Magnitude of vertical and horizonal componants

In summary: 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##no I think your right, I swapped them around.
  • #1
kaydis
23
1
Homework Statement
Find the magnitude of the vertical and horizontal components of this signal.
Relevant 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?
 
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  • #2
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?
 
  • #3
sorry, i just double checked and its 3π/4
 
  • #4
kaydis said:
sorry, i just double checked and its 3π/4

So, what do you know about components (of a vector)?
 
  • #5
i know that they're the parts of a vector and that you use trigonometry to find the componants?
 
  • #6
kaydis said:
i know that they're the parts of a vector and that you use trigonometry to find the componants?

Exactly!
 
  • #7
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?
 
  • #9
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.
 
  • #10
PeroK said:
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
 
  • #11
kaydis said:
if I do Vx=θcos(V)
i got -4.242640687

I've never seen anyone do that before. That's definitely a new approach.
 
  • #12
how would you do it? sorry I'm super confused :oldconfused:
 
  • #13
kaydis said:
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##
 
  • #14
no I think your right, I swapped them around
 

Related to Magnitude of vertical and horizonal componants

1. What is the difference between vertical and horizontal components in a scientific context?

The vertical and horizontal components refer to the two perpendicular directions in a 2-dimensional coordinate system. The vertical component is the measurement along the y-axis, while the horizontal component is the measurement along the x-axis.

2. How are the vertical and horizontal components related in a vector?

In a vector, the vertical and horizontal components are related through the Pythagorean theorem. The magnitude of the vector is equal to the square root of the sum of the squares of the vertical and horizontal components.

3. Can the magnitude of the vertical and horizontal components be negative?

The magnitude of the vertical and horizontal components can only be positive. Negative values can be represented by the direction of the vector, which can be either positive or negative along the vertical and horizontal axes.

4. How are the vertical and horizontal components used in physics?

The vertical and horizontal components are often used to analyze the motion of an object in a 2-dimensional plane. They can be used to calculate the velocity, acceleration, and forces acting on the object.

5. What is the significance of calculating the magnitude of the vertical and horizontal components?

Calculating the magnitude of the vertical and horizontal components allows for a better understanding of the overall movement and forces acting on an object. It also aids in the accurate prediction and analysis of physical phenomena.

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