The mathematics/physics/engineering study group thread

Saturos

Saturos

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As the name implies, this is a thread dedicated to the futhering of our Mathematics knowledge through the miracle of the internet(s) and Helplife2.net. I thought this was a good idea because I've noticed other members in the past posting threads just to ask one mathematics/physics/engineering question. Here's how it works:

  • Ask a mathematics question
  • Wait for someone to reply with a solution
  • If futher questions are needed, prvmsg. the member's reply instead of continually asking in the thread.
    This way others can have a chance to ask a question of their own. Or alternately, you can debate the solution amongst your own profiles.
  • If you have a solution to a problem, be sure to quote the asker's question so there's no confusion.
  • Lastly, no clowning around in this thread either please. This thread is strictly for those who need assistance.
    This means you should only post if you have either a question or solution or both.

I'd really like for this thread to get stickied moderators. You never know, it might come in handy yourselves one day! If anyone has any suggestions on ideas to improve this thread, feel free to prv msg. me and I'll update the initial post accordingly. :)
 
Solaris said:
does 0.999...=1?
Click to expand...

There was this girl called Pandora, see. And she had this box.

Regardless, the answer is yes, because the "..." bit means infinitely recurring. That allows you to take the geometric sum from n=0 to infinity of 0.9/(10^n) {This is the value of each digit in the term} with the exact answer being equal to 1. It is an equality, the only way it is an approximation is if you truncate the sum which means you're doing it wrong.
 
Does 1+1 infact = window? Show your work.
 
i'm building a two story brothel i was wondering how thick the support walls have to be?







just kidding, i've already figured out since i'm an engineer, lol

but i could use some help getting the ho's, any professional pimps around?
 
Avoidist said:
There was this girl called Pandora, see. And she had this box.

Regardless, the answer is yes, because the "..." bit means infinitely recurring. That allows you to take the geometric sum from n=0 to infinity of 0.9/(10^n) {This is the value of each digit in the term} with the exact answer being equal to 1. It is an equality, the only way it is an approximation is if you truncate the sum which means you're doing it wrong.
Click to expand...

the much easier way to prove it is:

x = .999...
10x = 9.99...
10x - x = 9.99... - x
9x = 9
x = 1
 
Hazar said:
the much easier way to prove it is:

x = .999...
10x = 9.99...
10x - x = 9.99... - x
9x = 9
x = 1
Click to expand...

wwaaiittt a second, that......makes perfect sense? WTF!?

The laws of time, space, and mathematics must be falling apart at their seems for 10x-x not to = 9x...
 
Or because the human brain can't comprehend infinity, just cheat anyway and say 0.999 recuring is 1.
 
i can't think of any situation where it would actually matter if 0.999999999999999999=1 or not, even in astronomy.

so for practical reasons 0.999999999999999999999=1




oh and

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9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
=1

i always wanted to do that :)

god damn 16000 character limit
 
Can somebody explain CMOS - TTL logic interfacing? I need the answer now. :(
 
I always thought the special relativity questions are fun to think about so heres one I remember doing in a tutorial:

A pole 20 metres long lies next to a barn 15 metres long. An athelete picks up the pole and walks away from the barn, then turns and runs at the barn with a velocity of 0.8c. The atheletes friends remains at rest next to the barn door.

a) How long is the pole as measured by the friend as it approaches the barn?

b) The barn door is initially open and immediately after the runner and the pole are entirely inside the barn he closes the door. How long after the door is shut does it take for the pole to hit the back of the barn in the reference frame of the atheletes friend?

c)According to the friend, both the pole and the runner are inside the barn. How do you reconcile this with the experience of the athelete?

Oh and another interesting one I liked, to do with gravity is:

Imagine an astroid where miners have dug a hole from end all the way to the other. One of the miners decides to jump into the hole but at the same time he throws a ball into orbit around the astroid. Does he catch the ball when he comes out the other end of the astroid?
 
TheHamster22 said:
Imagine an astroid where miners have dug a hole from end all the way to the other. One of the miners decides to jump into the hole but at the same time he throws a ball into orbit around the astroid. Does he catch the ball when he comes out the other end of the astroid?
Click to expand...

No, he is pulled to the center of the Asteroid by Gravity and does not fall all the way through. In addition, the ball would most likely not reach stable high orbit, and would collide with the asteroid very quickly.
 
