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Magnitude of vertical and horizonal componants

  • Thread starter kaydis
  • Start date
23
1
Homework Statement
Find the magnitude of the vertical and horizontal components of this signal.
Homework Equations
6∠(2pi/4) [W]
(∠ = angle)
I'm not too sure where to start, but 6 is the magnitude and (2pi/4) is the angle? So if I were to plot this on a graph I could get the value of the point on the horizontal and vertical axis. I've tried to plot is using graphing software but it wasn't working so I'm not sure if i was doing it correctly. Does anyone have any ideas?
 

PeroK

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I'm not sure I understand your notation, but if the angle is ##\frac{2\pi}{4}## then that's ##\frac{\pi}{2}##. Are you sure that's correct?

In any case, from what axis is the angle measured?
 
23
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sorry, i just double checked and its 3π/4
 

PeroK

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i know that they're the parts of a vector and that you use trigonometry to find the componants?
 

PeroK

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so if:
horizontal: Vx = Vcos
1567416180938.png

verticle : Vy = Vsin
1567416204645.png


this means that:
Vx = Vcos
1567416180938.png

= (3π/4)cos6
= 2.262347939

Vy = Vsin
1567416204645.png

= (3π/4)sin6
= -0.658357257

is this correct?
 

PeroK

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Delta2

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Btw to insert theta like this θ there is a toolbar above the text box we are writing and one of the options of the toolbar is Insert Symbol which you can use to insert various mathematical symbols.

Alternatively you got to learn to use Latex and type things like this
Code:
 ##\theta##
which will give ##\theta## as output text.
 
23
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Are you sure you haven't calculated ##v_x = \theta \cos(v)##?
for Vx=Vcosθ
I got 2.262347939

if I do Vx=θcos(V)
i got -4.242640687
 

PeroK

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how would you do it? sorry I'm super confused :oldconfused:
 

PeroK

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for Vx=Vcosθ
I got 2.262347939

if I do Vx=θcos(V)
i got -4.242640687
If you are confused, so am I. I have no idea how you get those answers. I get:

##v_x = v \cos \theta = -4.24##
 
23
1
no I think your right, I swapped them around
 

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