Consider the three vectors shown in the figure. They have magnitudes |A| = 38, |
ID: 1653600 • Letter: C
Question
Consider the three vectors shown in the figure. They have magnitudes |A| = 38, |B| = 13, and |C| = 32.4, and the labeled angles are A = 40°, B = 20°, and C = 15°. Note that the figure shows the definitions of the angles, but the arrows in the figure may not be to scale.What is the magnitude of the vector A + B + C?
I need Part C, D, E, & F. thanks
(10%) Problem 9: Consider the three vectors shown in the figure. They have magnitudes A = 38. B 13, and C = 32.4. and the labeled angles are a = 40°Cg = 20°, and ac = 15°. Note that the figure shows the definitions of the anales, but the arrows in the figure may not be to scale. Assignment Status Click here for detailed view Problem Status Completed 2 Completed 3 Completed Completed " 17% Part (a) In what quadrant is the vector A-B-C? 38 17% Part (b) What is the magnitude of the vector A + B-C? >17% Part (c) what the angle between the posrtive x-axis and the vector, measured clockwise in degrees? Partial Grade Summary 6 Completed 7Completed Completed Deducticns 2 Potential 98% 9 Partial Attempts remaining:- per attampT) cotanasi acos0 atan0 co sink0 cosh0 tanh0 cotanh0 e DegreesO Radians 10Completed ailed vie 190 106 Submit I give up Hints2 oe a deduction. Hints remaining: Feedback 196 deduction per feedback Use the formula for the angle betareen a vector and the positive axii, earured corclockhwiza, in tarma of tha vactors componet but mote that the reqaested angle is measured clockwise -Use the c vou found in part (a) 17% Part (d) In what quadrant is the vector-A-B-C? 17% Part (e) what is the magnitude ofthe vector-A-2B-C? 1796 Part (f) what is the angle between the negative x-axis and this vector, measured counterclockwise in degrees?Explanation / Answer
here,
A = 38 * ( cos(thetaA) i + sin(thetaA) j)
A = 29.1 i + 24.4 j
B = 13 * ( - cos(thetaB) i + sin(thetaB) j)
B = - 12.2 i + 4.4 j
C = 32.4 * ( - sin(thetac) i - cos(thetac) j)
C = - 8.4 i - 31.3 j
c)
A + B + C = 8.5 i - 2.5 j
theta = arctan(2.5/8.5)
theta = 16.4 degree
d)
R = - A + 2B + C = - 3.7 i + 1.9 j
this vector lies in 2nd quadrent
e)
magnitude , |R| = sqrt(3.7^2 + 1.9^2)
|R| = 4.2 units
f)
the angle between negative x axis and this vector , theta = arctan(1.9/3.7)
theta = 27.2 degree