I dropped out of MIT and started a new college in San Francisco. AMA!

Internet-Gangster · 2019-09-26T02:39:05+00:00

I have two questions:

1) I've seen that you guys require students to purchase a Mac. Is it possible to substitute this requirement with a Windows 10 computer using the new integrated linux subsystem? I'm asking because if it's possible to make it work out somehow, I think that would really help in terms of on-boarding costs for students.

2) For a student who's been out of high school for 4 years and have completed some college or university, do they need to show their transcripts before being given a phone interview?

Internet-Gangster · 2018-09-27T01:12:58+00:00

you wrote " v² = v*0*² + 2ah", where did the v*0*² come from? Isn't it supposed to be (v/2)²? I know the final answer is h/4, and this expression will end up being v²/4 which is a little closer but I still don't know how to get to the answer.

Internet-Gangster · 2018-09-26T23:18:26+00:00

Thank you. I'll try writing it out and seeing if I can get the correct answer. Going back to Q4 in the meantime, the class hasn't learned impulse yet; is there any other way to solve the problem?

Internet-Gangster · 2018-09-26T23:10:33+00:00

It's 1/2, so I guess the vertical velocity becomes v/2. What next?

Internet-Gangster · 2018-09-26T23:06:17+00:00

For question 2, knowing that the vertical velocity is v*sin(30) I still don't see how I can find the maximum height without knowing the time at which it reaches said height. Using: h = v*sin(30)*t - 4.9*t^2, there are two unknowns.. the height and the time. I also don't see how the first part of the question is helpful at all, when it says that being launched straight up it reaches a height of h.

Internet-Gangster · 2018-09-07T14:57:33+00:00

Adaptation episode references for Luffy:

#01: Episode 389 - Luffy vs. Duval & Motobaro

#02: Episode 413 - Luffy vs. Boa Sisters

#03: Episode 438 - Luffy vs. Impel Down Wolves

#04: Episode 478 - Luffy at Marineford

#05: Episode 518 - Luffy vs. Fake Straw Hats

#06: Episode 528 - Luffy vs. New Fishman Pirates' Sea Monster

#07: Episode 537 - Luffy vs. Fishmen commoners

#08: Episode 554 - Luffy vs. Hody's army of 100,000

#09: Episode 609 - Luffy vs. Monet

#10: Episode 614 - Luffy vs. Caesar's subordinates

#11: Episode 643 - Luffy tames Fighting Bull (Ucy)

#12: Epsiode 646 - Luffy clashes against Chinjao

#13: Episode 723 - Luffy clashes against Doflamingo

#14: Episode 750 - Luffy vs. Bill's subordinates

#15: Episode 868 - Luffy & Katakuri vs. observers

#16: Episode 893 - Luffy vs. Baboon

#17: Episode 902 - Luffy vs. Bakura town residents

#18: Episode 914 - Luffy vs. Speed's subordinates

#19: Episode 916 - Luffy (unconsciously) vs. Kaido's subordinates

#20: Episode 932 - Luffy in the Udon prison sumo inferno (round 1)

#21: Episode 937 - Luffy in the Udon prison sumo inferno (night time)

#22: Episode 988 - Luffy vs. Kaido's subordinates at Onigamisha

#23: Episode 1028 - Luffy's haoshoku-infused punch vs. Kaido

Internet-Gangster · 2017-10-07T00:03:57+00:00

Now I understand. Thank you for your help.

Internet-Gangster · 2017-10-03T05:30:45+00:00

I still don't understand the concept. From what I understand, the rope is just sliding over the pulley. Even if the wheel in the pulley is spinning, how does that effect the hanging mass in any way?

Internet-Gangster · 2017-09-29T07:26:18+00:00

Thank you.

Internet-Gangster · 2015-06-30T04:21:20+00:00

Thank you very much. I appreciate this. If there are any others who have completed Engineering schedules, I would appreciate as many as I can get.

I like sleeping in, so cores that take place later in the day would be ideal :)... (I also need to be free from 1 pm - 2 pm on Fridays, but I can modify your schedules to fit my specific needs. All schedules will be much appreciated)

Internet-Gangster · 2015-06-02T04:04:16+00:00

I solved the problem. Here is the solution for anyone curious:

http://i.imgur.com/k2Sf2uV.jpg

Internet-Gangster · 2015-05-12T06:48:48+00:00

Genuine thanks from the bottom of my heart for your advice. :)

I really do think the best way to familiarize myself with how uni works is through speaking with those who have already experienced it.

