There are significant challenges to overcome if we are to deploy something similar to Israel’s acclaimed air defense network because the threat to America is quite different.

Solution

Black body radiation, spectral density: ε (ν) dν = hνρ (ν) n (ν)

The photon energy, E = hν where h is Planck’s constant and ν is the photon frequency.

The density of states, ρ (ν) = Aνⁿ⁻¹ where A is a constant independent of the frequency and the frequency term is the scaling of surface area of an n-dimensional sphere.

The Bose–Einstein distribution,

n(v)$=\frac{1}{e\frac{hv}{kT}-1}$

where k is the Boltzmann constant and T is the temperature.

We let

$x=\frac{hv}{kT}$

and get

ε(x)$\propto =\frac{x^n}{e^x-1}$

We do not need the constant of proportionality (which is not simple to calculate in 4D) to find the maximum of ε (x). Working out the constant just tells us how tall the peak is, but we are interested in where the peak is, not the total radiation.

$\frac{d\epsilon }{dx}\propto \frac{n{x}^{n-1}}{e^x-1}-\frac{x^n{e}^x}{{\left(e^x-1\right)}^2}$

We set this equal to zero for the maximum of the distribution,

$\frac{x^{n-1}{e}^x}{{\left(e^x-1\right)}^2}\left(n\left(1-e^{-x}\right)-x\right)=0$

This yields x = n (1 − e^−x) where

$x=\frac{h{v}_{max}}{kT}$

and we can relate

${\lambda }_{max}=\frac{c}{v_{max}}$

and c being the speed of light.

This equation has the solution x = n +W (−ne⁻ⁿ) where W is the Lambert W function z = W (y) that solves ze^z = y (although there is a subtlety about which branch of the function). This is kind of useless to do anything with, though. One can numerically solve this equation using bisection/Newton–Raphson/iteration. Alternatively, one could notice that as the number of dimensions increases, e^−x is small, so to leading approximation x ≈ n. One can do a little better iterating this, x ≈ n − ne⁻ⁿ which is what we will use. Note the second iteration yields

$x\approx n-n{e}^{\left(n-n{e}^{-n}\right)}$

Number of dimensions, n	Numerical solution	Approximation
2	1.594	1.729
3	2.821	2.851
4 (the one we want)	3.921	3.927
5	4.965	4.966
6	5.985	5.985

Using the result above,

${\lambda }_{max}=\frac{hc}{kT{x}_{max}}=\frac{6.63 \times {10}^{-34}·3\times {10}^8}{1.38\times {10}^{-23}·6\times {10}^3·3.9}=616 nm$

616 nm is middle of the spectrum, so it will look white with a green-blue tint. Note, we have used T = 6000 K for the temperature here, as given in the question.

It would also be valid to look at ε (λ) dλ instead of ε (ν) dν.

Solution

First we will set up the equations of motion for our system. We will set one mass to be at position −x and the other to be at x, so the masses are at a distance of 2x from each other. Starting from Newton’s law of gravity:

$F=-\frac{G{m}^2}{{\left(2x\right)}^2}$

we can then use Newton’s second law to rewrite the LHS,

$m\ddot{x}=-\frac{G{m}^2}{4x^2}$

which we can simplify to

$\ddot{x}=-\frac{Gm}{4x^2}$

It is important that you get the right factor here depending on your choice for the particle coordinates at the start. Note there are other methods of getting this point, e.g. reduced mass.

We can now solve the second order ODE above. We will not show the whole process here but present the starting point and key results. We can write the acceleration in terms of the velocity. The initial velocity is zero and the initial position

$x_i=\frac{d}{2}$

So,

$v\frac{dv}{dx}=-\frac{Gm}{4x^2}\to \underset{0}{\overset{v}{\int }}vdv=-\frac{Gm}{4}\underset{x_i}{\overset{x}{\int }}\frac{dx}{x^2}$

and once the integrals are solved we can rearrange for the velocity,

$v=\frac{dx}{dt}=-\sqrt{\frac{Gm}{2}}\sqrt{\frac{1}{x}-\frac{1}{x_i}}$

Now we can form an expression for the total time taken for the masses to meet in the middle,

$T=\sqrt{\frac{2}{Gm}}\underset{0}{\overset{x_i}{\int }}\frac{dx}{\sqrt{\frac{1}{x}-\frac{1}{x_i}}}$

There are quite a few steps involved in solving this integral, for these solutions, we shall make use of the following (but do attempt to solve it for yourselves in full).

