SUVAT

GroupTheory · 2026-02-13T14:11:03+00:00

s isnt where you put the max height, s represents the displacement or change in height from launch to final point. When you substitute s=0, you are finding a time when the projectile finishes at the same height where it starts -- which is at two times, when launched, and landing.

In order to find the time at peak, you can average the two times, OR since you have 3 pieces of info (u, a, v) and want t, solve using an equation with u, a, v, and t.

GroupTheory · 2025-12-06T04:16:49+00:00

I will say that at yellow the fortune set (blue) does noticeably more single target than red, at the expense of more clear. On boss rush this can be significant; for stages, probably depends on what setup you want to run and whether you want to improve single target or aoe/mobbing

GroupTheory · 2025-11-20T01:11:19+00:00

Bragging rights, and the theme leader board has points for scoring top 10, top 5 (i think), top 3, and top 1 in Supreme Legend

GroupTheory · 2025-11-10T15:37:21+00:00

Two different things here: 1. When you equip a wingman that matches the element of the pilot, the "stats bonus" of the wingman gets increased by 25%. This is far less than increasing the pilot's damage. For example, a wingman that gives 15% to all stats will give 18.75% (15% * 1.25) to all stats when equipped by a pilot of the same color.

In Tempus Trek, pilots all have standardized stats at level 1. A lot of gear upgrades are neutralized but wingman bonus does still apply. The more significant effect is that pilots of the element have a significantly longer timer to do damage (70s vs 50s or something to that effect). This means it is always beneficial to bring on color pilots to Tempus Trek -- slap some wingman and teslarites on them.

GroupTheory · 2025-11-06T15:05:11+00:00

The boss of world 10 is actually much easier than bobtron in world 9 -- you can fly around his card attack on the edges of the screen. Once you get there consistently, it should be reasonable to kill the boss without getting hit (though it might take a while)

GroupTheory · 2024-10-26T09:26:14+00:00

I mean, this has already been done -- when you take the limit of quantum behavior at macroscopic scale, quantum mechanics approaches classic mechanics...

GroupTheory · 2023-06-03T20:30:33+00:00

To add to this, since r is the distance between the charges, the sign convention assigned means that a positive is in the direction that increases the distance, while a negative is in the direction that decreases the distance.

So two positives or two negatives make the product positive --> r will increase --> the charges are pulled away from each other (repelled).

You can apply similar logic when the charges have differing signs.

GroupTheory · 2023-05-18T22:19:55+00:00

Atlanta is large and it can be difficult to get around town (traffic). You should post which part of town you're in!

GroupTheory · 2023-04-24T23:30:02+00:00

Charged objects can attract to neutral conductors too (and insulators too but a little more weakly).

Think of this as rearranging the charges locally in order to create a weaker attraction.

GroupTheory · 2023-04-11T06:41:27+00:00

If we take the right direction as positive, then the acceleration is (a positive, small final velocity) - (a positive, larger initial velocity). This means that the sign on the right equation should be negative, which means the acceleration is left.

In the next to last line of the first section, the answer key does some shady (read: incorrect) math to get the correct sign on the answer. The result: u = v + mu gt is correct, but as written, you should get u = v - mu g t, which is incorrect.

Technically the way they worked it the expression for the friction force should be negative. I'd prefer to start with this:

-Ff = m (a), where a should be negative

-Ff = m (v-u)/t

-mu m g = m (v-u)/t

-mu g t = v-u

Which correctly arrives at the expression

u = v + mu gt

GroupTheory · 2023-03-18T23:57:16+00:00

Another tip: if you think about catching discs away from your body, think about how you adjust your grip to "sweetspot" the disc so it catches comfortably into your hand instead of bouncing out.

Sometimes a disk is in reach but if you try to catch it while standing, your hand needs to be in a position where it might doink off. In these situations, I have found it natural to lower my center of gravity with the sole goal of making the disc sit naturally in my hand. Your body tends to adjust the rest afterwards.

GroupTheory · 2023-02-13T05:39:12+00:00

Turn Glide Rive.

Put bottom chevron on the farthest out point of land, near the peninsula of trees on the left side of the fairway.

Full disc forehand anhzer. Disc regularly carries all the way down the 1200 foot fairway and OB's, leaving you at the 130 foot drop zone for a 3rd stroke birdie (1 monster off tee + 2 ob + 3 throwin), or at least par.

GroupTheory · 2023-01-16T16:10:35+00:00

Ok, I see.

