Okay, i've been trying to solve this equation for an hour already.. And my answer is different from others.. I don't know which is the right one.. So maybe you guys can help? Please?:)
V1 = 20 m; 10° East of North
V2 = 30 m; 10° South of East
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Vr = ????
2007-09-17 20:48:06 · 3 answers · asked by mariaastrud 1 in Science & Mathematics ➔ Physics
Vrx = 20cos(80) + 30cos(350)
Vrx = 33.02
Vry = 20sin(80) + 30sin(350)
Vry = 14.49
Vr = 36.06 m; 23.69° North of East
2007-09-17 21:00:31 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Choose a 'zero' reference (usually due East). Then
v1=20m, 80°
v2=30m, -10°
Calculate the x and y components of each
x1=20cos(80)=3.473
y1=20sin(80)=19.696
x2=30cos(-10)=29.544
y2=30sin(-10)=-5.209
Now add up the x and y components to get
xr=3.473+29.544=33.017
yr=19.696-5.209=14.487
and the resultant magnitude is
vr=â(xr²+yr²)=â(33.017²+14.487²)=36.055
the angle is
arctan(yr/xr)=arctan(14.487/33.017)=23.7° so
vr=36.055, 23.7° or
vr=36.055, at 23.7° north of east.
HTH
Doug
2007-09-17 21:04:31 · answer #2 · answered by doug_donaghue 7 · 1⤊ 0⤋
they're orthogonal.. isn't Vr just the square root of the sum of the squares of V1 and V2?
i.e.:
Vr=sqrt(20^2+30^2)
2007-09-17 20:55:38 · answer #3 · answered by m_tassadar 2 · 0⤊ 0⤋