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I think it's 10^56kg is the mass of the visible universe.
Founder of Jaco society

[22:08:23] <Confusius> I wish I was a bassist
[22:08:26] <Confusius> you fuckers look cool

Want to know how to play bass in jazz? Read this.
Quote by sinan90
I think it's 10^56kg is the mass of the visible universe.

Dark Matter weight > Visible matter weight. Dark Matter ftw.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
Quote by darkstar2466
Dark Matter weight > Visible matter weight. Dark Matter ftw.

It doesn't weigh anything if it has no gravity acting upon it. Weight=/=mass. Matter has different weights depending on where it is weighed but always has the same mass.
Founder of Jaco society

[22:08:23] <Confusius> I wish I was a bassist
[22:08:26] <Confusius> you fuckers look cool

Want to know how to play bass in jazz? Read this.
Quote by sinan90
I think it's 10^56kg is the mass of the visible universe.

only?
hue
Quote by sinan90
It doesn't weigh anything if it has no gravity acting upon it. Weight=/=mass. Matter has different weights depending on where it is weighed but always has the same mass.

F*ck, I got pwnd by saying weight. First rule of physics: weight =/= mass. I think dark matter is more massive than normal matter, is it not?

Quote by sinan90
I think it's 10^56kg is the mass of the visible universe.

It's much much much more than that. Heck, just our sun weighs about 2 x 10^30 kg.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
Last edited by darkstar2466 at Sep 10, 2007,
Quote by darkstar2466
F*ck, I got pwnd by saying weight. First rule of physics: weight =/= mass. I think dark matter is more massive than normal matter, is it not?

Isn't there still an open debate on whether or not dark matter/energy even exists? Or if not, what it is?
Quote by guywithguitar
Isn't there still an open debate on whether or not dark matter/energy even exists? Or if not, what it is?

yeah I think so, but its so hard to prove due to them being W.I.M.P (weakly interacting molecular particles)
Quote by Rocking-Rob
yeah I think so, but its so hard to prove due to them being W.I.M.P (weakly interacting molecular particles)

I don't know man... I've read the pro-arguments and the con-arguments and I am in favor that dark matter exists. The gravitational effects are too large to discredit it.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
is dark matter the same as anti matter? i've only done some very brief reading on the topic...

theories that push the boundaries of science and are plausible are pretty damn cool
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No, dark matter and anti matter are not the same thing. Anti matter is exactly that: the opposite of normal matter, and when the two come together they annhialate. Dark matter is basically just non-reflective matter that we can't see from Earth, and so can't be proved or disproved other than inferring it's existence from gravitational anomalies.
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Last edited by Yakult at Sep 11, 2007,
^ interesting.. in that case i've never heard of dark matter before, that's cool
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^ Yeah, personally I don't believe there's as much dark matter as people hark on about, and that the real problem is our lack of a solid understanding of gravity.
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Sorry I got my info wrong, the mass of the observable universe is 10^52, but thats from 23 year old text book, and what counts as observable I don't know.
Founder of Jaco society

[22:08:23] <Confusius> I wish I was a bassist
[22:08:26] <Confusius> you fuckers look cool

Want to know how to play bass in jazz? Read this.
Or so the story goes that elementary particles and their counterparts, the antiparticles, fought out a battle as the primordial soup exploded (the Big Bang), and normal matter won because there was an abundance of normal matter, compared to the antimatter. No one knows why, but hey, we would be anti-people if the antimatter won, huh?

Naturally occurring radioactive decay produces some small quantities of antimatter by the way.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
These problems are intended to help you practice calculations with large numbers of conducting particles and small amounts of charge found in a variety of examples of electrical conduction.
X-ray machines
An x-ray machine works by accelerating electrons from a hot cathode through a high voltage in a vacuum tube. The fast electrons crash into a metal target, producing a lot of heat and some x-rays, which can be used for medical purposes. The tube current (more electrons per second) determines the quantity of x-rays produced and their penetration is determined by the tube potential difference (more energy per electron). The current, potential difference and time of exposure are varied by the radiographer to examine different parts of the human body.
Some typical values are:

``Examination	Voltage / kV	Current / mA	Time / s	pelvis	            65	           350	              0.8	hand	             40	           80	               0.1	``

1. How much charge and energy do the electrons in the tube deliver during each exposure?

for this i got:
pelvis - q=0.28C e=18200J
hand - q=8x10^-3C e=320J

any help appreciated
Last edited by iamtompublic at Sep 11, 2007,
Quote by iamtompublic
any help appreciated

