Asked by Biggz
A hiker starts at his camp and moves the following distances while exploring his surroundings: 65.0 m north, 2.00 ✕ 102 m east, 1.10 ✕ 102 m at an angle 30.0° north of east, and 1.50 ✕ 102 m south.
(a) Find his resultant displacement from camp. (Take east as the positive x-direction and north as the positive y-direction.
(a) Find his resultant displacement from camp. (Take east as the positive x-direction and north as the positive y-direction.
Answers
Answered by
Henry
D=65m[90o] + 200[0o] + 110[30o] + 150[270].
X = 200 + 110*Cos30 = 295.3 m.
Y = 65 + 110*sin30 - 150 = -30 m., Q4.
Tan A = Y/X = -30/295.3 = -0.10159.
A = -5.80o = 5.8o S. of E. = Direction.
D = X/Cos A = 295.3/Cos5.8 = 297 m.
X = 200 + 110*Cos30 = 295.3 m.
Y = 65 + 110*sin30 - 150 = -30 m., Q4.
Tan A = Y/X = -30/295.3 = -0.10159.
A = -5.80o = 5.8o S. of E. = Direction.
D = X/Cos A = 295.3/Cos5.8 = 297 m.
Answered by
Jeremy
Your answer made no sense idiot henry
Answered by
Jeremy
herny is an idiot
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