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Vector is 6.0 units long and points along the negative axis. Vector is 8.5 units long and points at 40 to the positive axis
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Vector V1 is 6.6 units long and points along the negative x axis. V2 is 8.5 units long and points at +55 degrees to the positive
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I don't now
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Vector V1 is 6.29 units long and points along the negative x axis. Vector V2 is 4.23 units long and points at +35.0° to the +x
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i need help on this too!
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Vector V1 is 9.00 units long and points along the negative x axis. Vector V2 is 4.72 units long and points at +50.0° to the +x
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Go back and answer my reply to the previous question.
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Vector is 6.3 units long and points along the negative axis. Vector is 9.0 units long and points at 55 to the positive axis.
1-
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1. V1 = (6.3,180deg). X1 = hor. = 6.3cos180 = -6.3 Units. Y1 = ver. = 6.3sin180 = 0 Units 2. V2 =
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Vector is 6.3 units long and points along the negative axis. Vector is 9.0 units long and points at 55 to the positive axis.
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that's right here is the question Vector (V1) is 6.3 units long and points along the negative (X)
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Vector V1 is 6.29 units long and points along the negative x axis. Vector V2 is 4.23 units long and points at +35.0° to the +x
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see my last post.
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Vector V1 is 8.58 units long and points along the -x axis. Vector V2 is 4.45 units long and points at +40.0° to the +x axis.
(a)
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(a) To find the x and y components of each vector, we can use trigonometry. For Vector V1: Since it
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Vector V1 is 8.98 units long and points along the negative x axis. Vector V2 is 4.00 units long and points at +50.0° to the +x
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To find the x and y components of each vector, we can use trigonometry. Keep in mind that the
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Vector V1 is 6.29 units long and points along the negative x axis. Vector V2 is 4.23 units long and points at +35.0° to the +x
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vx1=-6.29 Vx2=4.23cos35 vy2=4.24sin35 assuming it is from the x axis counterclockwise. add them.
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Vector V1 is 6.29 units long and points along the negative x axis. Vector V2 is 4.23 units long and points at +35.0° to the +x
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you can do this. You have the x,y component of them added. magnitude=sqrt(x^2+y^2) angle=arctan y/x
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