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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Answer: Option B
The wavelength for a radio wave with a frequency of 2 × 10⁴ Hz is
1.5 × 10⁴ m.
Explanation:
Wavelength is the measure of distance between two successive crests or troughs in a standing wave. Also wavelength can be measured as the ratio of velocity of light to frequency. It is like this because wavelength is inversely proportional to the frequency.
As c = 3 × 10⁸ m/s and the frequency is 2 × 10⁴ Hz, then the wavelength will be
So, the wavelength for a radio wave with a frequency of 2 × 10⁴ Hz is
1.5 × 10⁴ m.