Answer:
d = 2021.6 km
Explanation:
We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them
Airplane 1
Height y₁ = 800m
Angle θ = 25°
cos 25 = x / r
sin 25 = z / r
x₁ = r cos 20
z₁ = r sin 25
x₁ = 18 103 cos 25 = 16,314 103 m
= 16314 m
z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m
2 plane
Height y₂ = 1100 m
Angle θ = 20°
x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m
z₂ = 20 103 without 25 = 8.452 103 m = 8452 m
The distance between the planes using the Pythagorean Theorem is
d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2
Let's calculate
d² = (18126-16314)² + (1100-800)² + (8452-7607)²
d² = 3,283 106 +9 104 + 7,140 105
d² = (328.3 + 9 + 71.40) 10⁴
d = √(408.7 10⁴)
d = 20,216 10² m
d = 2021.6 km
From a balistics pendulum as an example, which is probably where you are at...
Triangles, L = 12m, x_0 = 1.6, we need to find the angle (theta)
sin (theta) = 1.6/12 = 0.1333....
theta = ArcSin(0.1333...) = 0.1337 rad
Then, this is the height that the mass vertically raises in it's arc
y_2 = L-L*cos(theta) = 0.107 m
use y_2 in a kinematic swing...
<span><span>v=sqrt(<span><span>2g<span>y_2)</span></span></span>=1.45m/s</span></span>
A. Reduced greenhouse gas emissions.
The answer is C guide and inspire good conduct
12.5 times 14 and convert to meters its 1.75 meters per second
