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
Given below the arrangement of loading on the larger boat by two tug boats.
F₁ = 5 N
F₂ = 4 N
Angle between them θ = 90⁰
Resultant between two vectors, 
Substituting
So magnitude of the net force on the block = 6.403 N
V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s
v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
