Answer:
Explanation:
As we know that amplitude of forced oscillation is given as
here we know that natural frequency of the oscillation is given as
here mass of the object is given as
angular frequency of applied force is given as
now we have
Two airplanes leave an airport at the same time. The velocity of the first airplane is 650 m/h at a heading of 60.5 ◦ . The velo
city of the second is 560 m/h at a heading of 85◦ . How far apart are they after 2.9 h? Answer in units of m.
1 answer:
First, illustrate the problem as shown in the attached picture. Next, let's find the distance traveled by planes A and B after 2.9 h.
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
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