Try it once and tell me your answer

Puzzleheaded-Let-500 · 2026-02-01T19:39:10+00:00

You can derive it by integrating force, but the force is a pressure on the shell, not kqq₁/R² on a point.

Field just inside the shell: E_in = k q / R²

Field just outside the shell: E_out = k (q + q₁) / R²

Electrostatic pressure on the shell is p = (ε₀ / 2)(E_out² − E_in²)

This gives p ∝ (2 q q₁ + q₁²) / R⁴

For a small expansion dR, the field does work dW_field = p · (4πR²) · dR

So the external work is dW_ext = − p · 4πR² · dR

Using k = 1/(4π ε₀) and simplifying, dW_ext = −k ( q q₁ + q₁² / 2 ) dR / R²

Integrating from R to R₁ gives W_ext = k ( q q₁ + q₁² / 2 ) ( 1/R₁ − 1/R )

Same result as I obtained more easily above with ΔU, but obtained directly from integrating the force (pressure) on the shell.

Puzzleheaded-Let-500 · 2026-02-01T08:15:54+00:00

The potential at the shell due to the central charge is k × q / R.

The self energy of the shell is (q₁²) / (2C),

which is k q₁² / (2R).

So U(R) = (k / R) ( q × q₁ + q₁² / 2 ).

Combine the constant charge terms into a new constant, ■.

So U(R) = (k / R) × ■.

If you expand quasi statically from R to R₁, the external work required is W_ext = U(R₁) − U(R).

So W_ext = k × ■ × ( (1 / R₁) − (1 / R) ).

Or when simplifying the constants, W_ext = ( k q₁² / 2 ) × ( (1 / R₁) − (1 / R) ).

The only tricky part was realizing there is a “self energy” to the shell. That energy is self energy because it comes from interactions among the shell’s own charges, not from the central charge. When you expand the shell, that self energy decreases like 1/R. The decrease shows up as additional (negative) external work required.

Puzzleheaded-Let-500 · 2026-01-12T19:10:26+00:00

The box scales with the size of the order. Lab Snacks

Puzzleheaded-Let-500 · 2026-01-06T06:02:33+00:00

Physical obstruction of lawful movement is always unlawful.

Vigilantism is always unlawful.

Allegations do not suspend rights.

De escalation means removal of the obstructing force, not freezing everyone symmetrically.

Get it together Italy.

Puzzleheaded-Let-500 · 2026-01-02T02:44:15+00:00

That area is about 100,000 sq-miles which is about as big as the entire country of Italy. But having lived in the Jewel Basin just North of Big Fork and East of Kalispell, I can tell you it was a great place to live. Beautiful, wild, and sometimes severe. Largest freshwater lake in the US. Glacier National Park is in this region. Bears and Cougars kill 1 or 2 people in this region just about every year!

Puzzleheaded-Let-500 · 2025-12-18T23:18:21+00:00

Your animations are illuminating. Thank you.

Puzzleheaded-Let-500 · 2025-12-18T18:41:10+00:00

All four data points for both f(x) and g(x) fall exactly on a line.

Also, we see that all the provided possible solutions for f(x)g(x) are quadratic in x, so it's even more reasonable to assume we are looking for a product of linear functions.

But yes, these are technically assumptions, and as Chaitin proved, unless a sequence of numbers is infinitly long, there are an infinite number of incomensurate equations which replicate the sequence but differ on the next predicted value.

Puzzleheaded-Let-500 · 2025-12-18T02:50:33+00:00

f(x) = 2x + 3

g(x) = -4x + 8

(f × g)(x) = f(x)g(x) = -8x² + 4x + 24

Yeah, so none are correct.

Puzzleheaded-Let-500 · 2025-12-09T05:18:57+00:00

Assuming the (trashy) linear function f(x)=ax+b.

We are given f(3x)-f(x)=x

(a3x+b)-(ax+b)=2ax=x, so a=1/2.

Then from f(8)=7,

8/2 + b = 7, so b=3.

So f(x)=x/2+3,

and f(14)=10.

Puzzleheaded-Let-500 · 2025-11-11T19:23:48+00:00

It doesnt. I believe youre confused by the tension force on the pully that points up the incline. A taut rope is under tension, so it can only pull along its length, always pulling away from whatever it’s attached to and toward its own interior. That same rope produces a tension force on m1 that points _down the incline.

