface, and then sinks slowly, exactly as it does in rainclouds. Therefore, just because you can see fog gathering on the floor does not mean that the$CO_2$is confined there. The$CO_2$molecules have a speed, in random ... 10 Nice theoretical answers (I can certainly appreciate them, I'm a mathematician). But why delve into theory when experiment is available? In this video you can see a skier jump from more than 200 feet and get head first into the snow, without a helmet. The video starts with the aftermath, if you want to see the jump right away fast forward to about 1 ... 4 In Scalar Field Theory one defines creation and annihilation operators as: $$a_p = E(p)\phi(p) + i\pi(p)$$ and $$a^\dagger_p = E(−p)\phi(−p) − i\pi(−p)$$ combining together to give the momentum space representation of the quantum field: $$\tilde{\phi}(p) = \frac{a_p + a^\dagger_{-p}}{2E(p)}$$ By taking the Fourier transform of this momentum space ... 4 What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2? The equation you wrote doesn't mention anything about the linear acceleration. Is the center of mass located at its top and not the center? Or is there something else I am missing? The center of mass of the pencil is in the middle, not the top. There is likely something else ... 3 If the cable of the "elevator" is not connected to a point on earth, then the satellite must be in a geostationary orbit (or it will float away); this implies that if you now attach something to the platform (increasing the pull on the cable) you will pull the satellite down to earth. And as @lionelbrits pointed out, the pulling part of a space elevator ... 3 "Equation that is all over the internet"... You started at http://thatsmaths.com/2014/06/26/balancing-a-pencil/ and from there, you linked to http://arxiv.org/pdf/1406.1125v1.pdf which was the source for the former. In the third paragraph of that paper, it states We model the pencil as an inverted simple pendulum with a bob of mass m at one end of ... 3 Your equation is wrong. The flux is not$2EA$. Rather, is$-EA + EA = 0$. The reason is because Gauss's law involves the dot product of$E$and$dA$. The direction of$dA$is the way out of the cylinder. When the electric field goes through the bottom of the cylinder, the electric field is in the opposite direction of the area vector, so the electric flux is ... 3 Given the imprecision in these numbers, that means that you can lift anywhere between 0 and 0.1 kg per m^3 of air. Per Wikipedia, a typical hot air balloon holds 2,800 m^3 of air in the envelope, so it can suspend something between 0 and 280 kg in the basket. A typical human weighs under 100kg, so you could probably lift between one and three people with a ... 3 Dimensional analysis shows the book's answer is wrong. Let's work it out. I should obviously get length because this is a position operator. Since neither$e^{\text{i}jka}$nor the creation annihilation operators have units, I can ignore those terms. This reduces to \sqrt{\frac{1}{m}s\frac{m \cdot L^2}{s^2}}\sqrt{s} = ... 3 For$d=3$the group theoretic meaning of total angular momentum is that it is the Casimir operator of$SO(3)$. For$SO(d)$where$d>3$you have more than one Casimir operator, so it's not clear what you mean by "total angular momentum" In particular the number of Casimir operators is$[d]/2$, where$[d]=d$or$d-1$depending whether$d$is even or odd. 3 Why do we boil water to cook food? It's not actually because there's anything magic about the boiling of water, or that the physical process of boiling in particular does anything. Usually it's because we want a constant-temperature heat bath. Say you are boiling vegetables. You boil water, and you know that water is at 100 degrees. Water actually cannot get ... 3 It is by lepton number and charge, but you can't get energy/momentum to balance. In the$\nu_e$rest frame there isn't enough energy to make the products. If there is a nucleus around, you can imagine the$\nu_e$emitting a virtual$W^+$making the$e^-$, the$W^+$scattering electromagnetically off a nucleus to deal with the momentum, then decaying into ... 3 The lagrangian you're dealing with is$\mathcal{L}= \frac12 (\partial_{\mu} \phi)^2 - \frac12 m^2 \phi^2$. When you take the partial with respect to$\partial_{\mu}\phi$, you should be getting$2 * (\frac12 \partial^{\mu}\phi)$. This would make the first term in your expression$\dot{\phi}^2$instead of$\frac12 \dot{\phi}^2$and things would work out. If ... 