Leah.J
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You’re supposed to find thisName:  3ED13F6D-B5FF-4671-81EA-35778C2A9808.jpeg
Views: 4
Size:  77.2 KB
Given this
Name:  DAB966D0-2084-415D-A6D2-467A2C24760C.jpeg
Views: 4
Size:  73.9 KB
Here’s what I did, is my answer correct ?
Name:  7A7C2B8F-F620-43B5-B2C6-97AAC91401DA.jpeg
Views: 4
Size:  50.8 KB
Last edited by Leah.J; 4 days ago
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Muttley79
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(Original post by Leah.J)
You’re supposed to find this
Given this

Here’s what I did, is my answer correct ?
We need to see the question and your working to help you.
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Leah.J
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(Original post by Muttley79)
We need to see the question and your working to help you.
everything is uploaded now
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Muttley79
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(Original post by Leah.J)
everything is uploaded now
Was the diagram given to you? How does A = B or do you mean the magnitude is equal?
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Leah.J
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(Original post by Muttley79)
Was the diagram given to you? How does A = B or do you mean the magnitude is equal?
Only magnitude yes
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Muttley79
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(Original post by Leah.J)
Only magnitude yes
Sorry I really can't follow what you are trying to do here.
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Leah.J
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(Original post by Muttley79)
Sorry I really can't follow what you are trying to do here.
Find the magnitude of the 1st picture given in the info in the 2nd picture. This is exactly how we were given the question in the quiz.
The magnitude of A= that of B
There’s a 45 degree angle between them and A is parallel to the x axis.
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ghostwalker
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(Original post by Leah.J)
Find the magnitude of the 1st picture given in the info in the 2nd picture. This is exactly how we were given the question in the quiz.
The magnitude of A= that of B
There’s a 45 degree angle between them and A is parallel to the x axis.
Your general method looks fine - pythagoras on the "x" and "y" components - except for the same error in each part:

(A+A\cos 45)^2=???
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Leah.J
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(Original post by ghostwalker)
Your general method looks fine - pythagoras on the "x" and "y" components - except for the same error in each part:

(A+A\cos 45)^2=???
Oh , Name:  image.jpg
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Size:  68.1 KB ?
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ghostwalker
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(Original post by Leah.J)
Oh , Name:  image.jpg
Views: 8
Size:  68.1 KB ?
:party:

With that correction you should get the correct result, 1+\sqrt{2} or its decimal equivalent.
Last edited by ghostwalker; 4 days ago
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ghostwalker
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(Original post by Leah.J)
...
Now that you have the result, an alternative method is to look at this geometrically. A diagram is often helpful with vectors.

I've not filled in all the details (left as an exercise), but refering to the diagram, the result you're looking for is \tan x, where x is 67.5 degrees.

Name:  Untitled.jpg
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Size:  42.5 KB
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