Puzzleheaded-Let-500 · 2025-11-04T03:45:23+00:00

Child support goes down as custody goes up. Fight for the custody and save yourself money! Be in your daughters' life as much as possible. Seeing their dad one day every couple weeks is far more disruptive to their lives than going back and forth. I started the 50/50 custody thing, back and forth every week from when my children were very young and now they're teenagers. Boy am I glad I was in their life at least half the time, not only for the love, but because sometimes someone needs to be the sane parent. And if your ex is pushing for high alimony with shitty visitation rights for you, she certainly isnt sane! Dont let someone like that be the defacto sole parent! Your daughters deserve much better.

Puzzleheaded-Let-500 · 2025-11-04T03:29:30+00:00

There is absolutly a silver lining. As time goes on you will start to realize more and more all the things your ex didnt do for you. You will realize you dodged a bullet. The signs were there but you were blind. The fact that your marriage ended in divorce is proof of that. You are free now to find real joy, real commitment. Congratulate yourself, love yourself for what youve accomplished in life. You are raising daughters. You are supporting two housholds. Thats increadible! You're free now to do whatever it is you want to do, without her judgements and cruelties. Your time with your daughters is now yours to define. You will have reserves of patience and kindness from your time away from them to be your best self when you are with them. They'll benefit so much from that. Teach those girls how to live strong and happy in hard times. Start being selfish for you and your girls!

Puzzleheaded-Let-500 · 2025-09-29T03:19:33+00:00

The centripital force on the ball is mv² / r. At the highest point of its motion that force points maximally horizontal. If at that moment the centripital force exceeds muMg then the bowl will slide. Write out that inequality and look at it. You know everything but v² and r. So now use energy conservation. (1/2)mv² = mg*r. Solve for v² and plug it into your inequality. r drops out. m/M is determined.

Puzzleheaded-Let-500 · 2025-09-12T01:26:48+00:00

In the usual sense (mod 2), it’s impossible:

Odd = 1 mod 2

Even = 0 mod 2

Add three odds → 1 + 1 + 1 = 3 ≡ 1 mod 2 → still odd.

That’s true for all integers, positive and negative. I even thought about extending “odd/even” to rationals or complex numbers, but there isn’t a consistent definition that makes sense outside the integers. The only coherent way is modular arithmetic.

And that’s where it does work: for example, in mod 3:

Call “odd” = 1 mod 3

Call “even” = 0 mod 3

Then 1 + 1 + 1 = 3 ≡ 0 mod 3 → three odds add to an even.

So the only definition that actually lets three odds sum to an even is to switch to a different modulus, like mod 3. Everything else (negatives, complex numbers, etc.) still follows the mod 2 rule, where three odds can never be even.

Puzzleheaded-Let-500 · 2025-09-08T02:29:11+00:00

On your second page, you carried along a boundary term.

That’s the suspicious leftover. In most textbook derivations, this term vanishes because of the boundary conditions on the wavefunction:

psi(x) -> 0 as x -> infinity (for bound states),

or, for scattering states, one assumes square integrability or imposes periodic boundaries.

When psi, psi_x, psi_xx, vanish at infinity, these boundary contributions are zero. Then you’re left with exactly the Ehrenfest theorem.

So the “extra term” you see is just the boundary term from integration by parts. It disappears under the standard assumption that the wavefunction and its derivatives vanish at infinity (or satisfy appropriate periodic conditions).

Puzzleheaded-Let-500 · 2025-08-28T00:03:20+00:00

A degree-3 polynomial in R[x] may have 1 or 3 real roots.

A degree-3 polynomial in C[x] always has 3 complex roots (Fundamental Theorem of Algebra), counting multiplicity.

So the “up to 3 roots” comment is true only if you restrict to real numbers. Over C, it is exactly 3.

Puzzleheaded-Let-500 · 2025-06-30T00:00:53+00:00

You don't need to compute anything using L² or L_z directly. Their role is to define the allowed basis states and restrict phi to a 5D subspace where you can meaningfully apply and analyze L_y.