TheHamster22 said:
A pole 20 metres long lies next to a barn 15 metres long. An athelete picks up the pole and walks away from the barn, then turns and runs at the barn with a velocity of 0.8c. The atheletes friends remains at rest next to the barn door.

a) How long is the pole as measured by the friend as it approaches the barn?
Click to expand...
L' = L * (1 - (0.8)^2)^0.5

b) The barn door is initially open and immediately after the runner and the pole are entirely inside the barn he closes the door. How long after the door is shut does it take for the pole to hit the back of the barn in the reference frame of the atheletes friend?
Click to expand...
If the athlete instantly comes to rest from 0.8c, won't the bar instantaeneously go back to it's original length?

c)According to the friend, both the pole and the runner are inside the barn. How do you reconcile this with the experience of the athelete?
Click to expand...
The barn appears longer to the athlete.

Imagine an astroid where miners have dug a hole from end all the way to the other. One of the miners decides to jump into the hole but at the same time he throws a ball into orbit around the astroid. Does he catch the ball when he comes out the other end of the astroid?
Click to expand...
The miner would start oscillating between the two ends of the hole. So when he comes out the other end, he will only go as high as he was when he jumped in. If the ball is in a high orbit, he cannot catch it.
 
If a cat says hello does the rabbit come or go?
 
TheHamster22 said:
An athelete picks up the pole and walks away from the barn, then turns and runs at the barn with a velocity of 0.8c.
Click to expand...
Who ****ing cares about any of the other questions just have this man disected immediatly!
 
99.vikram said:
If the athlete instantly comes to rest from 0.8c, won't the bar instantaeneously go back to it's original length?
Click to expand...

Ahha but when the barn door is closed, from the perspective of the friend the stick is still shorter than the barn so there is a short time in which the athelete is still running while he and the stick are inside the barn.

You'd see this if you number crunched your length contraction equation :p The stick in the rest frame (barn and friend frame that is) is only 12m long.

99.vikram said:
The miner would start oscillating between the two ends of the hole. So when he comes out the other end, he will only go as high as he was when he jumped in. If the ball is in a high orbit, he cannot catch it.
Click to expand...

Aye true enough, but he threw the ball into orbit at a tangent to the surface of the astroid, true he'd have to throw it really fast to get it to orbit so close to the astroid but its probably more possible than athelete that runs at 0.8c :D (Also since the astroid has no air resistance you don't have to worry about that for very close orbits.)

Essentially if worked this problem you'd find that the time to fall through an astroid = time for half an orbit.

Techercizer said:
No, he is pulled to the center of the Asteroid by Gravity and does not fall all the way through. In addition, the ball would most likely not reach stable high orbit, and would collide with the asteroid very quickly.
Click to expand...

Think about it, when he jumps into the hole in the astroid he has maximum potential energy (no velocity) but when hes at the centre he has maximum velocity, theres no way he would just sit at the centre of the astroid unless there was another force acting there to stop him. This is essentially an example of simple harmonic motion.
 
Here is a physics question....

If a 6ft man who weight 70kg could travel at 8.6c where c is the speed of light off of earth where the direction of motion is on his vertical axis, how tall would they be according to a stationary observer? How tall would he be relative to himself?

How much momentum would he have?

If they traveled to a foreign planet 6 light years away, how much time would pass relative to the man? How much time would pass relative to a stationary observer on earth?

We just got done with this stuff in physics. Someone already went through it :(. Dang I should read more of the thread!
 
Is the speed there meant to be 0.6c?

Anyway yours is slightly more interesting in that you've used momentum which is always a shitter, esspically when your not dealing with eV's!

Here's my guesses:

Height according to earth: 1.463m

Height according to him: It doesn't change.

Momentum he has: 1.87x10^10 Ns (lol that was a pain to put in my calculator)

He thinks he takes 6.57x10^7 seconds to get there and from earth we all think it takes him 5.26x10^7 seconds.

I would convert it but my heads suddenly gone dead and i keep making mistakes D:
 
No he goes 8.6 * the speed of light!! Magic!

yeah it was meant to be 0.6c
I don't have a calculator off hand but... I can use Google's online calculator (I find it's actually easier to use than a regular calculator)
L = 6 * sqrt(1-0.6^2) = 4.8ft or 1.463m. Correct.

Momentum = (70 * 0.6 * 3 * 10^8)/(sqrt(1-0.6^2)) = 15,750,000,000 kg m/s
I think this would be the correct answer if you round off the speed of light to 3x10^8.

