Recent content by nathan17

Haha! I just got the email saying there was a reply.. I was just about to post saying I give up, but finally, lol.. I have to say thanks to everyone here because, I really couldn't do it without the help. I'm starting to get it now, hopefully by the time the test comes I will understand it more!

Okay, sorry about the late reply, been busy with study and everything.1(x-4)-2(x+5)=3(4-x) (x-4)-(2x+10)=12-3x x-4-2x-10=12-3x -x-14=12-3x -x = 26 - 3x 2x = 26 x = 13 Is that the solution to the problem?

Haha. Okay, just to clarify. (x - 4) - (2x + 10) = -x + 6. That is because it's divided by 2?

Haha, not only do I manage to not get the answer, but I get worse :P Yeah, it's x-4 and 4-x. I'm kind of afraid to post what I got because I know it's going to be wrong again lol. (x - 4) - (2x + 10) = 12 - 3x -2x + 6 = 12 - 3x -2x = 6 - 3x x = 6 If that's not right, maybe I should...

That's okay, I don't mind the feedback, good or bad, it all helps in the end! Heh, spend a few hours doing maths, you wouldn't think you would slip up on something so small..Sometimes it is the small things that get past though I guess. What annoyed me a bit was that the book chapter on linear...

Homework Statement Solve these linear equations: Homework Equations \frac{x-4}{6} - \frac{x+5}{3} = \frac{4-x}{2} The Attempt at a Solution \frac{x-4}{6} - \frac{x+5}{3} = \frac{4-x}{2} Multiplied by 6 \frac{6x-18}{6} - \frac{6x+30}{3} = \frac{24-6x}{2} Simplified x-18-2x+30 =12-6x Group...

Radioactivity Justification 1. The problem statements, all variables and given/known data 1a. Justify the appropriate charge and mass number values for an electron, a positron, a neutrino and an antineutrino. 1b. Justify the appropriate charge and mass number values for a gamma ray. 1c...

Hmm, thanks for the explanation guys! Yeah, it's the things like that, that I really need to be careful of when working these out when it comes exam time.

Oh. Heh, thanks. so it should be: K = 1.76 * 1.6x10-19 = 2.8x10-19 So, Vs = K / e = 2.8x10-19 / 1.6x10-19 = 1.76 V Cheers for the help, really appreciate it! Nathan

Homework Statement The work function of caesium is 1.35eV. A photoelectric cell has a caesium surface and is illuminated with monochromatic light of wavelength 400nm. What reverse potential difference must be applied to the tube to just stop a current passing through it? Homework Equations...

Ah, Cool. Thankyou! I only recently got good at rearranging equations (I have a physics tutor) so I didn't really think of that, makes it a lot easier. Cheers

Cool, thanks! Yeah, it's a separate question, I will have a crack at it and see how I go and start a new thread if needed. Cheers

Ok, I think I got it :D W = hf0 W/h = hf0/h Cancel out the h on the right side f0 = W/h where W = 4.31eV = 4.31 * 1.6x10-19 = 6.9x10-19J f0 = 6.9x10-19 / 6.63x10-34 = 1.04x1015Hz So, v = fλ v/f = fλ / f Cancel out the f on the right side λ = v/f λ = 3.0x108 / 1.04x1015 = 2.88x10-7m Yay, I...

Homework Statement The work function of zinc is 4.31eV. What is the maximum wavelength of light that will cause electrons to be emitted from a zinc surface? Homework Equations I'm not sure because, we are given the Work Function (W) which is 4.31eV and we need to find the max. wavelength...