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Can Vectors with an angle 180(degrees+) have a negative magnitude? 
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#1
Sep1611, 05:05 PM

P: 10

1. The problem statement, all variables and given/known data
Can vectors with 180+(degrees) have a negative magnitude? I'm trying to find components of a vector that is going 260 km, 48 (degrees) south of east. So I'm confused whether the 260 is positive or not because of the 48 degrees. 2. Relevant equations ([itex]\vec{V}[/itex]) (sin[itex]\theta[/itex]) ([itex]\vec{V}[/itex]) (cos[itex]\theta[/itex]) 3. The attempt at a solution (260km)(sin(48)) or (260km)(sin(48)) =[itex]\vec{V}[/itex][itex]_{y}[/itex] (260km)(cos(48)) or (260km)(cos(48)) =[itex]\vec{V}[/itex][itex]_{x}[/itex] Which one?? 


#2
Sep1611, 05:11 PM

P: 140

sin(x) = sin(x)
cos(x) = cos(x) Hope that helps. Just use the convention that counterclockwise is postive and clockwise is negative in terms of measuring an angle. 


#3
Sep1611, 05:18 PM

P: 10




#4
Sep1611, 05:20 PM

P: 10

Can Vectors with an angle 180(degrees+) have a negative magnitude?
and since it is in the 4th quadrant, y is negative (sin) and x is positive (cos) 


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