Answer:
Explanation:
Using the given values
Let
Px= 2.3 x 10^6, Py=0, Qx=1.5x10^6cos(70),
Qy= 1.5x10^6sin(70)
So let
Hx= Px+Qx=2.813 x 10^6, Hy=Py+Qy=1.401 x 10^6
Magnitude = √ ((2.813 x 10^6)^2+(1.401 x 10^6)^2)= 3.14 x 10^6
To find direction
tan စ = Hy/Hx = 1.401/2.813= 0.4980
စ = 26.5°
The magnitude and direction of the path of the spaceship if it had traveled to the space station through the nebula are 3.14 x 10⁶ m and 26.5°.
A nebula is a huge cloud consisting of dust and gas in space. Some nebula come from the gas and dust thrown out by the explosion of a dying star known as a supernova.
The captain orders the ship to travel in horizontal direction, Px= 2.3 x 10⁶ m, Py=0,
After turning 70° and it travels to reach the space station, Qx = 1.5x10⁶ x cos70° in horizontal direction.
and Qy= 1.5x10⁶sin70°
So, total horizontal distance moved is
Hx= Px+Qx = 2.813 x 10⁶
Total vertical distance moved is
Hy=Py+Qy = 1.401 x10⁶
Magnitude of the path = sq rt [(2.813 x 10⁶)²+(1.401 x 10⁶)²]
Path = 3.14 x 10⁶ m
Direction of the path,
tan α = Hy/Hx
tan α = 1.401/2.813
tan α = 0.4980
α = 26.5°
Thus, the magnitude and direction of the path of spaceship is 3.14 x 10⁶ m and 26.5°.
Learn more about nebula.
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