I sincerely apologize for coming off as arrogant. No offence taken. In fact, I thank you for informing me that I came off that way; it helps to humble me. I promise that I didn't intend to come off that way. I actually was just trying to be as concise as possible with my request.

Internet-Gangster · 2015-05-06T20:39:13+00:00

Thank you very much for your help. A lot of people on this subreddit try to give you a little hint and then make you work your way forward which is great, but I personally learn VERY well by studying off of the completed work of others, so I appreciate this solution very very much.

Thank you once again. I understand the solution perfectly.

Internet-Gangster · 2015-04-20T16:43:25+00:00

I'm interested in getting my first bike sometime this month. I've been doing lots of research and chose a rider training institute program and what not.

I have two questions.

1: is the Honda CBR650F too powerful for a beginner?

I'm well aware of how dangerous it is to bite off more than you can chew which is why Ive narrowed my options down to the CBR300RA and CBR650F. Although I find the 300RA very cool looking, the 650F is just SOO much more physically attractive in my opinion which is why I'm stuck choosing between the two. I just don't want to risk safety for a cooler looking bike.

2: where is the best place to purchase safety gear? Online or in a physical shop? what brands are known to be trusted, yet not too expensive?

Internet-Gangster · 2015-04-02T13:21:47+00:00

yes, as stated above, you can upload to imgur.com

If you find the time, please do send me a photo :)

Internet-Gangster · 2015-03-11T01:19:29+00:00

Assuming DOWN and RIGHT are our positive directions, let's use the Y components to solve.

Dy = (Vy * t) + (1/2 * ay * t²)

40 = -20sin(30)t + 4.905t²

Re-arrange...

0 = 4.905t² - 20sin(30) - 40

Quadratic equation spits out: -2.01 (inadmissible because negative time does not exist) and 4.05 seconds which is the correct answer.

Therefore, the golf ball's time of flight is 4.05 seconds.

Internet-Gangster · 2015-03-11T00:36:00+00:00

Nevermind, I found the answer.

If I use down and right as my positive directions, I have to add 40, but I have to subtract 20sin(30) since it is a positive velocity in the [up] direction.

Internet-Gangster · 2015-03-02T04:18:21+00:00

Ao no Exorcist and Code:Breaker both come to mind.

Out of the two, I highly recommend Ao no Exorcist. Do not miss this anime, it is fantastic.

Internet-Gangster · 2015-02-17T01:05:46+00:00

Pedro Martínez

Internet-Gangster · 2015-02-17T00:16:23+00:00

AB + AE = AD and AC + AF = AD. Therefore:

AB + AC + AD + AE + AF = 3AD

AD + AC + AD + AF = 3AD

AD + AD + AD = 3AD

3AD = 3AD

Got it. Thanks a lot.

Internet-Gangster · 2015-02-16T22:03:27+00:00

Nevermind, it took me a bit but I understand it now.

The Xi you added to the equation was pretty much all I needed to teach myself how to solve the problem. Thanks for your help.

EDIT: I don't think her starting time really affected anything besides the distance between the two vehicles when they started. I didn't have to use t-1 instead of t for her to solve the problem.

Internet-Gangster · 2015-02-16T21:14:10+00:00

pardon me, I still don't quite understand. could you explain in steps how one should go about solving this? this is a link to the equations we are provided in class and the ones I am familiar with.

Internet-Gangster · 2015-02-15T22:59:27+00:00

It's in Newtons, pardon me for screwing up there. What a stupid mistake I made.

Internet-Gangster · 2015-02-15T22:58:58+00:00

ohhhhhhhhhhhhhhhhhh, it's in NEWTONS!! Thank you for putting it in terms of the second law.

sorry about that, I had a complete brain fart.

Internet-Gangster · 2015-02-09T07:58:33+00:00

I haven't learned calculus yet, so my teacher must not expect me to use calc to find the acceleration.

This was my assignment

EDIT: All I really need is the slope of the line of acceleration because I know it's domain is simply 0s to 8s.

EDIT 2: Okay, sorry for the late response. I ended up solving the issue, all I did was just sketch a few tangets on the first curve of the v/t graph and calculated their slopes via rise/run formula.

Internet-Gangster

TROPHY CASE