$\underset{0}{\overset{1}{\int }}\sqrt{\frac{y}{1-y}}dy=si{n}^{-1}\left(1\right)=\frac{\pi }{2}$

Hence,

$T=\frac{\pi }{2}\sqrt{\frac{2x_i^3}{Gm}}=\frac{\pi }{2}\sqrt{\frac{d^3}{4Gm}}$

We can now rearrange for G and substitute in the values given in the question, don’t forget to convert the time into seconds.

$G=\frac{d^3}{4m}{\left(\frac{\pi }{2T}\right)}^2=6.67\times {10}^{-11} m^{3}k{g}^{-1}{s}^{-2}$

This is the generally accepted value for the gravitational constant of our universe as well.

Solution

We will write the period of oscillation of the pendulum at the surface of the Earth to be

$T_0=2\pi \sqrt{\frac{l}{g_0}}$.

At a height h above the surface of the Earth the period of oscillation will be

$T_h=2\pi \sqrt{\frac{l}{g_h}}$,

where g₀ and g_h are the acceleration due to gravity at the surface of the Earth and a height h above it respectively.

We can define τ₀ to be the total duration of the day which is 8.64 × 10⁴ seconds and equal to N complete oscillations of the pendulum at the surface. The lag is then τ_h which will equal N times the difference in one period of the two clocks, τ_h = NΔT, where ΔT = (T_h − T₀). We can now take a ratio of the lag over the day and the total duration of the day:

$\frac{{\tau }_h}{{\tau }_0}=\frac{N\left(T_h-T_0\right)}{N{T}_0}\Rightarrow {\tau }_h={\tau }_0\frac{T_h-T_0}{T_0}={\tau }_h={\tau }_0\left(\frac{T_h}{T_0}-1\right)$

Then by substituting in the expressions we have for the period of a pendulum at the surface and height h we can write this in terms of the gravitational constant,

${\tau }_h={\tau }_0\left(\sqrt{\frac{g_0}{g_h}}-1\right)$

[Award 1 mark for finding the ratio of the lag over the day and the total period of the day.]

The acceleration due to gravity at the Earth’s surface is

$g_0=\frac{GM}{R^2}$

where G is the universal gravitational constant, M is the mass of the Earth and R is the radius of the Earth. At an altitude h, it will be

$g_h=\frac{GM}{{\left(R+h\right)}^2}$

[Award 1 mark for finding the expression for the acceleration due to gravity at height h.]

Substituting into our expression for the lag, we get:

${\tau }_h={\tau }_0\left(\sqrt{\frac{{\left(R+h\right)}^2}{R^2}}-1\right)={\tau }_0\left(\sqrt{1+\frac{2h}{R}+\frac{h^2}{R^2}}-1\right)={\tau }_0\left(\frac{\sqrt{R^2+2hR+h^2}}{R}-1\right)={\tau }_0\left(\frac{R+h}{R}-1\right)$

This simplifies to an expression for the lag over a day. We can then substitute in the given values to find,

${\tau }_h=\frac{{\tau }_{0}h}{R}=\frac{8.64\times {10}^4 s·300 m}{8.3781\times {10}^6 m} =4.064 s\approx 4 s$

[Award 2 marks for completing the simplification of the ratio and finding the lag to be ≈ 4 s.]

Solution

We can imagine this as an inverted pendulum, with gravity acting from the centre of mass $\left(\frac{l}{2}\right)$ and at an angle θ from the unstable equilibrium point.

[Award 1 mark for a suitable diagram of the system.]