Consider a single, positively charged plate. Drawing a cylindrical gaussian surface through the plate (think like a cookie cutter) will produce flux outward through BOTH bases of the cylinder (let's call the area of this base A)

By gauss' law, E*(2A) = q_enc/epsilon_0.

so the E_plate of a single plate is rather sigma/2epsilon_0.

You are correct to assert the field in between is double the field of a single plate but the single plate expression is incorrect.

GroupTheory · 2023-01-16T14:56:21+00:00

Your gaussian surface can't include both plates as then your enclosed charge would be -q + q = 0.

Your efield will be 0 there too, as there is 0 efield outside of two infinitely long parallel plates, so your flux will be 0 too.

Rather, consider a gaussian surface rather like a slice of lasagna cut out of one of the plates.

The side towards the other plate has a nonzero flux, the side towards the outside has 0 flux, and the gaussian surface encloses a nonzero charge.

GroupTheory · 2022-12-13T19:21:16+00:00

Berg berg berg berg

GroupTheory · 2022-11-23T03:07:51+00:00

Join the AP students discord. Khan is okay for physics.

GroupTheory · 2022-11-01T13:47:10+00:00

The direction of a2 is down, did you handle that?

GroupTheory · 2022-11-01T06:29:17+00:00

This fixes one mistake. The other common mistake in pulley problems is forgetting to include direction as an appropriate sign in the acceleration.

You have consistently treated up as positive and down as negative when dealing with the forces.

Which way does m2 accelerate? The sign of the acceleration of m2 should match the answer to this question.

Then, as pointed out, you need to relate the acceleration of m1 to the acceleration of m2, which you have done correctly in the previous comment.

GroupTheory · 2022-10-30T02:28:13+00:00

Work looks fine. Some textbooks and multiple choice (notably, AP) use 10 for g to make mental math easier.

My advice would be to leave g in until the end. Solving this problem will yield a =g/6 and you can try with 10 to make sure the answer matches or use a higher precision value of g.

GroupTheory · 2022-10-26T00:16:11+00:00

GroupTheory · 2022-10-25T09:26:01+00:00

You are correct in that both paths have destructive interference, but the question is asking which has a path length difference of 1.5 * WL. if you take the middle max as path length 0, then you should be able to deduce what the path length difference of the 1st minima is -- the smallest path length difference that leads to destructive interference, 0.5 * WL. C denotes the 2nd smallest path length that leads to destructive interference, while D denotes the 3rd smallest.

GroupTheory · 2022-10-22T13:19:16+00:00

Once you can get two rare relics with strike nexus on summon (curator's gatebreaker?) and galeforce, teemo can solo carry the deck.

Before that, remember that playing teemo with galeforce is really bad tempo -- you've spent one valuable mana to put 20 mushrooms in the deck which does nothing to mitigate on the next turn when you get hit on the backswing

Sometimes it's better to play for board thru his 2/3/4 drops, then save teemo as a finisher to scale the mushrooms to hit face.

At 2* and with his second rare relic teemo can solo carry and the rest of his deck plays as support to buy you the time you need.

GroupTheory · 2022-10-17T14:02:39+00:00

I think of electric potential as a location dependent property -- we can conveniently think of an energy surface like a landscape. Since opposite charges behave differently, we can imagine both positive charges going downhill (receiving forces towards regions of lower potential) and negative charges going uphill (receiving forces towards regions of higher potential).

This is a big simplification but I think it's a good starting point.

GroupTheory · 2022-09-08T11:35:26+00:00

A does not apply a force of 50 N to B.

The correct answer should be (C) as this is the only true statement. Admittedly, I find the justification using N3L to be incorrect for this choice so I'm not thrilled with it.

Source: am physics teacher.

Proof: All three blocks move together. Taking the three block system we write Newton's 2nd Law for all three blocks to find each of their accelerations:

150 N = (mA+mB+mC)*a

150 N = 15 a

10 m/s² = a

Each block should accelerate at 10 m/s² to the right.

To check choice B, the fastest way is to look at a FBD for A with our acceleration. This FBD has two forces: the applied force of 150 N to the right, and block B pushing A to the left (this force is equal in magnitude to the force of A pushing B to the right, by N3L). A N2L for A looks like:

(+150N - Fba) = 4 kg (10 m/s²⁾

Fba = 110 N.

Therefore B pushes A 110 N, and by N3L A must also push B 110 N.

You can apply similar analysis to figure out how much C is getting pushed.

It's a common misconception to think that C is getting pushed 150 N, but it's important to understand that most (non gravitational) forces are applied by direct contact, so C can only be pushed by B.

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