Snail13579 Posted May 6, 2008 Share Posted May 6, 2008 Its the Ontario Association of Physics Teachers. They make contests for grade 11 students taking physics 11. The contest was today somewhere between morning to noon. If anyone else has taken the contest please discuss here(i need some answers now>) btw i can post the question sheet but im not really sure if im allow or not. Just in case one of you who took OAPT today wants to share some of your answers...please help, and do so =D here are previous questions:http://www.oapt.ca/grade_11_contest/past_exams.html its pretty interesting Link to comment Share on other sites More sharing options...
timo Posted May 6, 2008 Share Posted May 6, 2008 Why would you think you were not allowed to post the questions after the exam was taken? Are you doing some secret nuclear weapon research with your 11th-graders ? Link to comment Share on other sites More sharing options...
Snail13579 Posted May 6, 2008 Author Share Posted May 6, 2008 its jus that we have different time zones, and other countries like US who have different time zones are also writing this i believe *people from china are also writing it Link to comment Share on other sites More sharing options...
timo Posted May 6, 2008 Share Posted May 6, 2008 Makes more sense than I expected . I would assume that the Canadian teacher association in charge has taken care of the problem, e.g. by starting the tests at the physically same time everywhere (meaning earlier hours in the east, later hours in the west) or making different tests. But it was certainly kind of you if respected that possible loophole in the system and waited a few hours (e.g. number of time-zones minus time it takes to take the test) and then post the questions - you won't have that many time-zones, I'd think. I'd bet that the chances that you'll meet someone on this board who took the same test are more or less zero. But on the other hand judging from the questions from the last years I am pretty sure there is a lot of people interested in the questions (and trying themselves on them) and also several people that can answer them or answer additional questions about them. So please post the questions here at some time that you consider appropriate. EDIT: Oh, and welcome to sfn, btw. Link to comment Share on other sites More sharing options...
Snail13579 Posted May 6, 2008 Author Share Posted May 6, 2008 there are a few "hard" questions i didnt understand, it would be best if i had a scanner(although i got a friend who can scan it but isnt here atm). Basically you got 66 turbines to general 99MW of electricity. Each turbine got 3 blades, ea are 37m in length. Assume that avg density of air is 1kg/m3, what is minimum wind speed required for energy to pass the blades to generate the desired power(99MW) if its 50% efficiency. it gives like 5 options but i want to know how to get the answer without just selecting C(a general choice if you dunno i guess) Link to comment Share on other sites More sharing options...
timo Posted May 6, 2008 Share Posted May 6, 2008 Not sure if I really understand the question but the approach I'd try would have been something like that: - Energy per turbine required: 99 MW/66 = 1.5 MW/turbine. - Since each turbine has 50% effectivity you want 3 MW/turbine to be transferred by the wind. - Energy transferred to the blades is the kinetic energy of the wind: E = mv²/2. - The mass of wind hitting the rotors each second equals A*v * 1 kg/m³ (volume passing through times density), where A is the area of the blades. This is the point where some information, the size of a blade, seems to be missing. Or I don't really understood the question. - Anyways, from this you get E/1s = Av³/2 * kg/m³. This shall equal 3 MW => 3 MW = A v³/2 kg/m³ => 6,000,000/A Wm³/kg = [math] \frac{6,000,000}{A} \ m^5/s^3[/math] = v³. Please note that I might have misunderstood the question, but I think the three blades were supposed to be on one rotor and that some width for them is missing in what you posted. Link to comment Share on other sites More sharing options...
Snail13579 Posted May 6, 2008 Author Share Posted May 6, 2008 answers are 15km/h, 20km/h, 25 km/h, 30km/h, 40km/h, i got 40km/h, btw ur doing waaaay too much i jus changed 40km/h to m/s and multiplied and decided what i thought was right and i got 0.99 which should be multiplied by 100(error somewhere) Link to comment Share on other sites More sharing options...
ctedison12 Posted May 7, 2008 Share Posted May 7, 2008 so..... I have question that confused me... i think it's about a 160 hockey ball have a initial velocity 2m/s , and one guy do a work 25 over 0.8 m. it ask you the velocity after it being hit , i have forgotten the answer they give us... so what will be the answer of the wind power one.... ? By the way can u post question? the day after the testing date? I am really interested in it. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted May 7, 2008 Share Posted May 7, 2008 Why would you think you were not allowed to post the questions after the exam was taken? Are you doing some secret nuclear weapon research with your 11th-graders ? I've taken exams in which I've been told that I am never allowed to discuss any question from the exam. It's not unusual. Link to comment Share on other sites More sharing options...
Mr Skeptic Posted May 7, 2008 Share Posted May 7, 2008 there are a few "hard" questions i didnt understand, it would be best if i had a scanner(although i got a friend who can scan it but isnt here atm). Basically you got 66 turbines to general 99MW of electricity. Each turbine got 3 blades, ea are 37m in length. Assume that avg density of air is 1kg/m3, what is minimum wind speed required for energy to pass the blades to generate the desired power(99MW) if its 50% efficiency. it gives like 5 options but i want to know how to get the answer without just selecting C(a general choice if you dunno i guess) Here's how I would do it. First, assume that the area in question is the minimum size, so that its area would be [math]A = \pi r^2 = \pi 37^2[/math]. The power of the air flow per turbine would be [math]Power = 1/2 * 1 kg/m^3 * A * v * v^2 = 2*99/66 MW[/math] (to generate 99MW at 50% efficiency with 66 turbines), so then [math]v = (\frac{4*99000000}{66 * \pi 37^2})^{1/3} m/s = 11.17 m/s = 40.2 km/hour [/math]. I'm ignoring that the turbine has 3 blades since we know its efficiency. Overall, a very nice, fun, practical, and eco-friendly math problem. Link to comment Share on other sites More sharing options...
Snail13579 Posted May 7, 2008 Author Share Posted May 7, 2008 Here's how I would do it. First, assume that the area in question is the minimum size, so that its area would be [math]A = \pi r^2 = \pi 37^2[/math]. The power of the air flow per turbine would be [math]Power = 1/2 * 1 kg/m^3 * A * v * v^2 = 2*99/66 MW[/math] (to generate 99MW at 50% efficiency with 66 turbines), so then [math]v = (\frac{4*99000000}{66 * \pi 37^2})^{1/3} m/s = 11.17 m/s = 40.2 km/hour [/math]. I'm ignoring that the turbine has 3 blades since we know its efficiency. Overall, a very nice, fun, practical, and eco-friendly math problem. well im not sure if im just good at guessing or what, what i did took like 2 minutes: i picked the largest of the answers(40km/hr) and changed it into m/s and multiplied that by 66*66*37/2*10=99 which i thought was accidental(probably was) explanation for that plz? Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now