I would help but I'm back at uni in 1.5 weeks and so I'm putting off physics for as long as possible... Sorry
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The formulas are Q=It, so 0.35*0.8, and 0.08*0.1. E=VIt, so the sums would be 65000*0.35*0.8 and 40000*0.08*0.1. Whatever those equal are the answers I think.
Founder of Jaco society

[22:08:23] <Confusius> I wish I was a bassist
[22:08:26] <Confusius> you fuckers look cool

Want to know how to play bass in jazz? Read this.
Just something I was wondering how would you give a formula to find any number term in the Fibonacci sequence. Eg for odd numbers the sequence is 2n-1 etc.
Founder of Jaco society

[22:08:23] <Confusius> I wish I was a bassist
[22:08:26] <Confusius> you fuckers look cool

Want to know how to play bass in jazz? Read this.
Hey are any Aussies on here doing yr 12 Specialist Mathematics? preferrably in Victoria, because i'm not sure if the curriculum is the same everywhere...

I've been writing my notes for the end of year exam, and i'm pretty sure i've covered the theory of all the topics, but i haven't included any examples yet... so if you've got some good ones that have the answers completed, please post away!

Here are the topics:

Vectors, Complex Numbers, Trigonometric Functions, Differentiation, Integration (Antidifferentiation), Kinematics, Dynamics... I think thats all of them

So yeh, any help with some good exam style examples would be greatly appreciated!
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Quote by sinan90
Just something I was wondering how would you give a formula to find any number term in the Fibonacci sequence. Eg for odd numbers the sequence is 2n-1 etc.

Well the obvious way is F(n+2)=F(n+1)+F(n), where F(n) is the nth Fibonacci number.
Quote by porcupineoris
Well the obvious way is F(n+2)=F(n+1)+F(n), where F(n) is the nth Fibonacci number.

yeah, thats the only way of doing it. there isn't an nth term for fibonacci like "2n-1" gives you odd numbers.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
Quote by Sol9989
yeah, thats the only way of doing it. there isn't an nth term for fibonacci like "2n-1" gives you odd numbers.

Well, there is F(n)=[p^n-(1-p)^n]/rt(5), where p is the golden ratio (1+rt(5))/2 but im not sure if that gives the nth term straight off or whether you have to use the floor function (i.e. take nearest integer to it, rounding down).
Quote by porcupineoris
Well, there is F(n)=[p^n-(1-p)^n]/rt(5), where p is the golden ratio (1+rt(5))/2 but im not sure if that gives the nth term straight off or whether you have to use the floor function (i.e. take nearest integer to it, rounding down).

orly? i've not seen that before.... interesting.

edit: it actually works with no rounding. i've tried it with n=2 to n=5 so far, and its given me, 1,2,3 and 5.
"And after all of this, I am amazed...

...that I am cursed far more than I am praised."
Last edited by Sol9989 at Sep 25, 2007,
Quote by Sol9989
orly? i've not seen that before.... interesting.

edit: it actually works with no rounding. i've tried it with n=2 to n=5 so far, and its given me, 1,2,3 and 5.

yes, ive just looked at it again and that rounding is necessary for more generalised seris of that nature i think.

I found that on Wiki, and the proof is quite simple (post A-level though). Its worth a look if you're interested.
bag A contains 1 red ball and 1 black ball, and bag B contains 2 red balls; all four balls are indistinguishable apart from their colour. one ball is chosen at random from A and is transferred to B . one ball is then chosen at random from B and is transferred to A
Draw a tree diagram to illustrate the possibilites for the colours of the balls transferred from A to B and then from B to A

anyone help out my mate...?
bump?
Well, if you transfer a black ball to the second bag, you have a 1/3 chance of transferring a black ball from the second bag back to the first bag (and a 2/3 chance of transferring a red ball.) If you don't transfer the black ball to the second bag, you have a 1/1 chance of transferring a red ball (and a 0 chance of transferring a black ball.) Now take that and put it into a tree diagram. (Or just think about it for a little bit. It's not that hard.)

So:
You have a 1/6 chance of transferring the black ball to the second bag and then back.
You have a 1/3 chance of transferring the black ball to the second bag and then a red on back.
You have a 1/2 chance of transferring the red ball to the second bag and then a red one back.
METAR KTIK 040043Z COR RMK TORNADO 1W MOV NE. EVACUATING STATION
Last edited by aprescott_27 at Sep 25, 2007,
Is captain fallows on here? He's one of the physics teachers at SSFC, and he frequents a fair few forums to help people with physics work. he answers to the name of colin.
okay just checking:

problem:

simplify

x(2x+5)-3y

2x'+5x-3y

is that right?
the apostrophe means its squared
Yep, that's right. You can write squared as x^2 also.
The sum of the first two terms of an arithmetic series is 47. The thirtieth term of this series is -62. Find:
a) The first term of the series and the common difference
b) The sum of the first 60 terms of the series

I can do part b) but seeing I don't know how to attack the first section outside of trial and error.

