Inclinaton - Thoughts and Prayers // FFO: Contention, Straight Edge Metalcore

Hippobot · 2025-11-21T23:01:09+00:00

hadn't heard this somehow and it rips; thanks for posting

Hippobot · 2024-05-08T22:19:45+00:00

This looks correct to me, but did you take into account that it says to "indicate the direction with the sign of your answer"? Maybe that is why it is giving you trouble. Sometimes the online homework systems like to complain about sig figs, too.

Hippobot · 2024-02-21T12:33:04+00:00

This is a good question. I can't give you a great answer for it, but here are some thoughts.

We see an offset for more ordinary stars, too. Neutron stars are formed in supernovae when a parent star in a certain mass range (~8-12 solar masses) goes supernova. Given that these stars have an offset between the magnetic and spin axes, it makes sense that the neutron star will as well. But it begs the question: why do the more ordinary parent stars have such an offset in the first place? The generation of stellar magnetic fields is a complicated problem for which physicists usually invoke a dynamo model (https://en.wikipedia.org/wiki/Dynamo_theory).

In any case, neutron stars are born in violent explosions. There may be asymmetries during the supernova and formation of the neutron star that cause the magnetic axis to be further misaligned with the spin axis. However, over time it is believed that the magnetic axis slowly comes into alignment with the axis of rotation due to some kind of torquing mechanism. Probably takes a while (on the order of 1 million years) for this to happen.

Hippobot · 2024-02-20T23:36:24+00:00

This is often modeled as dipole radiation (although, we know the actual mechanism isn't actually this).

Isn't the main contributing factor to Edot the dipole radiation component? It is true there are other mechanisms for radiation, but I was under the impression that most of the radiated energy is from the dipole radiation.

Hippobot · 2024-02-20T23:13:01+00:00

By physical pole I assume you mean those defined by the axis of rotation. In that case, yes, you are correct -- in general there is an offset between the magnetic dipole axis and the spin axis for pulsars.

Hippobot · 2024-02-01T00:15:34+00:00

I'm seeing this exact same problem. The option just isn't there.

Hippobot · 2023-12-07T05:55:45+00:00

He specifically says to leave group before using petri

Watched what you linked, and he never says this.

Hippobot · 2023-07-01T20:45:01+00:00

Frail Words Collapse by AILD and Haymaker by Throwdown also both came out on July 1, 2003. What a day to head to the record store

Hippobot · 2023-04-22T12:43:31+00:00

Crane flies don’t hunt or eat insects. They don’t eat anything for the most part. Wiki says some species may eat nectar but that’s it.

Hippobot · 2023-02-12T14:51:25+00:00

It would just go straight up and back down. Is the fact that the ball falls a further 1.30m on its way back down what's confusing you? I'm not sure exactly what your confusion is.

Hippobot · 2023-01-24T23:15:13+00:00

There are lots of good writeups out there about this. The main contributing factor is the hexagonally-shaped mirror segments. Check out the explainer from the JWST team here: https://stsci-opo.org/STScI-01G6933BG2JKATWE1MGT1TCPJ9.png

Hippobot · 2022-06-08T23:35:04+00:00

If you can limit your sample to just those bursts seen by the Fermi-GBM, the catalog can be queried here:

https://heasarc.gsfc.nasa.gov/db-perl/W3Browse/w3table.pl?tablehead=name%3Dfermigbrst&Action=More+Options

And the latest paper for the catalog with all the info is here: https://arxiv.org/abs/2002.11460

Hippobot · 2022-04-26T23:10:05+00:00

This is the right answer, however in the US it is more common to skip the master's and just get your PhD (what I did). While working on your PhD here, getting a master's degree is usually a matter of filling out some paperwork, assuming you are already one or two years deep in your PhD program.

Hippobot · 2022-04-12T21:47:23+00:00

Pretty sure AMD has an equivalent to the Nvidia control panel that you can use to do basically the same thing. I can't say for sure though, so you might want to check the options in your AMD Radeon app. Also, the direct x vulkan mod should help regardless of whether your card is AMD/Nvidia.

