Asked by oluwatoyin
1. the pilot of a private plane flies 20km in a direction 60 degrees north of east,then 30km straight east,then 10km straight north.how far and in what direction is the plane from the starting point.
2. two forces acts on a point object as follows 100N at 170 degrees and 100N at 50 degrees.find their resultant.
3. A car is traveling eastwards at a speed of 40m/s.But a 3m/s wind is blowing southward.What are the direction and speed of the car relative to its original direction.
2. two forces acts on a point object as follows 100N at 170 degrees and 100N at 50 degrees.find their resultant.
3. A car is traveling eastwards at a speed of 40m/s.But a 3m/s wind is blowing southward.What are the direction and speed of the car relative to its original direction.
Answers
Answered by
Henry
1. d = 20km @ 60deg. + 30km @ 0 deg. + 10km @ 90 deg.
X = hor. = 20cos60 + 30km + 0,
X = hor. = 10 + 30 = 40km.
Y = ver. = 20sin60 + 0 + 10sin90,
Y = ver. = 17.32 + 0 + 10 = 27.32km.
tanA = Y / X = 27.32 / 40 = 0.4330,
A = 23.4 Deg.
d = X / cosA = 40 / cos23.4 = 43.6km @
23.4 Deg. N of E.
2. R = 100N @ 170 Deg. + 100N @ 50 deg.
X = hor. = 100cos170 + 100cos50,
X = hor. = -98.48N + 64.28 = -34.20N.
Y = ver. = 100sin170 + 100sin50,
Y = ver. = 17.36 + 76.60N.
tanA = Y / X = 76.60 / -34.20 = -2.2399
A = -65.94 deg.,CW.
A = -65.94 + 180 = 114 deg.,CCW.
R = X / cosA = -34.20 / cos114 = 84.1N.
@ 114 deg.
3. Vc=40m/s @ 0 deg. + 3m/s @ 270 deg.
X = hor. = 40cos(0) + 3cos270,
X = hor. = 40 + 0 = 40m/s.
=
Y = ver. = 40sin(0) + 3sin270,
Y = ver. = 0 + (-3) = -3m/s.
tanA = Y / X = -3 / 40 = -0.075,
A = -4.29 deg.,CW. = 4.29 deg.,S of E.
A = -4.29 + 360 = 355.7 deg.,CCW.
Vc = X / cosA = 40 / cos355.7 = 40.1m/s
@ 355.7 deg.
X = hor. = 20cos60 + 30km + 0,
X = hor. = 10 + 30 = 40km.
Y = ver. = 20sin60 + 0 + 10sin90,
Y = ver. = 17.32 + 0 + 10 = 27.32km.
tanA = Y / X = 27.32 / 40 = 0.4330,
A = 23.4 Deg.
d = X / cosA = 40 / cos23.4 = 43.6km @
23.4 Deg. N of E.
2. R = 100N @ 170 Deg. + 100N @ 50 deg.
X = hor. = 100cos170 + 100cos50,
X = hor. = -98.48N + 64.28 = -34.20N.
Y = ver. = 100sin170 + 100sin50,
Y = ver. = 17.36 + 76.60N.
tanA = Y / X = 76.60 / -34.20 = -2.2399
A = -65.94 deg.,CW.
A = -65.94 + 180 = 114 deg.,CCW.
R = X / cosA = -34.20 / cos114 = 84.1N.
@ 114 deg.
3. Vc=40m/s @ 0 deg. + 3m/s @ 270 deg.
X = hor. = 40cos(0) + 3cos270,
X = hor. = 40 + 0 = 40m/s.
=
Y = ver. = 40sin(0) + 3sin270,
Y = ver. = 0 + (-3) = -3m/s.
tanA = Y / X = -3 / 40 = -0.075,
A = -4.29 deg.,CW. = 4.29 deg.,S of E.
A = -4.29 + 360 = 355.7 deg.,CCW.
Vc = X / cosA = 40 / cos355.7 = 40.1m/s
@ 355.7 deg.
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