Any tips?
Founder of Jaco society

[22:08:23] <Confusius> I wish I was a bassist
[22:08:26] <Confusius> you fuckers look cool

Want to know how to play bass in jazz? Read this.
Well, they all have the same difference, so just take a stab at a common difference. It has to be odd because you're taking the sum of two consecutive terms and getting an odd number. It also has to be negative because the series starts out big and gets smaller. I'll say -3 because it can't be too big, but -1 is obviously not going to work. So if the 30th term is -62, that means the first is -62 - (-3 * 29) = 25. You multiply the common difference by 29 (one less than 30) because you want the 1st term, not the 0th term. So 25 seems reasonable as a first term. Let's check it. If 25 is the first term and -3 is the common difference, that means the second term is 22. 25 + 22 = 47. Sweet.

So: First term is 25 and the common difference is -3.

To approach that problem, I thought about what kind of number the common difference must be (negative and odd) and then made a guess at what it was. I think it's trial and error, but you kind of have an educated guess in there.
METAR KTIK 040043Z COR RMK TORNADO 1W MOV NE. EVACUATING STATION
Last edited by aprescott_27 at Sep 28, 2007,
(a) By completing the square, find in terms of k, the roots of the equation x^2+2kx-7=0
(b) Show that, for all the values of k, the roots of x^2+2kx-7=0 are real and different.

Help!
a) x^2+2kx-7=0

x^2+2kx+k^2-7-k^2=0 * Completing the Square *

(x+k)^2-7-k^2=0

(x+k)^2=7+k^2

x+k=(+or-)(7^(0.5)+k) * 7^(0.5) is the square root of 7 *

x+k=7^(0.5)+k and x+k=-7^(0.5)-k

x=7^(0.5) and x=7^(0.5)i-2k

If that's what you were asking for I'm pretty sure it's right

but i don't understand what you meant by question b)
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Quote by Zackie EL
Bingo.

Bingo?!

Pi over two is about 1.7

what the hell does this have to do with bingo?

whats next, the root of e is Ludo?
go to gaol.
go directly to gaol.
do NOT pass go
do NOT collect \$200

bahaha!
Quote by cammo_1234
a) x^2+2kx-7=0

x^2+2kx+k^2-7-k^2=0 * Completing the Square *

(x+k)^2-7-k^2=0

(x+k)^2=7+k^2

x+k=(+or-)(7^(0.5)+k) * 7^(0.5) is the square root of 7 *

x+k=7^(0.5)+k and x+k=-7^(0.5)-k

x=7^(0.5) and x=7^(0.5)i-2k

If that's what you were asking for I'm pretty sure it's right

but i don't understand what you meant by question b)

I've already handed it in, I think you got it right though. Thanks anyway.
I know it's a big bump but I need help. :/

Find the sums of the series:
a) 5 + 7 + 9 + ... + 75
b) 34 + 29 + 24 + 19 + ... + -111
Physics, I know the answer but I just can't write it in less than a paragraph...

"What do we mean when we say that motion is relative? What is everyday motion usually relative to?"

"Explain the difference between instantaneous speed and average speed"
Grrr!!! Rrrr!!! Rawr!!! Meow???
^

"What do we mean when we say that motion is relative? What is everyday motion usually relative to?"

(Galilean) Motion is relative to a reference frame. Everyday motion (as seen on Earth) is usually relative to the Earth, which we claim is "at rest" for the purposes of most macroscopic, nonrelativistic calculations. Most accurately, the 'universal' reference frame is that of light.

"Explain the difference between instantaneous speed and average speed"

Instantaneous speed refers to the speed of an object at a given moment. Average speed is the ratio of the distance an object travels to the amount of time it takes to travel that distance. Instantaneous speed IS an average speed, in the special case where the distance traveled gets infinitely small.

Clear?

Quote by iamtompublic
I know it's a big bump but I need help. :/

Find the sums of the series:
a) 5 + 7 + 9 + ... + 75
b) 34 + 29 + 24 + 19 + ... + -111

To answer these, use Gauss's method. Say you would like to add all of the integers between 1 and 100.

1+2+3+...+100

You could use a calculator and add them conventionally, OR you could add them a more convenient way:

(1+100)+(2+99)+(3+98)+(4+97)+...+(50+51) = 101 + 101 + 101 + 101 + ...

The answer is the same, but in this case, you only need to multiply 101 x 50 to get 5050.