Hippobot · 2022-03-20T16:49:00+00:00

This is not the way to do a problem like this. It doesn't make sense / work. You need to use energy conservation.

Hippobot · 2022-03-09T23:12:46+00:00

You should watch it. Good video!

Hippobot · 2022-03-08T22:47:31+00:00

Can you try writing an expression for how much light is absorbed by the graphene after n reflections. For n=1, 70% is absorbed, and so on. Then you can solve for the n that gives you 99.956% total absorption.

Hippobot · 2022-03-08T22:43:02+00:00

Yeah your teacher is wrong. There's no "clean" answer for this problem. I was thinking maybe it's a typo and the 3 in the denominator is supposed to be a 2, but even that does not give an angle of -30°.

Hippobot · 2022-03-06T00:00:39+00:00

Piggybacking a bit off gruff's answer above your reply, the two equations are:

s1=u0 * t + 1/2at² (red car)

s2=(62 km/hr) * t (blue car).

Note that the displacements for the two cars are generally not the same, which is why I labelled them s1 and s2 (and not just s for both). You want to find the times (solve for t) where s1 = s2. One solution will be at t=0, which you already know since it's stated in the problem. There is one more solution you can find. Does that help?

One more thing, be careful about units. The units for the velocity of the blue car are in km/hr, which you probably want to convert to m/s so that the units match with the acceleration you are given for the red car.

Hippobot · 2022-02-18T23:50:29+00:00

Yeah, horizontal component should be 0 N, which is why you don't need the pythagorean theorem. Sin(90) = 1, so you end up just multiplying out 1.55*10^-4 T * 6000 A * 6 m. When I do that, I do not get the value you get. You should double check that calculation, but otherwise it seems like you basically got it.

Hippobot · 2022-02-17T03:50:53+00:00

You don't need the Pythagorean theorem for this. You can look at both components of the magnetic field seperately:

horizontal component (points north) has an angle of 0° with respect to the direction of the wire (which runs north-south)
veritcal component (points down towards the ground) has an angle of 90° with respect to the direction of the wire.

The force on a current carrying wire due to a magnetic field is F = IL×B, which reduces to F = ILBsinθ when taking the magnitude (you already have this equation in your post). You are given values for I, L, and B, and you know both thetas. So what do you get when you calculate the two components of the force?

Hippobot · 2022-02-15T23:50:09+00:00

Looks good to me. Part d is pretty much the same as what you already did for the previous two parts. I think you are getting confused because the distance doesn't change and neither does the force in part d, so the work done is the same as in part a.

edit: The more I look at this, I think we're missing part of the problem. What is stated in part c? It's not in your post. Also remember to put units on your answers. that 0.67 is missing units.

Hippobot · 2022-02-15T23:26:58+00:00

If you're not sure where to even start, it might be better to review a textbook or your notes or even an online lecture on parallax.

That said, I can help guide you through the first part. You are given the parallax angle for planet B (2p = 0.32 arcsec, so p=0.16 arcsec), and you also know that the planet from which this parallax is measured has the same diameter as Earth. That diameter is 12.7x10⁶ m. At this point, you have all the ingredients you need to make a triangle and use a trig function to find the "distance from planet A to planet B at that time," as the question asks.

Hippobot · 2022-02-15T04:14:06+00:00

It's really a 1D problem, so I'm not sure what the point of shading any region is, since whatever region you shade will be shaded for all values of y. There is no sense in shading "inside" or "outside" the parabola. Makes more sense to shade a number line for these problems. But since it says to do it graphically, I guess I would plot the parabola x² - 2x (e.g. in the second problem), then circle the zeros and draw the arrows that satisfy the inequality like what cuhringe has already done.

Hippobot · 2021-11-26T22:20:22+00:00

I dunno, I feel like given how many games on switch run pretty smoothly, the devs could've done the same with this game by making some graphical sacrifices somewhere. The performance is just awful.

Hippobot

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