We must now find the equations of motion of the system. For this we can use Newton’s second law $\overrightarrow{F}=m\overrightarrow{a}$ in its rotational form τ = Iα (torque = moment of inertia × angular acceleration). We have another equation for torque we can use as well

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left|\overrightarrow{r}\right|\left|\overrightarrow{F}\right|\sin \theta \widehat{n}$

where r is the distance from the pivot to the centre of mass $\left(\frac{l}{2}\right)$ and F is the force, which in this case is gravity mg. We can then equate these giving

$\left|\overrightarrow{r}\right|\left|\overrightarrow{F}\right|\sin \theta =I\alpha$

Substituting in the given moment of inertia of the stick and that the angular acceleration

$\alpha =\frac{{\delta }^2\theta }{\delta t^2}=\ddot{\theta }$

We can cancel a few things and rearrange to get a differential equation of the form:

$\ddot{\theta }-\frac{3g}{2l}\sin \theta =0$

we then can take the small angle approximation sin θ ≈ θ, resulting in

$\ddot{\theta }-\frac{3g}{2l}\theta =0$

[Award 2 marks for finding the equation of motion for the system and using the small angle approximation.]

Solve with ansatz of θ = Ae^ωt + Be^−ωt, where we have chosen

${\omega }^2=\frac{3g}{2l}$

We can clearly see that this will satisfy the differential equation

$\dot{\theta }=\omega A{e}^{\omega t}-\omega B{e}^{-\omega t} and \ddot{\theta }={\omega }^{2}A{e}^{\omega t}+{\omega }^{2}B{e}^{-\omega t}$

Now we can apply initial conditions to find A and B, by looking at the two cases from the uncertainty principle

$\Delta x\Delta p=\Delta xm\Delta v\frac{\hslash }{2}$

Case 1: The stick is at an angle but not moving

At t = 0, θ = Δθ

θ = Δθ = A + B

At t = 0, $\dot{\theta }=0$

$\dot{\theta }=0=\omega A{e}^{{\omega }_0}-\omega B{e}^{-{\omega }_0}$, A=B

This implies Δθ = 2A and we can then find

$A=\frac{\Delta \theta }{2}=\frac{2\Delta x}{2l}=\frac{\Delta x}{l}$

So we can now write

$\theta =A\left(e^{\omega t}-e^{-\omega t}\right)=\frac{\Delta v}{\omega }\left(e^{\omega t}-e^{-\omega t}\right)$ or $\theta =\frac{2\Delta v}{\omega l}\sin h \omega t$

Case 2: The stick is at upright but moving

At t = 0, θ = 0

This condition gives us A = −B.

At t = 0, $\ddot{\theta }=\frac{2v}{l}$

This initial condition has come from the relationship between the tangential velocity, Δv which equals the distance to the centre of mass from the pivot point, $\frac{l}{2}$ and the angular velocity $\dot{\theta }$. Using the above initial condition gives us $\dot{\theta }=2\omega A$ where $A=\frac{\Delta v}{\omega l}$

We can now write

$\theta =A\left(e^{\omega t}-e^{-\omega t}\right)=\frac{\Delta v}{\omega }\left(e^{\omega t}-e^{-\omega t}\right) or \theta =\frac{2\Delta v}{\omega l}\sinh \omega t$

[Award 4 marks for finding the two expressions for θ by using the two cases of the uncertainty principle.]

Now there are a few ways we can finish off this problem, we shall look at three different ways. In each case when the stick has fallen on the ground $\theta \left(t_f\right)=\frac{\pi }{2}$.

