Paj! Posted January 19, 2009 Share Posted January 19, 2009 I wonder if the artists on here could siphon me their painting/drawing skills for the art exam? kthxbai Link to comment Share on other sites More sharing options...
Ashley Posted January 20, 2009 Share Posted January 20, 2009 I made this for my mixcd for work (remember, I work at Blockbusters)... I'm game if you are :p Link to comment Share on other sites More sharing options...
Gentleben Posted January 20, 2009 Share Posted January 20, 2009 Right I need some help!Just doing an A2 physics past paper and came upon this question: State and explain 2 differences between Gamma and X-radiation. I don't have any notes on me so if someone could help me with that, it'd be much appreciated Gamma turns you into the Incredible Hulk X-Radiation doesn't:yay: Link to comment Share on other sites More sharing options...
Daniel Posted February 15, 2009 Share Posted February 15, 2009 Anyone know mechanics? Got a beam supported at one end only, where do the reactions go? Below i have my diagram and below that the reactions for a normal simply supported beam. Normally i would take moments about A to get RB then find RA from that. What do i do when its only suported at one end? Link to comment Share on other sites More sharing options...
Supergrunch Posted February 15, 2009 Share Posted February 15, 2009 I'm pretty sure it just acts straight upwards at the joint. Link to comment Share on other sites More sharing options...
Daniel Posted February 15, 2009 Share Posted February 15, 2009 Your right, had to work out shear forces and bending moments too with some forces and my answers are coming out right Thanks! Link to comment Share on other sites More sharing options...
chairdriver Posted February 16, 2009 Share Posted February 16, 2009 Just shove arrows everywhere, and if there isn't a force there it will come out as 0. That's what I did in M2 alot when I had no clue where the forces were acting. It usually meant I ended up with alot of unsolvable variables, but often you could just use moments and get over it. Link to comment Share on other sites More sharing options...
Supergrunch Posted February 16, 2009 Share Posted February 16, 2009 Just shove arrows everywhere, and if there isn't a force there it will come out as 0. That's what I did in M2 alot when I had no clue where the forces were acting. It usually meant I ended up with alot of unsolvable variables, but often you could just use moments and get over it. Lolz, nice approach. I really don't like mechanics much as I can never get the answer right - it caused me to change from physics last year, actually. Pure and higher level statistics always seemed so much more straightforward. (and interesting, come to that...) Link to comment Share on other sites More sharing options...
chairdriver Posted February 16, 2009 Share Posted February 16, 2009 I hate applied maths. I wish my school had allowed people to do Triple Maths, where you do all 18 modules. Quite a few people I spoke to at Oxford did that. I'm in no mood for Chemistry/Physics. Link to comment Share on other sites More sharing options...
Daniel Posted February 23, 2009 Share Posted February 23, 2009 This is the next question. Is same as all my other examples in my book except its not supported at the end... How does this change shear force and bending moments? Does it need an extra force at the left end pulling it down? Normally if it was supported at both ends i would calculate total downward force (length x Wx x 0.5) then take moments about left hand support to find reactions. Work out equations for shear force and bending moments then calulate max bending moment. But moving this support has completely thrown me :P I apologise for poor paint skills! Link to comment Share on other sites More sharing options...
The fish Posted February 23, 2009 Share Posted February 23, 2009 This is the next question. Is same as all my other examples in my book except its not supported at the end... How does this change shear force and bending moments? Does it need an extra force at the left end pulling it down? Normally if it was supported at both ends i would calculate total downward force (length x Wx x 0.5) then take moments about left hand support to find reactions. Work out equations for shear force and bending moments then calulate max bending moment. But moving this support has completely thrown me :P I apologise for poor paint skills! All the moving support does, IIRC, is cancel out, in this case, any horizontal reaction forces it would normally apply if said support were fixed. Link to comment Share on other sites More sharing options...
Daniel Posted February 23, 2009 Share Posted February 23, 2009 well there are no horizontal forces in this as far as i know. Ill keep pondering, thanks Link to comment Share on other sites More sharing options...
MoogleViper Posted February 28, 2009 Share Posted February 28, 2009 Does anybody know how to calculate spearman's rank using a graphical calculator? (I know how to work it out I'm just lazy.) I'm using a Casio FX-9750G. I know how to work out PMCC but not spearman's rank. It's nothing major to work it out but I really am that lazy. Link to comment Share on other sites More sharing options...
Konfucius Posted February 28, 2009 Share Posted February 28, 2009 This is the next question. Is same as all my other examples in my book except its not supported at the end... How does this change shear force and bending moments? Does it need an extra force at the left end pulling it down? Normally if it was supported at both ends i would calculate total downward force (length x Wx x 0.5) then take moments about left hand support to find reactions. Work out equations for shear force and bending moments then calulate max bending moment. But moving this support has completely thrown me :P I apologise for poor paint skills! They used a moving support in that case because otherwise you would have four unknown forces but you only get three equations. What exactly do you have to calculate? If you have to calculate shear-forces throughout the whole beam I'd recommend starting on the far left end because there only the horizontal force applies without any supports which makes it very easy to calculate. Link to comment Share on other sites More sharing options...
