anonymous
  • anonymous
Consider a particle with initial velocity that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis. What is the x component Vx of V? What is the y component Vy of V?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Hunus
  • Hunus
|dw:1329692762479:dw| The x component Vx of V is the velocity vector V projected onto the x axis. We obtain this magnitude as follows \[v_{x}=12\cos(120)=-12\cos(60)=-12(\frac{1}{2})=-6\] The y component Vy of V is the velocity vector V projected onto the y axis. We obtain this magnitude as follows \[v_{y}=12\sin(120)=12\sin(60)=12\frac{\sqrt{3}}{2}=6\sqrt{3}=10.4\] |dw:1329693129338:dw|
anonymous
  • anonymous
why does sin have a positive value?
Hunus
  • Hunus
y has a positive value in the top two quadrants

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Hunus
  • Hunus
|dw:1329693437661:dw|
anonymous
  • anonymous
so to figure out positive and negative i have to look at 12
Hunus
  • Hunus
You have to look at the direction it's pointing in. For theta equal to 120 or '60 above the negative x axis' the vector is pointing in the negative x direction and in the positive y direction
anonymous
  • anonymous
so what is u have something like this :
anonymous
  • anonymous
srry i mean
anonymous
  • anonymous
|dw:1329693963427:dw|
anonymous
  • anonymous
now its moving clockwise
Hunus
  • Hunus
The vector A you described is at an angle of (180+60) 240 degrees with respect to the positive x axis as you can see from your picture it is directed toward the negative x negative y direction, so both x and y will be negative.
anonymous
  • anonymous
why do i have to add 180?
Hunus
  • Hunus
|dw:1329694073557:dw| Because we're making the angle with respect to the positive x axis to make it simple to take the sine and cosine of the vector
Hunus
  • Hunus
By adding 180 to 60 we're changing it from your original picture |dw:1329694180018:dw| and making it |dw:1329694218137:dw|
anonymous
  • anonymous
or u can say the angle is -60
Hunus
  • Hunus
If we're taking it to be with respect to the positive x axis |dw:1329694331261:dw| then it would be -120 degrees. -60 degrees looks like this |dw:1329694414386:dw|
anonymous
  • anonymous
|dw:1329694584288:dw|
anonymous
  • anonymous
thats how u show it
anonymous
  • anonymous
cause its moving clockwise
Hunus
  • Hunus
Yes.
anonymous
  • anonymous
so in this question is the 60 degrees going clockwise
Hunus
  • Hunus
It typically doesn't matter how you arrive at the angle, but it appears that the angle was rotated counter clockwise 60 degrees. The vector is at 240 degrees with respect to the positive x axis.
anonymous
  • anonymous
so qhen do i know when to add 180 or minus 180
Hunus
  • Hunus
Look at the positive x axis and see how many degrees it takes to get to your vector. This is only showing whether a vectors components are positive of negative.
Hunus
  • Hunus
or negative*
anonymous
  • anonymous
can u please give me an example of each type if itsnt too much trouble
Hunus
  • Hunus
|dw:1329695075597:dw| Is vector M's x component positive or negative? What about its y component?
anonymous
  • anonymous
the x compoinent is negative while the y component is positive
Hunus
  • Hunus
\[v_x=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]so we know that the value of Vx is negative while the value of Vy is positive. From this we could say that \[v_{y}=|M| \sin(30)=\frac{|M|}{2}\] and \[v_{x}=-|M|\cos(30)=-|M|\frac{\sqrt{3}}{2}\] Where |M| is the magnitude of the vector M However we could also say that, to get to vector M from the positive x axis |dw:1329695464157:dw| we would have to go through an angle of 150 degrees. Thus would make our calculations \[v_y=|M|\sin(150)=\frac{|M|}{2}\] As you can see we arrive at the same answer which is why it doesn't matter which way you get to the angle.
Hunus
  • Hunus
\[v_{x}=|M|\cos(150)=|M|(\frac{-\sqrt{3}}{2})=\frac{-|M|\sqrt{3}}{2}\]
anonymous
  • anonymous
what does 150 degrees tellus ?
anonymous
  • anonymous
are we saying the angle 150 is equal to 30 degrees
Hunus
  • Hunus
150 degrees tells us what the angle is from the positive x axis. That's what we usually like to use as a reference point when using sines and cosines.
Hunus
  • Hunus
What we are saying is that we only need the reference angle (how many degrees it is from the closest x axis) if we know if x is positive or negative or y is positive or negative
anonymous
  • anonymous
so basically we are saying the angle is 150 degrees away from the orgin
Hunus
  • Hunus
From the positive x axis
anonymous
  • anonymous
what about the negative x axis
Hunus
  • Hunus
150 degrees tells us how far we would have to rotate a vector from the positive x axis (0 degrees) to get to where the vector is.
anonymous
  • anonymous
ohhh i get it
Hunus
  • Hunus
Good :)
anonymous
  • anonymous
and to find the negative value what would i have to do
Hunus
  • Hunus
Well, the degrees from the positive x axis tells us whether it's positive or negative. For example, between 0 and 90 degrees, x and y are negative, between 90 and 180 degrees x is negative and y is positive, between 180 and 270 degrees both x and y are negative and from 270 to 360 degrees x is positive and y is negative. If you, for example, put in 6sin(340) you will get a negative value because y is negative in that quadrant. If you put in 12cos(320) you will get a positive value because x is positive in that quadrant.
Hunus
  • Hunus
between 0 and 90 x and y are positive*** sorry
anonymous
  • anonymous
so it doesnt matter whether u say 30 degreees or 150 degrees
anonymous
  • anonymous
thanks . could u please help me with another question
Hunus
  • Hunus
It matters, but only as far as the dirrection goes. This is because the MAGNITUDE of cos(30) is the same as the MAGNITUDE of cos(150), but one is negative and one is positive
Hunus
  • Hunus
Sorry for the delay on that response
anonymous
  • anonymous
and that has to do withe quadrant whether is positive or negative

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