Ok, for a vecor with two dimensions we depict it as (xi+yj) and the absolute value is 100. So Sqrt(x^2 + y^2) = 100 that is what is given. Now to have 20 vectors, you need to have 20 sets of x,y which satisfies this equation Sqrt(x^2+y^2)=100.. and I think that is what your physics teacher asked you to do...
Ok, now we know
Sqrt (X^2+Y^2) = 100
or X^2+Y^2 = 100^2 = 100,00
Now we will just need 10 sets of X,Y and rest 10 sets will be negative vectors of the same set we will get and that will give the 20 sets
Now take a calculator and set the value of x as 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 and determine the corresponding values of Y and we will get the following points that satisfy the equation above
(0, 100), (10, 99.49), (20, 97.98), (30, 95.39), (40, 91.65), (50, 86.6), (60, 80), (70, 71.41), (80, 60), (90, 45.59), (100, 0) = thus there are now 11 sets we got so the corresponding negative vectors sets would be (0, -100), (-10, -99.49), (-60, -80), (-80, -60), (-100, 0), (-20, -97.98), (-30, -95.39), (-40, -91.65), (-50, -86.6)... and we have got already 20 sets ...
Likewise we can get many more sets with many different values of x, y .... that will satisfy the equation....
I hope I have answered your question fully ... :)
2006-07-05 16:49:52
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answer #1
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answered by TJ 5
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Since the number of vectors is even you can form right triangles. Use the rules to calculate the distance of a leg on a right triangle, then compute how many right triangles would be required to reach 100 meters. You can make a right triangle leg any size to fit the 100 meter distance. The size of that leg will determine the size of the other legs. Those legs can be of any size since you are CROSSING not TRAVELING 100 meters.
You need to figure out the rest, and make sure that you show this answer to your teacher so you can explain it properly.
2006-07-05 23:33:26
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answer #2
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answered by Dan S 7
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This is trig with simultaneous equations, but your rules don't make any sense. Rewrite the quedstion. Write an equation tha describes each give rule, and solve for the unknowns. This method will work on any such problem, but you rules are very poorly stated.
2006-07-05 23:31:56
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answer #3
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answered by Chris W 2
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