Notices
Results 1 to 8 of 8

Thread: Significant digits

  1. #1 Significant digits 
    Forum Sophomore
    Join Date
    Sep 2005
    Posts
    106
    1) Place the answer in proper amount of significant digits and unit:
    K = [6.63x10^(-34) J.s] (7.09x10^14 s) - [2.17 x 10^(-19) J]

    Note: J.s = Joule x sceonds

    Can someone help me with this question, please? I don't know how to write the answer with the proper amount of significant digits and unit.....

    I got:

    K=4.70x10^(-19) J.s^2 - 2.17x10^(-19) J

    Note: J.s^2 = Joule x seconds^2

    They have different units, how can I subtract this and what are the final units? What would be the correct number of significant digits in the final answer?


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Radioactive Isotope mitchellmckain's Avatar
    Join Date
    Oct 2005
    Location
    Salt Lake City, UTAH, USA
    Posts
    3,112
    It is impossible. Units must match over + - and = signs. I suspect that the J.s is a typo and should be J/s


    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
    Reply With Quote  
     

  4. #3  
    Forum Sophomore
    Join Date
    Sep 2005
    Posts
    106
    I think so!

    I have another qeustion regarind sine law

    2) "The sine law can yield an acute angle rather than the correct obtuse angle when solving for an angle grater than 90 degrees. This problem occurs because for an angle A between 0 degrees and 90 degrees, sin A = sin (A + 90 degrees). To avoid this problem, always check the validity of the angle opposite the largest side of a triangle."

    Why sin A = sin (A + 90 degrees) if A is between 0 and 90 degrees? I don't get it......
    Reply With Quote  
     

  5. #4  
    Forum Radioactive Isotope mitchellmckain's Avatar
    Join Date
    Oct 2005
    Location
    Salt Lake City, UTAH, USA
    Posts
    3,112
    Well for one thing your formula is incorrect. It is not true that sin(A) = sin(A+90). The correct equalities are the following:
    sin(A) = sin(180 - A)
    cos(A) = cos(-A) = cos(360 - A)
    cos(A) = sin(A+90)
    sin(A) = cos(A-90)

    Anyway the point is that if you look at the values of the sin and cos on all the angles between 0 and 360 degrees you will see that there are two different angles which give the same result for every result other that 1 and -1.

    For example sin(45)=sin(135), sin(30)=sin(150), sin(60)=sin(120)
    for these same angles the cosines are not equal
    cos(45) = - cos(135), cos(30) = -cos(150), cos(60) = -cos(120)
    Likewise for the cosine: cos(45)=cos(-45)=cos(315) and so on
    but sin(45) = -sin(-45) = -sin(315)

    So if you have both the cosine and the sine of an angle then there is only one possible angle between 0 and 360 which matches both. But with only just the sine or just the cosine there are two possible angles which can match. For triangles the cosine is not a problem because you cannot have angles in a triangle which are negative or greater than 180 degrees, but there are triangles with angles greater than 90 degrees so the equality, sin(A) = sin(180 - A) is of particuar importance for triangles and means you cannot take the result of the inverse sin on your calculator as the correct solution for A in an equation like sin(A) = b.

    But when it is not just about triangles you have the same problem when solving cos(A)=b, and tan(A)=b, for the angle A. But when restricted to triangles your calculator inverse cos and inverse tan does give the correct result (except that it might give a negative answer in which case you need to add 180 degrees).
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
    Reply With Quote  
     

  6. #5  
    Forum Sophomore
    Join Date
    Sep 2005
    Posts
    106
    "The sine law can yield an acute angle rather than the correct obtuse angle when solving for an angle grater than 90 degrees. This problem occurs because for an angle A between 0 degrees and 90 degrees, sin A = sin (A + 90 degrees). To avoid this problem, always check the validity of the angle opposite the largest side of a triangle."

    So is this quote wrong? It is right off my textbook......


