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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 ...
@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} ... 16 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 ... 11 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 ... 5 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 ... 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 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 In the general case you want the Cartan-Karlhede algorithm. It is an algorithm for producing a complete set of classifying invariants for a metric, expressed as functions of the coordinates. Given the components of the metric$g$in the coordinates$x_1, x_2, \ldots, the algorithm produces a list \begin{align} \Lambda & = \Lambda(x_i) \\ \Psi_k & = ... 3 Why do we boil water to cook food? It's not actually because there's anything magic about the boiling of water, or that the physical process of boiling in particular does anything. Usually it's because we want a constant-temperature heat bath. Say you are boiling vegetables. You boil water, and you know that water is at 100 degrees. Water actually cannot get ... 3 It is by lepton number and charge, but you can't get energy/momentum to balance. In the\nu_e$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 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. 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 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 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 ... 2 What's wrong about my process? You're answer cannot be correct since it does not satisfy KVL. For your final charge solution, the final voltages across the capacitors are given by $$V_{C1} = \frac{16.08}{4.08}\mathrm V = 3.94\mathrm V$$ $$V_{C2} = \frac{24.39}{6.19}\mathrm V = 3.94\mathrm V$$ $$V_{C3} = \frac{13}{3.3}\mathrm V = 3.94\mathrm V ... 2 It is up to you where to define the zero potential (potential energy undefined up to a constant, so by adding any constant, the zero becomes anywhere you want it), but let's consider the point where the potential itself is minimum and subtract this value so that the potential is everywhere positive except at it's minimum (just a convention)$$\vec{F} = ... 2 There's no such thing as an imaginary point. In other words, you can certainly plug imaginary numbers into a formula, but those imaginary numbers don't represent points in space, and thus the results you get will not represent the conditions at any actual point. However, you don't need an actual point in space for this purpose. The potential energy field ... 2 If you have a good grasp on the relative scales of other things in physics, you may be able to relate the wavelengths to those. Otherwise, your best bet is just to memorize the wavelengths (or frequencies). Since$f = \frac{c}{\lambda}$for light, you'll be able to figure out the frequencies if you know the wavelengths. Here are some things I use to help ... 2 A good reference was given in an answer to a related question: Cserti 2000 (arXiv preprint, whose numbers I'll be referring to) solved a number of generalizations of the 2D lattice problem. For a$d$-dimensional lattice, the resistance between the origin and the point$(l_1, \ldots, l_d)$is given by eq. 18 in that paper: $$R(l_1, \ldots, l_d) = R_0 ... 2 If you were to Wick rotate t \rightarrow i \theta, the metric would be ds^2 = dr^2 + r^2 d\theta^2, which is just flat space in polar coordinates. The standard cartesian coordinates can be obtained by x=r\cos\theta, y=r\sin\theta. The same procedure works in the original Lorentzian signature metric, but with hyperbolic trig functions instead of sines ... 1 I've come across an answer in Peskin's Introduction to Quantum Field theory where he looks at how to make the 3-momentum delta function invariant that might satisfy you. Look at a boost in the p_3 direction so that p'_3= \gamma(p_3+\beta E),E'=\gamma(E+\beta p_3).$$ \delta^3(p-q)= \delta^3(p'-q')\frac{dp'_3}{dp_3} = ... 1 The typical galaxy is small. The argumentation (which is not my own thinking) goes as follows: The distribution of the galaxies' luminosities is given by the luminosity function$\phi(L)$. Assuming a relation between the luminosities and sizes$R \propto L^{0.4}$(Holmberg 1975), this can be translated to a size distribution$\phi(R)$. The typical size of ... 1 You can also do the following, which may not be as general as you want it, but the idea might be usefull for other problems. You already know that the given metric is Minkowski metric in different coordinates, so look at the null geodesics. In the usual coordinates$(t',x')$they are given by$x'\pm t'=const$. Then find the null geodesics in the given ... 1 After the mass exits the pipe, the tube will start to rotate from the recoil. At$t=0$, there is zero angular momentum,$L=0$. Let's take it that the pipe rotates about its centre-of-mass, and use that point as the origin from which to calculate the angular momentum. At time$t+dt$, the puff of gas has angular momentum$L_{gas} = dm \times l/2 \times v0$. ... 1 As a hint: remember that $$\sigma_i^{2n}=\mathbb I$$ and that $$\sigma_i^{2n+1}=\sigma_i$$ so when you taylor expand $$e^{i\sigma_i\theta}$$ you will have the sum split in just two terms, one proportional to the identity and one to the sigma matrix. EDIT: an operator function of the form $$F=e^{\alpha M}$$ can be expanded as$$F=\sum \frac{(\alpha ... 1 Try to compute$\sigma_2^2$and$\sigma_2^3\$ and see what you get. I think you can take it from here.