UziT Posted April 13, 2009 Share Posted April 13, 2009 Heres a question I got from IT teacher, I have no idea how to answer it :S A ball-sorting robot sorts balls according to their colour. A line of balls each either red or blue is initially on a ramp. The robot takes each ball from the ramp, detects its colour, and then places it into one of two trays – one tray for the red balls and one for the blue balls. The robot continues sorting balls until they have all been assigned to the appropriate trays, write a set of instructions for the robot to sort the line of balls, using a repetition structure and a selection structure. anyone care to help? Link to comment Share on other sites More sharing options...
dwarf Posted April 13, 2009 Share Posted April 13, 2009 Aww man those processes are annoying I remember doing stuff like that for ICT for things like vending machine etc. I would help but don't know how to answer it though. Link to comment Share on other sites More sharing options...
ShadowV7 Posted April 17, 2009 Share Posted April 17, 2009 Holy crappers, i'm stumped at something which no doubt is incredibly easy. Java FTW! Anyway, to you programmers out there, part of my assignement includes converting some stuff to do with hexadecimal and binary with using RGB values, i.e 128,0,128 should give "800080" in hex and "100000000000000010000000", though I need to use padding to bring out the right amount of "0"s. Anyway, I have this for the padding: private String padOut(String s, int l) { if (s.length() >= l) return s; StringBuffer padding = new StringBuffer(); for (int p=0; padding.length() + s.length() < l; p++) padding.append("0"); return padding.toString() + s; } Now I need to get it into this method: private String getHexColourString(int red, int green, int blue) { String hex1 = Integer.toHexString(red); String hex2 = Integer.toHexString(green); String hex3 = Integer.toHexString(blue); String hexColour = (hex1 + hex2 + hex3); return hexColour(); } But I can't remember how Anyone have any idea? I'm sure it's something simple i've completely overlooked. Link to comment Share on other sites More sharing options...
MoogleViper Posted April 23, 2009 Share Posted April 23, 2009 Quick question; 3 (a) A long-term trial was carried out into the effectiveness of giving accident victims with serious head traumas a steroid drug in addition to other treatments. In the trial, 1061 victims were randomly assigned to be given the steroid drug and the remainder were given a drug with no active ingredient (a placebo). The victims either died as a result of their injuries or survived. The results of the trial are summarised in Table 1. Table 1 Additional treatment given Steroid drug Placebo Total Died 396 422 818 Survived 665 665 1330 Carry out a test, using the 5% level of significance, to investigate whether the survival of accident victims with serious head traumas is independent of the additional treatment given. (10 marks) Which test do I use? Link to comment Share on other sites More sharing options...
Supergrunch Posted April 23, 2009 Share Posted April 23, 2009 I'd use a chi-squared contingency test. Link to comment Share on other sites More sharing options...
MoogleViper Posted April 23, 2009 Share Posted April 23, 2009 Damn it. That's the chapter I haven't covered yet. Oh well I can't be arsed with it. Thanks anyway Grunchy. Link to comment Share on other sites More sharing options...
Supergrunch Posted April 23, 2009 Share Posted April 23, 2009 Damn it. That's the chapter I haven't covered yet. Oh well I can't be arsed with it. Thanks anyway Grunchy. No problem - it helps to learn the specific uses of each test before the exam, and you'll find that it's normally fairly obvious which is appropriate. For instance, contingency tests are used if they give you a table of data and you have to check that the results aren't merely a result of chance; regular chi-squared tests are used when checking the accuracy of a model against data; the basic t-test is used to find whether the sample mean is representative of the sample mean; two sample is used when two population means are being compared (though if there are merely two pieces of data for each member of the population, you should use a paired-sample t-test instead, or even Wilcoxon signed-rank if you want to be non-parametric about things); z-tests are used to check if your data is normally distributed; etc. (not sure which tests are on your course) It can still be easy to get confused though - it's best just to try and think clearly about what you actually want to find out. Link to comment Share on other sites More sharing options...
MoogleViper Posted April 24, 2009 Share Posted April 24, 2009 Thanks Grunch. I learned how to do chi squared today. I think I've learned them all. Done z, spearman's rank, wilcoxon paired, mann whitney U, (paired) sign, and now chi squared. Link to comment Share on other sites More sharing options...
Mr_Odwin Posted April 24, 2009 Share Posted April 24, 2009 As a statistician it's perhaps shocking how often I refer to this table. Link to comment Share on other sites More sharing options...
UziT Posted May 10, 2009 Share Posted May 10, 2009 any of you guys familiar with JavaScript? I cannot seem to separate the 'firstWord' and 'secondWord'! <HTML> <HEAD> <TITLE>Krazy Phrases</TITLE> <SCRIPT LANGUAGE = "JavaScript"> var firstWord; var secondWord; var thirdWord; firstWord = window.prompt('Please enter your first word',''); secondWord = window.prompt('Please enter your second world',''); thirdWord = window.prompt('Please enter your third world',''); document.write('The '+ firstWord + secondWord + 'the '+ thirdWord); </SCRIPT> </HEAD> <BODY> </BODY> </HTML anyone help? deadline is 2morrow Thanks! Link to comment Share on other sites More sharing options...
Mr_Odwin Posted May 10, 2009 Share Posted May 10, 2009 Not familiar but I would have thought something like document.write('The '+ firstWord + ' ' + secondWord + 'the '+ thirdWord); would work. Link to comment Share on other sites More sharing options...
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