key2thecup Posted November 8, 2013 Share Posted November 8, 2013 Yeahhhhh damn lol I don't get it. Link to comment Share on other sites More sharing options...
marleau_12 Posted November 8, 2013 Share Posted November 8, 2013 damn lol I don't get it. When this happens, I always just keep my answer and say the textbook's wrong. Probably why I blow at math lol Link to comment Share on other sites More sharing options...
NucksPatsFan Posted November 8, 2013 Share Posted November 8, 2013 Does anyone have a good source for scholalry articles/journals for biomechanics? Doing a lab experiment on the physics behind a basketball jump shot and SPORTdiscus sucks Looking to find good articles/journals on what makes a jump shot "good" in regards to joint angles, angle velocities, take off velocities, jump height, forces produced by angles ..etc. Link to comment Share on other sites More sharing options...
GoaltenderInterference Posted November 8, 2013 Share Posted November 8, 2013 Does anyone have a good source for scholalry articles/journals for biomechanics? Doing a lab experiment on the physics behind a basketball jump shot and SPORTdiscus sucks Looking to find good articles/journals on what makes a jump shot "good" in regards to joint angles, angle velocities, take off velocities, jump height, forces produced by angles ..etc. Do you have access to academic journals through your school? I quickly browsed through Academic Search Premier through my school and there are a few articles on jumpshots (though I'm not sure how useful it'll be to you). Like these ones: Alves Okazaki V, Rodacki A. Increased distance of shooting on basketball jump shot. Journal Of Sports Science & Medicine [serial online]. June 2012;11(2):231-237. Available from: Academic Search Complete, Ipswich, MA. Accessed November 7, 2013. Lindner M, Kotschwar A, Zsoldos R, Groesel M, Peham C. The jump shot – A biomechanical analysis focused on lateral ankle ligaments. Journal Of Biomechanics [serial online]. January 3, 2012;45(1):202-206. Available from: Academic Search Complete, Ipswich, MA. Accessed November 7, 2013. Link to comment Share on other sites More sharing options...
babych Posted November 8, 2013 Share Posted November 8, 2013 Amateur, need help, can someone help me with this: B=7.5 A=15.3 h=? h2 = a2 + b2 h2 = (15.3)2 + (7.5)2 h = √ (15.3)2 + (7.5)2 h = √ 234.09 + 56.25 h = 23.2 Okay, I'm super amateur, I'm guessing the square root of the h gets transferred into √.... how does this work? I don't know how it works so what I did was, 234.09+56.25=290.34 which has the square root of 17.03 (Wrong answer) +'s to anyone who can teach me in laymen terms lol I think you might have written the question down wrong. If b = 17.5 then the answer is 23.2. If b = 7.5 then h = 17 Link to comment Share on other sites More sharing options...
ckamo Posted November 8, 2013 Share Posted November 8, 2013 Your answer should be right, might be an error in the textbook/solution manual. Link to comment Share on other sites More sharing options...
NucksPatsFan Posted November 8, 2013 Share Posted November 8, 2013 Do you have access to academic journals through your school? I quickly browsed through Academic Search Premier through my school and there are a few articles on jumpshots (though I'm not sure how useful it'll be to you). Like these ones: Thanks for the reply. Yes I can access academic journals through the school library, but most of them focus on free throws, or look at female jump shots. In our experiment, we had a skilled male and unskilled male. Was just hoping someone knew of an overseas database that I could search. If not, guess I'll have to go old school and read books Link to comment Share on other sites More sharing options...
key2thecup Posted November 8, 2013 Share Posted November 8, 2013 I think you might have written the question down wrong. If b = 17.5 then the answer is 23.2. If b = 7.5 then h = 17 Your answer should be right, might be an error in the textbook/solution manual. The book: Page 5 out 20 http://www.surreyconnect.sd36.bc.ca/Secondary/Courses/Documents/SAs%20Jan%2027-2012/Grade%2011/Math%2011%20Apprenticeship%20and%20Worplace%20Reading%20Booklet.pdf I wrote it down right, It must be a error, that's ridiculous. Link to comment Share on other sites More sharing options...
marleau_12 Posted November 8, 2013 Share Posted November 8, 2013 The book: Page 5 out 20 http://www.surreycon...ing Booklet.pdf I wrote it down right, It must be a error, that's ridiculous. Lol that is redonculous Link to comment Share on other sites More sharing options...
key2thecup Posted November 8, 2013 Share Posted November 8, 2013 Lol that is redonculous I know dude, guy like me starts pulling my hair out thinking I don't know how to do it properly, these morons write the thing wrong lol. Link to comment Share on other sites More sharing options...
marleau_12 Posted November 8, 2013 Share Posted November 8, 2013 I know dude, guy like me starts pulling my hair out thinking I don't know how to do it properly, these morons write the thing wrong lol. Pretty freakin annoying when you can't figure out a math problem, I hear ya Link to comment Share on other sites More sharing options...
babych Posted November 8, 2013 Share Posted November 8, 2013 The book: Page 5 out 20 http://www.surreycon...ing Booklet.pdf I wrote it down right, It must be a error, that's ridiculous. Yeah, that's wrong. I would be willing to bet that the mistake the author made was typing in 17.5 instead of 7.5 into the equation and not double checking it. It does get frustrating when there are errors in the book. Link to comment Share on other sites More sharing options...
