Converting Polar Coordinates to Cartesian Coordinates

AI Thread Summary
To convert polar coordinates (r=6.00 m, θ=250°) to Cartesian coordinates, the formulas x = r * cos(θ) and y = r * sin(θ) are used. The user confirms the approach of calculating the cosine and sine of the angle to find the x and y values. The calculations involve substituting the given r and θ into the equations. The final Cartesian coordinates can be derived from these calculations. This method effectively transforms polar coordinates into Cartesian coordinates.
chocolatelover
Messages
238
Reaction score
0

Homework Statement


The polar coordinates of a point are r=6.00 m and theta 250. What are the Cartesian coordinates?

X=?
Y=?


Homework Equations



Cos=adj./hyp.
sin=opp./hyp

The Attempt at a Solution



Would I just need to calculate the cos and sin? That is, I would just do sin 250=y/adj and solve for y, right?

Thank you very much
 
Physics news on Phys.org
right.
x=rcos(theta)
y=rsin(theta)
 
Thank you very much

Regards
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top