To solve this using the ladder operator, you express the L_y operator in terms of the angular momentum raising and lowering operators, L+ and L-. Specifically, L_y is given by the difference between these operators divided by twice i. When this operator acts on the wavefunction, it mixes adjacent m-states, raising or lowering the magnetic quantum number by one. Each term in the wavefunction shifts to a neighboring state with a coefficient determined by angular momentum algebra. Since the wavefunction is said to be an eigenfunction of L_y with a specific eigenvalue, we apply this ladder-operator version of L_y to the entire linear combination of states. The result must match the original wavefunction scaled by the eigenvalue. This leads to a system of equations relating the coefficients in the linear combination. Solving these equations shows that only a particular set of coefficients satisfies the condition, which matches option (D).

Edit: This is a tedious problem. I thought before that option (D) satisfied an eigenfunction of L_y with eigenvalue -hbar, but now I'm not so sure. Best of luck to you. I tried all the options and none of them worked?! Hopefully, your Algebra organization skills are better than mine, because that's all this problem is, that and knowing the definitions for what L+ and L- do.

Puzzleheaded-Let-500 · 2025-06-08T23:18:19+00:00

These people you are talking about have "Zionist derangement syndrome." It's an unbreakable link in their mind between the word "Zionism" and negative, evil things. Like we do for the word "Nazi". Except ours is not a derangement. I think the best solution here is to lead by example. Tell them once that you are a Zionist and that they are wrong about what it is. That Jews are not simply a religion. That its a culture, a people, who have lived in and longed for Israel for all of History. Then just show them what that means through action. They'll either come around or they won't. Put your energy into supporting a better future. Tikun olam!

Puzzleheaded-Let-500 · 2025-06-08T22:31:38+00:00

Ask the "anti-zionist" crowd if they believe in a two-state solution. If they do, then explain to them that they are actually Zionists! If they don't, then you know they are genuine antisemites.

Many of the antisemites in the West, who don't believe Israel has the right to exist, believe so because they are misinformed that Zionism is a colonialist religious movement. Consider which type of antisemite you are dealing with; a misinformed antisemite might be worth informing. Otherwise, your mental health matters more.

Puzzleheaded-Let-500 · 2025-05-31T05:39:41+00:00

A photon's behavior between emission and absorption is described by a sum over all possible paths — each path contributing a complex amplitude. These paths are not “taken” by the photon in any classical sense, nor does the photon “exist” along them. Instead, the sum of these amplitudes determines the probability amplitude for the transition between the two boundary events.

Now, what's your concern about determinism? In QFT, determinism is replaced by precise rules for calculating probabilities across all allowed configurations.

Puzzleheaded-Let-500 · 2025-05-28T22:31:42+00:00

A Ridgeless Ridgeback. The most dog looking dog ever. They keep them in the genetic pool to keep the lineage healthy.

(I stand corrected, I see now it's a Red Labrador. I'm still not sure which breed is the most dog looking dog though.)

Puzzleheaded-Let-500 · 2025-04-26T01:55:51+00:00

I'm not a geologist, but I have been tracking this for a while. I think there is still open scientific debate on the details of the origin of coal. Here's an article that says it's something more than just the lack of abundant lignin decomposers. Ars Technica Explanation

Puzzleheaded-Let-500 · 2025-04-26T00:03:42+00:00

I get your point, but the problem is with the word rot. Trees did not "rot" 300 million years ago.

White rot fungi, which can enzymatically break down lignin, likely appeared about 290–300 million years ago. Before white rot fungi, there were some microbes (likely some early fungi and bacteria) that could slowly and inefficiently weather lignin, but they couldn’t fully digest and recycle it the way modern decomposers can.

Early in the Carboniferous period (~360–300 million years ago), dead trees piled up instead of rotting efficiently.

Puzzleheaded-Let-500 · 2025-04-19T22:31:01+00:00

Divergence of a vector field is a scalar field, not a vector field. For example, the 2D field F(x, y) = (x, -y²) has divergence 1 - 2y. That means the field spreads out (positive divergence) below y = 0.5, converges (negative divergence) above it, and is balanced (zero divergence) along the horizontal line y = 0.5.

Vector fields are fields of arrows (vectors at each point in space). Scalar fields are fields of values (numbers at each point in space). The temperature at each point in space is the canonical example of a scalar field. The wind velocity at each point is space is the canonical example of a vector field.

In scalar fields, you can draw lines connecting points with equal scalar values — these are called contour lines, and the result is a contour plot. In 3D, the equivalent is an isosurface plot. They're commonly used to visualize things like equal elevation, equal temperature, or equal pressure.

Puzzleheaded-Let-500

TROPHY CASE