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Sagot :
Vf=√30²+100²
Vf=104 m/s
|Ф|=tan inverse of |100÷30|
Ф= 73 degrees
Therefore the resulting velocity of the airplaine is 104 m/s [ 73 degrees N of W]
Vf=104 m/s
|Ф|=tan inverse of |100÷30|
Ф= 73 degrees
Therefore the resulting velocity of the airplaine is 104 m/s [ 73 degrees N of W]
Answer:
The resultant velocity of the airplane is 104.4 m/s.
Explanation:
It is given that,
Velocity of airplane, [tex]v_1=100\ m/s[/tex] (due north)
Velocity of wind, [tex]v_2=30\ m/s[/tex] (east to the west)
We need to find the resultant velocity of the airplane. The resultant velocity is given by :
[tex]v=\sqrt{v_1^2+v_2^2}[/tex]
[tex]v=\sqrt{(100)^2+(30)^2}[/tex]
v = 104.4 m/s
So, the resultant velocity of the airplane is 104.4 m/s. Hence, this is the required solution.
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