Answer:
Please check the explanation.
Step-by-step explanation:
Given the vector
v = 6i + 2√3j
The Magnitude of a vector:
![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D)
The Direction of a vector:
tan Ф = y/x
y=2√3
x = 6
tan Ф = y/x
= 2√3 / 6
= √3 / 3
Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m
time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s
the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s
Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.
Answer:
MP = 29
Step-by-step explanation:
MP = 17 + 3y
5y + 9 = 17 + 3y
-9 -9
5y = 8 + 3y
-3y -3y
2y = 8
2y / 2 = 8 / 2
y = 4
MP = 17 + 3 (4)
MP = 29
Minus the 21 from the other side of the equation making it 7x=-21 then u divide the 7 from the x to the -21 resulting in x=-3
Answer:
Step-by-step explanation:
