For the answer to the question above, so at the instant, the acceleration of the airplane is southward, the direction of the velocity is also southward. The direction should be the same because it is both a vector quantity and it does not make sense if the direction and acceleration have different direction.
help me w mine and ill try to help with yours
Answer:
In explanation
Explanation:
Pilots who dont use trim often like the feeling of holding constant back pressure because The heavier control forces makes it more difficult to over-control the airplane inside the turn, so it gives the sense of a more stable flight
Answer:
a) D_ total = 18.54 m, b) v = 6.55 m / s
Explanation:
In this exercise we must find the displacement of the player.
a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components
sin 45 = y₁ / d
cos 45 = x₁ / d
y₁ = d sin 45
x₁ = d sin 45
y₁ = 8 sin 45 = 5,657 m
x₁ = 8 cos 45 = 5,657 m
The second offset is d₂ = 12m at 90 of the 50 yard
y₂ = 12 m
x₂ = 0
total displacement
y_total = y₁ + y₂
y_total = 5,657 + 12
y_total = 17,657 m
x_total = x₁ + x₂
x_total = 5,657 + 0
x_total = 5,657 m
D_total = 17.657 i^+ 5.657 j^ m
D_total = Ra (17.657 2 + 5.657 2)
D_ total = 18.54 m
b) the average speed is requested, which is the offset carried out in the time used
v = Δx /Δt
the distance traveled using the pythagorean theorem is
r = √ (d1² + d2²)
r = √ (8² + 12²)
r = 14.42 m
The time used for this shredding is
t = t1 + t2
t = 1 + 1.2
t = 2.2 s
let's calculate the average speed
v = 14.42 / 2.2
v = 6.55 m / s
To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,
Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective
Replacing with our values
Therefore the focal length of th eobjective lenses is 27.75cm
