005 (part 1 of 2) 10.0 points Consider four vectors F1, F2, F3, and F4 with magn

Question

005 (part 1 of 2) 10.0 points Consider four vectors F1, F2, F3, and F4 with magnitudes F1 = 32 N, F2 = 26 N, F3 = 22 N, and F4 = 49 N, theta1 = 150 degree, theta2 = -150 degree, theta3 = 18 degree, and theta4 = -59 degree, measured from the positive x axis with counterclockwise positive. What is the magnitude of the resultant vector F = F1 + F2 + F3 + F4? Answer in units of N 006 (part 2 of 2) 10.0 points What is the direction of this resultant vec-tor F, within the limits of -180 degree and 180 degree as measured from the positive x axis with coun-terclockwise positive? Answer in units of degree 007 10.0 points Two airplanes leave an airport at the same time. The velocity of the first airplane is 750 m/h at a heading of 33.5 degree. The velocity of the second is 600 m/h at a heading og 135 degree. How far apart are they after 2.5 h? Answer in units of m

Explanation / Answer

For 005)

We need all the components of the four vectors

F1x = (32)(cos 30) = -27.7 (- since its in the second quadrant)

F1y = (32)(sin 30) = 16

F2x = (26)(cos 30) = -22.5 (- since its in the third quadrant)

F2y = (26)(sin 30) = -13 (- since its in the third quadrant)

F3x = (22)(cos 18) = 20.9

F3y = (22)(sin 18) = 6.8

F4x = (49)(cos 59) = 25.2

F4y = (49)(sin 59) = -42 (- since its in the 4th quadrant)

The net x = -27.7 - 22.5 + 20.9 + 25.2 = -4.1

The net y = 16 - 13 + 6.8 - 42 = -32.2

The net is from the Pythagorean Theorem

net2 = (32.2)2 + (4.1)2

net = 32.5 N

For 006)

The angle is from the tangent function

tan(angle) = 32.2/4.1

angle = 82.7o

That is -82.7o which is also 277.3o

For 007

The first plane distance

d = vt = 750 (2.5) = 1875 miles

The second plane distance

d = 600(2.5) = 1500 miles

The angle between the two is 180 - 45 - 33.5 = 101.5

Apply the law of cosines

c2 = a2 + b2 - 2abcos(C)

c2 = (1875)2 + (1500)2 - 2(1500)(1875)(cos 101.5)

c = 2624 miles

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