Question
Consider the three vectors shown in the figure. They have magnitudes |A| = 37, |B| = 7.5, and |C| = 33.9,....

Class Management Help HW 1 B egin Date: 1/13/2019 12:01:00 AMDue Date: 1/22/2019 11:59:00 PM End Date: 5/6/2019 11:59:00 PM (20%) Problem 9: Consider the three vectors shown in the figure. They have magnitudes AI-37, İBİ-7.5, and ICI-33.9, and the labeled angles are 0- 40°, 0-20°, and ac-15°, Note that the figure shows the definitions of the angles, but the arrows in the figure may not be to scale. 17% Part (a) In what quadrant is the vector A+B+C? 04 Correct 17% Part (b) What is the magnitude of the vector A-B + C? Grade Summary- Deductions 0% Potential 100% A+B+C sin) | cos() | tan() | π|( 71819 cotan asin acos(0 atan acotan) sinh coshO tanhocotanh(0 Submissions Attempts remaining: (2% per attempt) detailed view Degrees Radians ens
17% Part (a) Correct! In what quadrant is the vector A+B+C? Q4 17% Part (b) what is the magnitude of the vector A + B + C? A-B-C|-11 sinO tan cosO cotan asin) acoso atan acotan sinh0 cosh t cotanhO o Degrces O Radians 4 5 6 1 23 0 Submit Hint ive Hints: 1% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. ▲ 17% Part (c) What is the angle between the positive x-axis and the vector, measured clockwise in degrees? 17% Part (d) In what quadrant is the vector-A+2B+C? Q3 Correct! 17% Part (e) what is the magnitude of the vector-A +2B+ C? 厶17% Part (f) What is the angle between the negative x-axis and this vector, measured counterclockwise in degrees?
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Answer #1

here,

vector A , A = |A| * ( cos(thetaA) i + sin(thetaA) j)

A = 37 * ( cos(40) i + sin(40) j)

A = (28.3 i + 23.78 j)

vector B , B = |B| * ( - cos(thetaB) i + sin(thetaB) j)

B = 7.5 * ( - cos(20) i + sin(20) j)

B = (-7.05 i + 2.57 j)

vector C , C = |C| * ( - sin(thetaC) i - cos(thetaC) j)

C = 33.9 * ( - sin(15) i - cos(15) j)

C = (- 8.77 i - 32.7 j)

a)

R = A + B - C = 12.48 i - 6.35 j

it is in fourth quadrant

b)

the magnitude of (A + B - C) , |R| = sqrt(12.48^2 + 6.35^2) = 14 units

c)

the angle , theta = arctan((6.35)/12.48) = 26.97 degree clockwise from +X axis

d)

D = - A + 2B + C = - 5.43 i - 51.34 j

so, it is in third quadrant

e)

magnitude , |D| = sqrt(5.43^2 + 51.34^2)

|D| = 51.6 units

f)

the angle , phi = arctan(51.34/5.43) = 83.96 degree counterclockwise from -x axis

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