2 Noether theorem tells you that if you can find a (one parameter) group of infinitesimal transformations$\alpha$and$\beta$such that: $$t'=t+\alpha\epsilon$$ $$q'^\mu=q^\mu+\beta^\mu\epsilon$$ and your lagrangian is invariant under this group of transformations, then the quantity ... 2 I think the key point that you are missing is that the object is very far away compared to all of the other distances. Combine your final two equations to eliminate$y_0$and then let$d_0\rightarrow\infty$, i.e.$\tfrac{1}{d_0}\rightarrow0$. 2 If you start with the bag stationary at 300m then drop it the bag is going to fall straight down, and its maximum height would indeed just the 300m point it started from. However you're not starting with the bag stationary. You're starting with the bag moving upwards at 13 m/s. So the bag is going to start at 300m then move up, come to a halt, then start ... 2 Let's say car and bike be at rest at$1pm$so,$v_c=0$and$v_b=0$. Calculations for motion of car: Since car is moving with constant acceleration, At 1:00:00pm,$v_c=0m/s$,$S_c=0m$At 1:00:01pm,$v_c=4m/s$,$S_c=4m$At 1:00:02pm,$v_c=8m/s$,$S_c=12m$At 1:00:03pm,$v_c=12m/s$,$S_c=24m$At 1:00:04pm,$v_c=16m/s$,$S_c=40m$Calculations for motion ... 1 So yeah, there are these things called vectors which represent arrows; they are made up of components. For example, we would not say "velocity (horizontal)" and "velocity (vertical)" but we would add labels$v_x, v_y$where$x$is usually a horizontal-label and$y$is usually a vertical-label; then we would write the arrow as$\vec v = [v_x, v_y]$, packaging ... 1 In the ground state, both electrons are in the state with the lowest value of "n". E.g., in the case of an infinite potential well, the lowest quantum number is n=1. In this case, both electrons have n=1, but one electron is spin up and one electron is spin down, because of exclusion. The first excited state is one in which one of the electrons has n=1 ... 1 In this answer we will just make a general conceptional remark about variational/functional derivative (FD), which hopefully implicitly answers OP's specific questions. OP is apparently considering the 'same-spacetime' FD, $$\tag{A}\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} ... 1 The functional derivative \frac{\delta}{\delta \phi} acts on functionals, things that map functions to real numbers. That is, they act on actions S, not lagrangians L. I don't know where you got your original question, but there indeed should be a minus sign! Altogether, I think what you're asking is why:$$ \frac{\delta}{\delta \phi} \int d^4x ... 1 Your calculations are correct. They differ from your model (which uses ABS braking) however, because they don't take into account the duty cycle of the braking. If this is added to your calculations, then the two results should be similar. 1 I suppose the right function to consider for$R$is$R=\Vert\mathbf x_i-\mathbf x_j\Vert$and the usual differentiation operators for the$\nabla$s. So, if$\mathbf x_i = (x_i,y_i,z_i)$, then $$R = \sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}$$ and $$\nabla_i = \left(\frac\partial{\partial x_i},\frac\partial{\partial y_i},\frac\partial{\partial z_i}\right)$$ ... 1 The efficiency of the pumping source is$x$means that$x$amount of electrical power is converted to energy which is useful for pumping the laser medium. The absorption of the pump is$y$means that$y$amount of the energy from the pump source is actually pumped into the medium to generate the population inversion necessary for lasing. The total amount ... 1 So this is a bit tricky actually and perhaps the original question makes this more clear. The problem I'm having when thinking about this is what exactly does the function generator do. Does it take the original force of the string and add some force$F(t)$on top of it, or does it essentially fix the force on one side of the spring? So if the spring is ... 1 Think of the function generator producing prescribed displacements$X(t)$. What is the extension of the first spring? It is$x$, so its restoring force is$-k_1 x$. Now what is the extension of the second spring? It is$X(t)-x$, with its restoring force$-k_2 (X(t)-x)$. What is missing from the diagram is the direction of positive forces and displacements. ... 1 If you need the escape velocity alone, this can be determined from conservation of energy. You can also determine the velocity an object has at a certain position, but if you want to determine the motion diagram you have to solve Newton's equations $$m\ddot{\mathbf a} = -G\frac{mM}{r^3}\mathbf r$$ which in spherical coordinates simplify to$\$\ddot r = ...

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