Method 1

Take $\theta =\frac{2\Delta x}{l}\cosh \omega t$ and $\theta =\frac{2\Delta v}{\omega l}\sinh \omega t$, use $\theta \left(t_f\right)=\frac{\pi }{2}$ then rearrange for t_f in both cases. We have

$t_f=\frac{1}{\omega }\cos h{}^{-1}\frac{\pi l}{4\Delta x} and t_f=\frac{1}{\omega }\sinh {}^{-1}\frac{\pi \omega l}{4\Delta v}$

Look at the expression for cosh⁻¹ x and sinh⁻¹ x given in the question. They are almost identical, we can then approximate the two arguments to each other and we find,

$\Delta x=\frac{\Delta v}{\omega }$

we can then substitute in the uncertainty principle $\Delta x\Delta p=\frac{\hslash }{2}$ as $\Delta v=\frac{\hslash }{2m\delta x}$ and then write an expression of $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$, which we can put back into our arccosh expression (or do it for Δv and put into arcsinh).

$t_f=\frac{1}{\omega }\cosh {}^{-1}\frac{\pi l}{4\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

Method 2

In this next method, when you get to the inverse hyperbolic functions, you can take an expansion of their natural log forms in the tending to infinity limit. To first order both functions give ln 2x, we can then equate the arguments and find Δx or Δv in terms of the other and use the uncertainty principle. This would give the time taken as,

$t_f=\frac{1}{\omega }\ln \frac{\pi l}{2\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

Method 3

Rather than using hyperbolic functions, you could do something like above and do an expansion of the exponentials in the two expressions for t_f or we could make life even easier and do the following.

Disregard the e^−ωt terms as they will be much smaller than the e^ωt terms. Equate the two expressions for $\theta \left(t_f\right)=\frac{\pi }{2}$ and then take the natural logs, once again arriving at an expression of

$t_f=\frac{1}{\omega }\ln \frac{\pi l}{2\Delta x}$

where $\Delta x=\sqrt{\frac{\hslash }{2m\omega }}$ and $\omega =\sqrt{\frac{3g}{2l}}$.

This method efficiently sets B = 0 when applying the initial conditions.

[Award 2 marks for reaching an expression for t using one of the methods above or a suitable alternative that gives the correct units for time.]

Then, by using one of the expressions above for time, substitute in the values and find that t = 10.58 seconds.

[Award 1 mark for finding the correct time value of t = 10.58 seconds.]

• If you’re a student who wants to sign up for the 2025 edition of PLANCKS UK and Ireland, entries are now open at plancks.uk

The post PLANCKS physics quiz – the solutions appeared first on Physics World.

Le maestro Benoît continue d'affoler les compteurs et a atteint sa 100e victoire dans N'oubliez pas les paroles ce jeudi 30 janvier 2025 sur France 2. En fin d'émission, Nagui a annoncé une nouvelle surprenante…

Le Stade Brestois le redoutait. En barrages d'accession aux 8es de finale de la Ligue des Champions, le club finistérien affrontera le Paris Saint-Ger......

Les Bretons rencontreront le PSG lors des barrages de la C1. Comme durant la phase aller, Monaco croisera Benfica.

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Celtic will host the first leg as they compete in a Champions League knockout tie for the first time in 12 years

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Star posted cryptic message online, saying: ‘My mom and dad are gonna be so mad at me!!’

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A ‘Hervana’ performance at the FireAid benefit concert in California was met with a rapturous reception from fans of the Nineties rock band

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Hélène Miard-Delacroix, professeure d'histoire et civilisation de l'Allemagne contemporaine à Sorbonne Université, aborde le rapprochement entre les conservateurs et l'extrême droite en Allemagne. "Cette montée de l'extrême droite en Allemagne est relativement rapide", analyse-t-elle.

Le Paris SG, l’AS Monaco et le Stade brestois 29 sont fixés sur leur sort lors des barrages de la Ligue des champions.

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Tom and Ben Curry will start together for England in Dublin while Maro Itoje steps up as captain.

