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0

Got it! as $H=U+PV$ and $PV = n_{tot}RT=const$ throughout the process we get $\Delta(PV)=0 => \Delta U= \Delta H$


0

I saw this post by accident and was just amazed on how much effort people will put into theories! :) Plenty of people posted their opinions of velocity (falling person) and density (people/snow) but there's a piece that makes all the difference I haven't notice. If you take a cube the same density as a ball... which will sink lower into the snow below? ...


-1

logically in the frame of centre of mass the accn of body is zero so momentum is conserved and as mass has not changed initial velocity is equal to final velocity in frame of centre of mass


0

Multiplying vectorially with $\mathbf n$ your (2), and using the identity $$\tag A \mathbf A \times ( \mathbf B \times \mathbf C) = (\mathbf A \cdot \mathbf C) \mathbf B - (\mathbf A \cdot \mathbf B) \mathbf C $$ you obtain $$ \tag B (\mathbf B_{\text{above}} - \mathbf B_{\text{below}})\times \mathbf n = - \mu_0 \mathbf K. $$ Now use in both the terms in ...


0

OK,maybe I get where my mistake is. It's very important that $ T^{(1)}_{ab}=G^{(1)}_{ab}=0 $ but $ T_{ab}\not=0 $. From http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec14.pdf ,we can learn $$ T^{03}=-\frac{1}{16\pi}[(\frac{\partial h_{11}}{\partial t})^2+(\frac{\partial h_{12}}{\partial t})^2] $$ (c=1 and G=1.)Because $$ ...


0

See picture. The current equal -1 Ampere. It is easy to obtain, if we determine potential in each point. I will set $\phi=0$ in up right corner of your picture. I did mistake in a sign of current on the picture, but I lazily to remake picture.


0

In addition to Pentane's solution, this can be done as a projectile problem as well. The following is a basic kinematics equation $$v_f^2 = v_i^2 + 2 a \Delta x$$ At the top of your jump $v_f=0$, so you can solve for $v_i$ and get the answer that way.


0

Start with the two pertinent conservation laws for elastic collisions: kinetic energy and momentum. Remember that momentum is a vector. In the center of mass frame, the total momentum is zero. That will get you started. Do the work for two particles first. As an aside you should try to show the total momentum is zero in the CoM frame by example by taking ...


0

You can use conservation of energy: the kinetic energy when they land has to be the same as the kinetic energy when they left the ground. The kinetic energy when they land is also equal to the potential energy when the person is 60 cm off of the ground. So you can use: $$mgh=\frac{1}{2}mv^2$$ and the mass cancels out leaving: $$gh=\frac{1}{2}v^2$$


1

About 50 years ago in Reader's Digest there was an article about a Soviet airplane pilot who bailed out at high altitude. He fell into a snow-filled ravine and survived. If the angle of the snow is high enough it is no big deal. At Squaw Valley I have seen skiers do drops that might have been 100 feet. If the landing is steep enough it is OK. It is ...


1

$\newcommand{\ket}[1]{\lvert #1 \rangle}$There is no "reason" you can use the ladder operators. Rather, they are the reason angular momentum is quantized in integer steps. You can define them, there's no inconsistency, so you can use them, and using them leads you to conclude that the angular momentum is raised/lowered in integer steps by them, in the way ...


1

There's no uniform density of heated air. It depends on the temperature (higher T -> lower density) but also on the ambient air pressure. In Denver, cold air is less dense, because the ambient pressure is lower. But this same effect also increases the density of hot air, by the same percentage. So, the result is that the lift of a balloon decreases with ...


1

So I found the answer....I was not taking the final step of multiplying The difference in density between the surrounding air and the heated air and then multiplying by the envelope volume. I was just focusing on the difference and getting stuck there. I chose Denver because it's a mile above sea level with a known air density. I could have chosen Mt Everest ...


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 ...


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 ...


