The upward force acting on the wing of an airplane ir flight is called?
2 answers:
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Answer:
a) F = 64.30 N, b) θ = 121.4º
Explanation:
Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components
let's use trigonometry
Force F1
sin 170 = F_{1y} / F₁
cos 170 = F₁ₓ / F₁
F_{1y} = F₁ sin 170
F₁ₓ = F₁ cos 170
F_{1y} = 100 sin 170 = 17.36 N
F₁ₓ = 100 cos 170 = -98.48 N
Force F2
sin 30 = F_{2y} / F₂
cos 30 = F₂ₓ / F₂
F_{2y} = F₂ sin 30
F₂ₓ = F₂ cos 30
F_{2y} = 75 sin 30 = 37.5 N
F₂ₓ = 75 cos 30 = 64.95 N
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = -98.48 +64.95
Fₓ = -33.53 N
Y axis
F_y = F_{1y} + F_{2y}
F_y = 17.36 + 37.5
F_y = 54.86 N
a) the magnitude of the resultant vector
let's use Pythagoras' theorem
F = Ra Fx ^ 2 + Fy²
F = Ra 33.53² + 54.86²
F = 64.30 N
b) the direction of the resultant
let's use trigonometry
tan θ’= F_y / Fₓ
θ'=
θ'= tan⁻¹ (54.86 / (33.53)
θ’= 58.6º
this angle is in the second quadrant
The angle measured from the positive side of the x-axis is
θ = 180 -θ'
θ = 180- 58.6
θ = 121.4º
Answer:
The maximum height that the fish can jump is 2.19 m.
Explanation:
Hi there!
Please, see the attached figure for a better understanding of the problem.
The motion of the salmon is a parabolic one because when it jumps, it already has a horizontal velocity (see figure).
The position and velocity vectors of the salmon at a time t, can be calculated as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = position of the salmon at time t.
x0 = initial horizotal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration of gravity.
Looking at the figure, notice that at the maximum height, the vertical velocity is zero (because the velocity vector is horizontal). Using the equation of the vertical component of the velocity, we can obtain the time at which the salmon is at its maximum height:
vy = v0y + g · t
To find the initial vertical velocity, v0y, let´s look at the figure. Notice that the initial velocity is the hypotenuse of the triangle formed with the horizontal velocity and the vertical velocity. Then:
v0² = v0x² + v0y²
Solving for v0y:
v0y = √(v0² - v0x²)
v0y = √((6.75 m/s)² - (1.65 m/s)²)
v0y = 6.55 m/s
Now, using the equation of the vertical component of the velocity at the maximum height (vy = 0):
vy = v0y + g · t
0 = 6.55 m/s + (-9.8 m/s²) · t
-6.55 m/s / -9.8 m/s² = t
t = 0.67 s
Now, using the equation of the vertical position at t = 0.67 s, we can find the maximum height:
y = y0 + v0y · t + 1/2 · g · t²
y = 0 m + 6.55 m/s · 0.67 s + 1/2 · (-9.8 m/s²) · (0.67 s)²
y = 2.19 m
The maximum height that the fish can jump is 2.19 m.