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Bang played an abusive husband in the Apple TV+ series

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The Argentine forward could present an opportunity for United to boost their flexibility in the transfer market and fend off any problems from PSR, though his talent remains alluring after shining at Old Trafford since graduating from the academy in 2022

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The 1,119-foot-long Triton-class ship is perfect for everyone

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C’est le début de la fin pour l’ADSL en France. Orange a commencé les travaux visant à mettre au rebut ce réseau historique : pas moins de 162 communes et 210 000 locaux vont être débranchés à partir aujourd’hui. L’idée est évidemment de mettre l’accent sur la fibre optique, de plus en plus implantée. 829 communes seront déconnectées en janvier 2026, puis 2 150 au début 2027. L’objectif est d’abandonner totalement l’ADSL d’ici 2030. La liste des villes concernées par cette première vague est disponible sur le site d’Orange.

L’abandon de l’ADSL avait été annoncé par Orange en 2019 face à la montée en puissance de la fibre. 9 Français sur 10 sont raccordables tandis que quasiment 3/4 des foyers y sont déjà connectés. Si les habitants des communes concernées ont logiquement été préparés à cette migration forcée, l’opérateur a travaillé avec les maires pour identifier les lignes ADSL restantes et assurer une transition sans accrocs. « Personne ne sera laissé sur la route », a déclaré le patron d’Orange France auprès du Figaro.

Évidemment, quelques-uns ont tout de même été laissés sur le carreau. Orange estime que moins de 7 000 accès sur les 210 000 foyers débranchés n’avaient pas pris les mesures nécessaires. L’opérateur nuance en affirmant qu’il s’agit pour la plupart de locaux inactifs ou de gens en pleine migration. Il n’était plus possible de prendre une ligne ADSL depuis un an dans les villes concernées, et la fermeture du réseau cuivre ne peut pas être enclenchée dans une zone où la fibre n’a pas été déployée complètement.

Orange a tout intérêt à initier ce chantier. Premièrement, les abonnements aux offres ADSL sont en chutes libres. « Nous avons perdu 20 % d’abonnés sur le cuivre en un an. C’est un réseau de 1 million de kilomètres qui s’endort », a déclaré la directrice générale d’Orange. Deuxièmement, la fibre a l’avantage d’être plus économe tout en étant moins sensible aux perturbations électroniques. Troisièmement, le cuivre peut être recyclé et revendu à bon prix : Orange espère réussir à compenser le coût des travaux à la revente. L’intérêt écologique de l’opération est également évident.

Ce plan de démantèlement va continuer sur les prochaines années, Orange s’occupant de démonter les poteaux, câbles et autres armoires techniques devenus obsolètes. L’opérateur va devoir travailler minutieusement pour éviter les plaintes des retardataires. Il compte bien sur ses concurrents pour faire la retape de la fibre, SFR et consorts lui payant une commission pour l’utilisation de son réseau cuivre. « Nous avons besoin des élus locaux, l’État aussi a son rôle à jouer », ajoute la directrice générale d’Orange.

One of the important pieces of open-source software still working toward proper Wayland support is the Chromium Embedded Framework "CEF" that in turn is depended upon by software like Steam, OBS Studio, Spotify, and many other software packages for having an in-app browser-type experience. The good news is there has been some recent progress on native Wayland support for CEF...

Ce mois de février propose plusieurs sorties intéressantes avec des arrivées très attendues par les joueurs. On y retrouvera le retour de l’une des franchises phares de Capcom, mais également le spin-off version pirate d’une série bien appréciée par ses fans.

Février marque le véritable début d’année 2025 en ce qui concerne les sorties vidéoludiques : si le nombre total de nouveautés peut paraître assez faible, la qualité devrait être au rendez-vous. Tout d’abord, le mois commencera de façon très forte avec la sortie de deux titres très attendus, à savoir Kingdom Come Deliverance 2 et Civilization VII sur consoles et PC.

En revanche, la fin de mois sera marquée par l’arrivée de deux productions japonaises très attendues par les joueurs : le plus que prometteur Monster Hunter Wilds et l’imprévisible, mais totalement déjanté, Like a Dragon: Pirate Yakuza in Hawaii.

