anonymous
  • anonymous
Hi guys, Matt from Australia here. Brushing up on course material before beginning Bachelor of Science in Physics next year through an Australian university. With this subject, I found that they have chopped up the old lectures from 1999 a bit. I have instead downloaded all 20-something lectures and am watching them in order. It seems to be working fine. I'd like to get this subject knocked over in the next 6 weeks or so, I'd be interested to hear from anyone on the same timeline as me. Two questions at the moment...
OCW Scholar - Physics I: Classical Mechanics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1) in the challenge problems for vectors (topic 3), question 1.4 shows the answer to be 63 deg... I got 56 deg. Anyone have this issue also?
anonymous
  • anonymous
2) same topic (vectors, topic 3) challenge problems, problem 2.1. The notes show that the denominator of the function is magnitude of vector A * magnitude of vector B, yet when he calculates it out, he uses root3 * 2 (i.e. root3 + root 3, instead of root 3 * root 3 = 3). Hope this makes sense.
anonymous
  • anonymous
3) and another one... maybe I'm just not getting it... see attachment and red circled areas. Cross product formula for Bx should be -2?
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anonymous
  • anonymous
For your question 1) the challenge problems for vectors (topic 3), question 1.4 . I also get answer 56 degree. I think the solution in the pdf has mistake.
anonymous
  • anonymous
For your question 2 about vectors( topic 3) challenge problems, problem 2.1, I think the solution in the pdf has mistake again. The angle should equal to : arc cos ( -1 / ( 3^0.5 X 3^ 0.5) ) = arc cos ( -1/3 ) = 109 degree reference: http://en.wikipedia.org/wiki/Tetrahedral_molecular_geometry
anonymous
  • anonymous
I am programmer and computer science usually use ^ to represent power. e.g. square root of 3 = 3 power 0.5 = 3^0.5
anonymous
  • anonymous
For your question 3 about vectors( topic 3) challenge problems, problem 3.1, you are ture that the suggested answer has mistake. Bx should be -2. A x B = ((1)(3)-(-1)(-1)) i + ((-1)(-2)-(1)(3)) j+ ((1)(-1)-(1)(-2))k = 2i -j +k ............ (ans1) Check the intermediate answer ans1: Let C = A x B, then i) dot product of A and C is 0 ( becasue they are prependicular ) ii) dot product of B and C is also 0 ( becasue they are prependicular ) product of C and A = (Ax )(Cx) + (Ay)(Cy) + (Az)(Cz) = (1)(2) + (1)(-1) + (-1)(1) = 0 product of C and B = (Bx)(Cx) + (By)(Cy) + (Bz)(Cz) = (-2)(2) + (-1)(-1) + (3)(1) = 0 If you use the pdf's answer C = A x B = 2 i- 5 j- 3 k to calcualte product of B and C, you will find that it is not equal to 0 ! (that means they are not prependicular): = (Bx)(Cx) + (By)(Cy) + (Bz)(Cz) = (-2)(2) + (-1)(-5) + (3)(-3) = -8 !
anonymous
  • anonymous
Great, thanks Ricky, good to have my thoughts confirmed :)

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