0

The motorcycle is moving at $10\frac{m}{s}$, so his distance from a starting point of 0 each second can be modeled as: $$y = 10x$$ The car is moving at a constant acceleration, so that distance from an origin of 0 can be modeled as: $$y = 2(x^2 + x)$$ Therefore, when they are equal to each other, they are at equal distances from the starting point. ...


0

You made a mistake in the direction of your drag force (vector). You calculated the magnitude of this force correctly, then applied that magnitude both along your horizontal and vertical basis vectors for a total of $\sqrt{2}$ of the desired drag force, and pointing in the wrong direction! The missing direction is given by $-\vec{v}^0 = -\vec{v} / \left| ...


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 ...


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 ...


0

You mean only the gravity is considered? If so, here's the solution. Let's assume the mass of earth is $6 \times 10^{24} kg$. If your speed is much lower than the first cosmic velocity, regard it as the normal projectile motion. For higher speed($\leq v_{1stcosmic.}$) x(t) is like followings. where the f(t) means the distance to center of earth in meter. t ...


0

How the acceleration change may depends on the relation between the two bodies' mass and the coefficient of friction of the surfaces. I will assume the initial friction is exerted by ground. In the following I use U means upper body (your second body), L means lower (your first body), G means ground, 2 stands for 'to'(or any other prepositions). 1.If U is ...


0

The force of friction will increase with the addition of the second mass. However, the mass of the system will increase as well, and the net effect is that the acceleration does not change. To see this, note that: $$a={F \over m}={\mu _kmg \over m}=\mu _kg$$ Where $\mu _k$ is the coefficient of kinetic friction between the moving body and the surface it's ...


0

You got the correct answer. Maybe you want it in terms of $\cos^{-1}$ instead? In that case the expression looks like: $$ \theta_{\tt max}=\cos^{-1}(\sqrt{\frac{n^2-1}{8}}) $$ This is for the case of two internal reflections (the picture on the left). Edit: I think I just realized that you have written down the correct equation and the correct maxima in ...


0

Under the conditions as explained in the OP's comments, the question is indeterminate. Make BC a post and stick it to the ground at C. Make the anchor at C strong enough to resist the $30 N$ force and the $330 N-ft$ (there's a unit for you!) of clockwise torque about $C$. Then the support at $A-B$ is unnecessary. Change the force and torque at $A$ to any ...


1

The function $A_z$ is only a function of $r=\sqrt{x^2+y^2+z^2}$ and $\theta$ (the integral over $z'$ is not a function of $z$, but does depend on $\theta$). I.e., $$ A_z=\frac{\mu e^{-iBr}}{4\pi r}{\tt C(\theta)} $$ Since only (apparently) $A_z$ is non-zero we only need to be able to take the derivative w.r.t. $z$ to get $\nabla \cdot A_z$. $$ \nabla \cdot ...


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 ...


14

@SeƱor 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} ...


26

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 ...


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.


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 ...


0

When you're using the equation F=ma, the F is ALWAYS the total/resultant/net/unbalanced force, NOT one of the individual forces. It describes the effect (the acceleration) that happens due to the cause (the total force on an object). Here you happen to be right because (at least horizontally) there is only one force, the frictional force, so you should get a ...


0

So I'm assuming you're saying that the work done on the car in distance $d$ has to be equal to its kinetic energy $\frac{1}{2} m v^2$. Then, using $W = F d$: $$ F d = \frac{1}{2} m v^2 \\ m a d = \frac{1}{2} m v^2 \\ d = \frac{v^2}{2 a} $$ So, yes, this equation is correct. Your relation between the two forces is also correct. Since mass drops out, the ...


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 ...


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. $\tfrac{1}{d_0}\rightarrow0$.


0

As I expected, this simple question calls for no more than a little sleight of hand, as pointed out by the sole comment above. It looks as if @Meng Cheng won't make it an answer. Thanks to him. And here I confirm that it works well.


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.