I know your time is wrong because he is traveling through space very quickly and therefor he is starting to not travel through time as much. So to him the trip would seem quicker than on earth. In fact if you could travel at the speed of light, the trip would seem instantaneous and not a second of time would pass, but you can't.. so deal with it.
On earth we would pass through 5 years of time. He would pass through 4.
3 light years = 2.83815852 x 10^16 meters
(3 light years)/0.6c = 5 years (or 157,784,630 seconds)
5 = T/(sqrt(1-0.6^2))
T = 4 years (or 126,227,704 seconds)

Trip would take about 4 years for him, but it would seem like 5 on earth.

I think I love the Google's Calculator!
 
TheHamster22 said:
Think about it, when he jumps into the hole in the astroid he has maximum potential energy (no velocity) but when hes at the centre he has maximum velocity, theres no way he would just sit at the centre of the astroid unless there was another force acting there to stop him. This is essentially an example of simple harmonic motion.
Click to expand...

As he passes through the centre of gravity of the asteroid, wouldn't his body be completely crushed? The miner cannot pass through the centre of gravity instantaneously, so his body would be pulled in two directions.
 
Here is one. You are in a space ship in orbit over the earth. On the same orbit as you is a dead satellite that you want to catch up with. Say it's roughly 10 km ahead of you measured circumferentially. How should you fire your maneuvering jets to come into contact with the satellite?

And here is another one, simple geometry: A farmer has a barn and a hen house not very far from each other, but he wants to measure the length more precisely. He doesn't have a measuring tape, but he has two ladders. One is 30 feet long and one is 20 feet long. He only knows this: if he puts the base of the longer ladder at the edge of the hen house, and leans it against the barn, and he puts the base of the shorter ladder at the edge of the barn and leans it onto the hen house, the point where the two ladders cross each other is exactly his height. He is 6 feet tall. How far apart are the hen house and the barn?

P.S. I just got my mechanical engineering degree today
 
venturon said:
As he passes through the centre of gravity of the asteroid, wouldn't his body be completely crushed? The miner cannot pass through the centre of gravity instantaneously, so his body would be pulled in two directions.
Click to expand...

He wouldn't be crushed at the centre because once you are inside the mass the gravitational field no longer varies according to the 1/r^2 rule but more the gravitational field is directly proportional to r.

This is because the close to the centre he is, the less mass is underneath him pulling him into the centre, so at the centre it would be zero gravitational force pulling him in to the centre. Also at the centre you wouldn't have to worry about the gravitational force of all the mass above you as that would all be canceled out by symmetry.
 
I got better at maths by lowering my standards.
 
Can somebody tell me exactly what Algebra 2 is supposed to be? I'm almost finished with this course, and it seems at least thus far, to just be mostly a review of Algebra 1.

Finished 9 chapters out of 13, and thus far everything has been a review of what I already knew from Algebra 1... with the only chapters left being Radicals, Quadratics, Imaginary Numbers & Complex Numbers, and Quadratic Equations & Functions.

The only thing of those that I think I might not know is the Imaginary & Complex Numbers.
 
Raziaar said:
Can somebody tell me exactly what Algebra 2 is supposed to be? I'm almost finished with this course, and it seems at least thus far, to just be mostly a review of Algebra 1.

Finished 9 chapters out of 13, and thus far everything has been a review of what I already knew from Algebra 1... with the only chapters left being Radicals, Quadratics, Imaginary Numbers & Complex Numbers, and Quadratic Equations & Functions.

The only thing of those that I think I might not know is the Imaginary & Complex Numbers.
Click to expand...
The 2's and 1's of Algebra along with the "Pre-s" differ from school to school. I think however that college algebra is a universal higher grade form of algebra that combines trigonometry and basic algebra with more sophisticated, detailed formulas.

Likewise Calculus combines College algebra with even more detailed formulas relating to physics.

I'm not really sure myself though. Just an educated guess really.
 
So what kind of stuff does college algebra cover?

Throw out some topics!
 
Raziaar said:
So what kind of stuff does college algebra cover?

Throw out some topics!
Click to expand...
OK, so I was wrong, according to Wikipedia. There's "abstract" algebra too. Maybe this'll better help you understand the applications of algerbra and it's many functions and complexities.

Amazing too the timeline of algebraic mathematics that is provided. Damn Greek pioneers and they're super mathematical intelligence. :p *jealousy*
 
Why would you want to do maths out of school. This logic confuses me.
 
Shift said:
Why would you want to do maths out of school. This logic confuses me.
Click to expand...

Because burger flipping jobs are no fun. Not that I've ever had one... I just want to be better qualified.
 
Ok, so my penis is 5.5'' long and I'm jerking off at 0.4c. How long is my member as measured by CyberPitz who's watching on webcam?
 
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