Points:3 ) Two airplanes leave an airport at the same time, one gotng northwest
ID: 3019694 • Letter: P
Question
Points:3 ) Two airplanes leave an airport at the same time, one gotng northwest at 404 mph and the other going east at 34 mphHow f ar apart are the planes after 3 hours (to the nearest mile)? D) 691 mi A) 1765 mil C)2072 mi B)1630 mi Find the magnitude and direction angle (to the neareset tenth) for each vector. Give the measure of the direction angle as an angle in 10,3601 Points 3 /A)723 B) 5; 216.9 233.1 D) 5:53.1 Find the angle between v and w. Round your answer to one decimal place, if necessary. Polsts: 3 16)v-2.4j 5i.9 A) 55.6 B) 178 ) 278 D) 65.6 Solve. Points: 3 1) What is the minimum force requined to prevent a ball weighing 229 lb from rolling down a ramp with the horizontal inclined 29.0 A) 10 lb B) 20 b C) 56 Ib D) 11.1 Ib Points: 3 Find v w A) 30i+ 19j-30k B)-18 12j-5k C)-12i-25j 18k 12i 3j-1k For the point given in rectangular coordinates, find equivalent polar coordinates (r, 8) for r> 0 and 0° s0Explanation / Answer
14) Option C) is correct.
Assuming the "NW" one is going true NW, they angle between their paths is 135º.
After 3 hours, the eastbound plane is 343 * 3 = 1029 miles from the origin.
The NW-bound plane is 3 * 404 = 1212 miles from the origin.
The third leg of the triangle can be calculated using the Law of Cosines.
c^2 = a^2 + b^2 - 2abcosC
c^2 = 1029^2 + 1212^2 - 2*(1029 * 1212 * cos(135))
c^2 = 1058841 + 1468944 – (-1763733.62)
c = 2071.6 miles
15) Option C) is correct.
Magnitude = sqrt(3^2 + 4^2) = 5
Angle = arctan(-4/-3) = 53.1 degree = 53.1 + 180 = 233.1 degree (In 3rd quadrant)
16) v = 2i + 4j, w = -5i + 9j
|v| = sqrt(20), |w| = sqrt(106)
v.w = -10 + 36 = 26
cos(Theta) = v.w/(|v|*|w|) = 26/[sqrt(20)*sqrt(106)]
Theta = 55.6 degree