D.Doughty Posted November 11, 2013 Share Posted November 11, 2013 A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. The mass passes through the equilibrium point (y=0) with positive velocity at t=0. Positive direction of y-axis is downward. At what time will the spring stretch to its maximum length at first time? HELP PLEASE Link to comment Share on other sites More sharing options...
canucks2121 Posted November 11, 2013 Share Posted November 11, 2013 Is the DELF exam the French one? Because I'm pretty sure that it was taken out of high school at least. I haven't taken any university French so I can't help you there. Im in French Immersion and we have a choice to do the DELF Link to comment Share on other sites More sharing options...
key2thecup Posted November 21, 2013 Share Posted November 21, 2013 Ok need some help here again. Question asks for exact answer, this is a textbook example question. a = √20 At this point you can calculate an approximation for this exact expression with your calculator to get a answer of 4.47cm however this question has asked for an exact answer. Therefore we will try and find a perfect root that is a factor of 20. Perfect roots are found by multiplying an integer by itself. For example 2 multiplied by 2 is 4. Four is a perfect square root because the √4=2. Other perfect square roots include: 9,16,25,36,49....... In this case 4 is the largest perfect root that is a factor of 20. That allows us to write the answer in the following simplified form: a = √20 a = (√4) (√5) a = 2 (√5) = 2√5 cm http://www.surreyconnect.sd36.bc.ca/Secondary/Courses/Documents/SAs%20Jan%2027-2012/Grade%2011/Math%2011%20Apprenticeship%20and%20Worplace%20Reading%20Booklet.pdf (page 7) I don't understand where the 5 comes from?! or the answer for that matter...... Link to comment Share on other sites More sharing options...
Likewise Posted November 22, 2013 Share Posted November 22, 2013 Ok need some help here again. Question asks for exact answer, this is a textbook example question. I don't understand where the 5 comes from?! or the answer for that matter...... The 5 is a factor of sqrt 20. Sqrt 5 and sqrt 4 multiply together and end up equaling sqrt 20. However, sqrt 4 can be factored further as its factors are sqrt 2 and sqrt 2. When they are multiplied, you get 2. Since the sqrt 5 cannot be factored, your equation looks like 2 * sqrt 5, equaling 2sqrt5. Link to comment Share on other sites More sharing options...
babych Posted November 22, 2013 Share Posted November 22, 2013 ^what she said Link to comment Share on other sites More sharing options...
key2thecup Posted November 23, 2013 Share Posted November 23, 2013 Basic Trigonometry question: I'm being asked to find side "X", the missing side. How do I know when to use Sin, Cos or Tan? I know about SOHCAHTOA = Sin= Opp/Hyp .... Cos= Adj/Hyp ... Tan= Opp/Adj Here's the triangle: Am I suppose to use 62* as the angle, then 9.8 becomes ADJ, the side to the left is Hyp because its opposite of the right angle and then side X becomes Opp? So does that mean I use Tan? Link to comment Share on other sites More sharing options...
babych Posted November 23, 2013 Share Posted November 23, 2013 Basic Trigonometry question: I'm being asked to find side "X", the missing side. How do I know when to use Sin, Cos or Tan? I know about SOHCAHTOA = Sin= Opp/Hyp .... Cos= Adj/Hyp ... Tan= Opp/Adj Here's the triangle: Am I suppose to use 62* as the angle, then 9.8 becomes ADJ, the side to the left is Hyp because its opposite of the right angle and then side X becomes Opp? So does that mean I use Tan? Yes. When using SOHCAHTOA you determine which ratio to use based on what two sides you using. Edit: just to make sure you have the answer right for this question... tan(62) = x/9.8 so x = 9.8 * tan(62) = 18.431 Link to comment Share on other sites More sharing options...
key2thecup Posted November 23, 2013 Share Posted November 23, 2013 Yes. When using SOHCAHTOA you determine which ratio to use based on what two sides you using. Edit: just to make sure you have the answer right for this question... tan(62) = x/9.8 so x = 9.8 * tan(62) = 18.431 Thanks, this thread is awesome lol Link to comment Share on other sites More sharing options...
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