Any Physics People around?

[agahnim]

Does anyone know how to answer this question?

Find the vector's magnitude and direction.

A (vector) = 5.00 "i-hat" - 6.00 "j-hat"

A = ?

I can't find this in the book anywhere and the teacher doesn't teach worth a crap.

Any suggestions are welcome.

 

[Siva]

F = MA

 

[mdchesne]

magnitude: sqrt((5^2)+(6^2))

 

[mdchesne]

direction: 5x + 6y

so it's over 5, up 6

SoCaToa, physics boy

tan(theta) = 6/5

solve for theta

 

[mchammer187]

direction is just the vector/magnitude

 

[mdchesne]

Originally posted by: Siva
F = MA

stfu, lol. wrong reply

 

[Siva]

Originally posted by: mdchesne
direction: 5x + 6y

so it's over 5, up 6

SoCaToa, physics boy

tan(theta) = 6/5

solve for theta

damn, I took physics in high school and i'm retaking it in college, this is gonna suck

 

[l Xes l]

and i see dead people

 

[dds14u]

Originally posted by: mdchesne
direction: 5x + 6y

so it's over 5, up 6

SoCaToa, physics boy

tan(theta) = 6/5

solve for theta

I believe he said 5x - 6y.

So the vector would be pointing to the bottom right.
When you solve for theta, subtract it from 360.
Or use the negative of the theta.
Make sure you calculator is in degree mode and not radian mode.

 

[Vertimus]

How is this physics?

This is like pre-calculus math

 

[agahnim]

I figured out how to get the direction, not understanding how to get the magnitude.

 

[dighn]

Originally posted by: agahnim
I figured out how to get the direction, not understanding how to get the magnitude.

use pythagorus (sp) ie magnitude^2 = sum of (each component squared)

 

[mdchesne]

oh yea, right. wrong quadrant. I wouldn't use the negative of theta. too much work. define the correct quadrant using negatives and positives around the magnitude. so +/-(whatever the magnitude would be)cos(theta) +/- (magnitutde)sin(theta)

first mag +/- defines location of x (negative or positive in relation to Y axis)
second mag +/- defines verticle position with respect to x axis

it's not physics, BTW, it's calc

 

[agahnim]

doh forgot about that....now I got it thanks all

 

[agahnim]

Well its the introductory material to physics havent seen this stuff in years so I'm a little rusty.

 

[mdchesne]

Magnitude:

the square root of ((5^2) + (-6^2)+(0^2))
__________AXIS:___X____Y____Z____

you are not in a 3-dimensional plane, so there is no k axis

 

[IHateMyJob2004]

Originally posted by: agahnim
Does anyone know how to answer this question?

Find the vector's magnitude and direction.

A (vector) = 5.00 "i-hat" - 6.00 "j-hat"

A = ?

I can't find this in the book anywhere and the teacher doesn't teach worth a crap.

Any suggestions are welcome.

A = SQRT (i^2 + j^2)

 

[IHateMyJob2004]

Originally posted by: agahnim
Does anyone know how to answer this question?

Find the vector's magnitude and direction.

A (vector) = 5.00 "i-hat" - 6.00 "j-hat"

A = ?

I can't find this in the book anywhere and the teacher doesn't teach worth a crap.

Any suggestions are welcome.

And I hope this is a high school question, be casue if you are in college and this is problematic, you are so screwed.

 

[mdchesne]

agreed. but it's a simple equation, maybe he just didn't have it

 

[mdchesne]

oh, i actually have a question concerning vectors now. ironic. Was checking into some college books and i found this one problem involving a 3-dimensional vector problem.

A plane is heading due east and climbing at a rate of 80km/h. If it's airspeed is 480km/h, and there is a 100km/h wind blowing northeast, what is the groundspeed of the aircraft?

now, i'm thinking you can get 100cos(45)i + 100sin(45)j from the NE wind
now is 480i + 80k right for the plane? I just realized i forgot how to do 3D vectors after these years.

 

[dighn]

^ i would assume that 480 is the total combiend speed. so it would be |80k + horizontal| = 480. the horizontal speed should be sqrt(480^2-80^2)