Concernant les autres sorties à surveiller de près, la compilation Yu-Gi-Oh ! Early Days Collection sortira sur Nintendo Switch et PC et s’adressera davantage aux premiers joueurs de la série. Les fans de la belle aventurière Lara Croft seront ravis de la retrouver dans Tomb Raider IV-V-VI Remastered tandis que l’exclusivité Xbox et PC de ce nouveau mois sera Avowed, un titre développé par la talentueuse équipe de chez Obsidian Entertainment.

Notre calendrier complet pour février 2025 se trouve ci-dessous :

Date	Jeu	Plateforme(s)
04/02/2025	Kingdom Come Deliverance 2	PS5 - Xbox Series X|S - PC
05/02/2025	Rift of the NecroDancer	PC
06/02/2025	Candy Crush Solitaire	iOS - Android
06/02/2025	Momodora: Moonlit Farewell	PS5 - Xbox Series X|S - Nintendo Switch
06/02/2025	Sworn	PC
07/02/2025	Sugoro Quest: Dice Heroes	PS5 - Xbox Series X|S - PS4 - Xbox One - Nintendo Switch
11/02/2025	Civilization VII	PS5 - Xbox Series X|S - PS4 - Xbox One - Nintendo Switch - PC
12/02/2025	Legacy: Steel & Sorcery	PC
13/02/2025	Phantom Breaker: Battle Grounds Ultimate	PS5 - Xbox Series X|S - PS4 - Xbox One - Nintendo Switch - PC
13/02/2025	Urban Myth Dissolution Center	PS5 - Nintendo Switch - PC
14/02/2025	Afterlove EP	PS5 - Xbox Series X|S - Nintendo Switch - PC
14/02/2025	Date Everything !	PS5 - Xbox Series X|S - Nintendo Switch - PC
14/02/2025	Tomb Raider IV-V-VI Remastered	PS5 - Xbox Series X|S - PS4 - Xbox One - Nintendo Switch - PC
16/02/2025	Berserk Boy	PS5
18/02/2025	Avowed	Xbox Series X|S - PC
18/02/2025	Lost Records: Bloom & Rage	PS5 - Xbox Series X|S - PC
20/02/2025	Ninja Issen	Nintendo Switch
20/02/2025	Stories from Sol: The Gun-Dog	PS5 - PS4 - Nintendo Switch - PC
20/02/2025	The King of Fighters XIII Global Match	PC
21/02/2025	Like a Dragon: Pirate Yakuza in Hawaii	PS5 - Xbox Series X|S - PS4 - Xbox One - Nintendo Switch - PC
21/02/2025	RPG Maker WITH	PS5 - PS4
25/02/2025	Blasphemous	iOS - Android
25/02/2025	Ninja Five-O	PS5 - PS4 - Nintendo Switch - PC
27/02/2025	DanMachi – Fullland of Water and Light	Nintendo Switch - PC
27/02/2025	Lara Croft and the Guardian of Light	iOS - Android
27/02/2025	Yu-Gi-Oh ! Early Days Collection	Nintendo Switch - PC
28/02/2025	Monster Hunter Wilds	PS5 - Xbox Series X|S - PC
28/02/2025	Omega 6: The Triangle Stars	Nintendo Switch - PC
28/02/2025	PGA Tour 2K25	PS5 - Xbox Series X|S - PC

Alors, sur quels titres allez-vous craquer ?

Cet article Le calendrier des sorties jeux vidéo du mois de février 2025 est apparu en premier sur JVFrance.

Tens of thousands of people have gathered on a riverbank near Bangladesh’s capital to listen to sermons by Islamic scholars at the Biswa Ijtema or global congregation of Muslim devotees

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Le tirage au sort des barrages d’accession aux huitièmes de finale de la compétition a eu lieu vendredi 31 janvier, à Nyon (Suisse).

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The Gunners secured their place in the top eight after winning all four of their games at home

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A flight simulator has recreated the journey an American Airlines plane took before a fatal crash with a US Army helicopter on Wednesday, 29 January.

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After a freezing night out in the open, hundreds of striking Serbian students have resumed their 2-day anti-graft protest march from the capital, Belgrade to the northern city of Novi Sad

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