pierrezonra
06.10.2019 •
Physics
Aplane is flying at 290km/h [e 42° s] relative to the air when the wind velocity is 65km/h [e 25° n]. calculate the velocity of the plane relative to the ground
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Ответ:
320 km/h [E 31° S]
Explanation:
"E 42° S" means "east, 42° south", or 42° south from east.
"E 25° N" means "east, 25° north", or 25° north from east.
Taking east to be +x and north to be +y, the horizontal component of the velocity is:
vₓ = 290 cos (-42°) + 65 cos 25°
vₓ = 274 km/h
And the vertical component of the velocity is:
vᵧ = 290 sin (-42°) + 65 sin 25°
vᵧ = -167 km/h
The magnitude is found with Pythagorean theorem:
v² = vₓ² + vᵧ²
v² = (274)² + (-167)²
v = 320 km/h
And the direction is found with trig:
θ = atan(vᵧ / vₓ)
θ = atan(-167 / 274)
θ = -31°
Therefore, the plane's velocity relative to the ground is 320 km/h [E 31° S].
Ответ:
Newtons third law of motion says that for every action there is a(n) EQUAL and opposite reaction.
I hope this helped. :)