0

I(ring) = Idisk(R2) - Idisk(R1). The trick is figuring out the mass. Mass of R2-sized disk would be MR2 = M*(pi*R2*R2)/((pi*R2*R2)-(pi*R1*R1)) Mass of R1-sized disk would be MR1 = MR2*(pi*R1*R1)/(pi*R2*R2) So I(ring) = 1/2MR2*(R2*R2) - 1/2MR1*(R1*R1) = 1/2( (M*(pi*R2*R2*R2*R2) - M*(pi*R1*R1*R1*R1))/((pi*R2*R2) - (pi*R1*R1)) ) ) I guess the pis ...


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 \begin{equation} \sqrt{\frac{1}{m}s\frac{m \cdot L^2}{s^2}}\sqrt{s} = ...


0

Let us denote $\ P$ the point corresponding to the end of the rod with no force applied, then we apply a force $\ \overrightarrow{F} $ on the other end of the rod, such that is parallel to the plane and perpendicular to the rod. Now, with respect to the center of mass, the only force whose torque is different from zero is $\ \overrightarrow{F}$: ...


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 ...


0

Answer B is Correct An aeroplane passenger, given a proper vantage point, will see circular concentric rainbows Rainbows viewed from airplanes are created like rainbows viewed from the ground, except that you're simply viewing it from a different (flying) perspective. That is, the water droplets there use internal reflection and dispersion to shoot light ...


-1

You are supposed to calculate $$ \int_A dA r^2 \sigma = \int_0^a dy \int_0^{b\left(1-y/a\right)} dz \ y^2 \sigma\left(y,z\right) $$ where $r = y$ is the distance of the differential mass $\sigma\left(y,z\right) dy dz$ from the $z$ axis, and $\sigma\left(y,z\right) = 2 M / \left(ab\right)$ is the (constant) mass density of the triangle.


-1

Moment of inertia (with the x-axis as the axis of rotation): $I_x = \int (y^2+z^2) \rho dv$ Density $\rho$ is constant, volume element is $dv = dydzdx$. The integral over $x$ is simply the thickness $t$ of the triangle. Clearly it holds: $M= \rho t \frac{ab}{2}$. Integration interval: $z \in [0,b - \frac{b}{a}y], y \in [0,a]$ (the z-interval depends on the ...


0

The magnitude of the force between $i$ and $j$ is indeed $$ F_{ij} = \frac{Gm_im_j}{\lvert\vec{r}_{ij}\rvert^2}. $$ The direction of this vector is directed along the line connecting the two points, the same as $\vec{r}_{ij}$ (my notation for the vector difference between the positions of $i$ and $j$). In principle you can compute the magnitude for each pair ...


0

As I said in my comment I suggest you to use the Gauss theorem and the fact that inside each metal the field is zero. The picture shows the three metal plates (black), two of them connected to the ground for realizing potential zero. The closed surfaces (light-grey) pass through the metals, and I consider their vertical sides of surface equal to the unity. I ...


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 ...


0

in your answer you ignored the initial velocity the bag will rise a little until its velocity is zero then it will fall again , it is like he found the total distance


0

You are failing to do the integral correctly. In the integral $$ \int r^4\mathrm dV=\iint r^6\,\mathrm dr\,\mathrm d\Omega $$ the $r$ is a function of space that needs to be integrated over. The result will not be $R^4\times \tfrac{4\pi}{3}R^3$ (for $R$ the outer radius of the sphere). Your result is also plainly incorrect from dimensional analysis ...


0

You'll have a different integration to do than here because the volume element in spherical coordinates is r^2 sin(theta) dr d(theta) d(phi). Because there is no dependence on angle (at least none mentioned) the angular integral will give you 4*pi, and the r integral shoud give r^7/7, so that overall rho*(4/7)pir^7 = Q


1

I believe the equation you wrote down for $x(t)$ is not necessarily right depending on the acceleration. I think it assumes the decrease in velocity is simply $-at$, which is only true for constant acceleration. To do this right, you have to start with acceleration and work up to position. First, write down acceleration. $$ a(t)= \frac{dv}{dt} = ...



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