    ==========================================
    Besides, is the following also correct?

    cos(A) = sin(90-A)
    sin(A) = cos(90-A)
    Reply With Quote  
     

  7. #6  
    Forum Radioactive Isotope mitchellmckain's Avatar
    Join Date
    Oct 2005
    Location
    Salt Lake City, UTAH, USA
    Posts
    3,112
    Quote Originally Posted by kingwinner
    "The sine law can yield an acute angle rather than the correct obtuse angle when solving for an angle grater than 90 degrees. This problem occurs because for an angle A between 0 degrees and 90 degrees, sin A = sin (A + 90 degrees). To avoid this problem, always check the validity of the angle opposite the largest side of a triangle."

    So is this quote wrong? It is right off my textbook......


    ==========================================
    Besides, is the following also correct?

    cos(A) = sin(90-A)
    sin(A) = cos(90-A)
    The book is definitely wrong, but these two additional equalities are correct.

    You can check these yourself. All of these are derived from these two sum angle formulas:
    sin(A+B) = sin(A)cos(B) + sin(B)cos(A)
    cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
    by putting in a negative angle for B you can also get that
    sin(A-B) = sin(A)cos(B) - sin(B)cos(A)
    cos(A-B) = cos(A)cos(B) + sin(A)sin(B)

    So looking at your two equalities,
    sin(90-A) = sin(90)cos(A) - sin(A)cos(90) = cos(A) because sin(90)=1 and cos(90)=0
    cos(90-A) = cos(A)cos(90) + sin(A)sin(90) = sin(A)
    so they check out, but they are the standard equalities. Mine are a little more unusual, but they check out too. In particular,
    sin(A+90) = sin(A)cos(90) + cos(A)sin(90) = cos(A)

    One of the reasons I gave this is that it directly contradicts the one given by your book. If sin(A+90) = cos(A) then it cannot also be equal to sin(A). Besides just try A=0, sin(0)=0 and cos(0)=1, so is sin(0+90)=sin(0)? Obviously not since that would mean that sin(90)=sin(0) or that 1=0. So your book is definitely wrong. Try A=30 degrees on your calculator!

    PS. one of the things that makes it easier to remember the signs in the sum angle formula is by doing the following,
    Take the second two formulas,
    sin(A-B) = sin(A)cos(B) - sin(B)cos(A)
    cos(A-B) = cos(A)cos(B) + sin(A)sin(B)
    and replace the B with A.
    sin(A-A) = sin(A)cos(A) - sin(A)cos(A)
    cos(A-A) = cos(A)cos(A) + sin(A)sin(A)
    The first checks out since sin(0)=0. The second is another well known equality cos(A)^2 + sin(A)^2 = 1, since cos(0)=1.
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
    Reply With Quote  
     

  8. #7  
    Forum Sophomore
    Join Date
    Sep 2005
    Posts
    106
    "But with only just the sine or just the cosine there are two possible angles which can match. For triangles the cosine is not a problem because you cannot have angles in a triangle which are negative or greater than 180 degrees, but there are triangles with angles greater than 90 degrees so the equality, sin(A) = sin(180 - A) is of particuar importance for triangles and means you cannot take the result of the inverse sin on your calculator as the correct solution for A in an equation like sin(A) = b. "

    Thanks a lot for your detailed explanation! For triangle problems using sin and cos law, cos law is fine with the result on the calculator since the sum of the angles in a triangle is 180 degrees. But for sin law, say
    sin A=0.568
    A= sin^-1 (0.568)
    A=34.6 (from calculator) or 145.4 (180-the value on the calculator)

    And then pick the one that makes sense and matches the information from the question!

    Is this the way to deal with such sin law physics problems?
    Reply With Quote  
     

  9. #8  
    Forum Radioactive Isotope mitchellmckain's Avatar
    Join Date
    Oct 2005
    Location
    Salt Lake City, UTAH, USA
    Posts
    3,112
    Yes but sometimes both answers can be correct too. In an SSA problem (that is where you have two sides and an angle not between them), you can have two solutions, one solution or no solutions. So for example supose you have sides of 4 and 5 and an angle of 40 degrees opposite the side length 4.

    Then sin(40)/4 = sin(A)/5
    which gives sin(A)= .803
    the calculator gives 53.46 degrees for the inverse sin of .803
    but the angle 126.53 is another possible solution for A.

    If the side lenth 5 were length 6.222895307 instead then there would be only one solution for A, 90 degrees and if that same side were any longer than this then then there would be no solutions for A (because then you would have to take an inverse sine of a number greater than 1).
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •