Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
2000÷330=6.06 repatant so the answer would be about 6.06 seconds
Answer:
μ = 0.725
Explanation:
This problem refers to Newton's second law.
F = ma
Let's write the equations on each axis
Y Axis
N-W = 0
N = W
N = mg
X axis
F-fr = ma
With the body not started moving its acceleration is zero
F-fr = 0
F = fr
The friction force equation is
fr = μ N
fr = μ m g
Let's replace and calculate
F = μ m g
μ = F / mg
μ = 321 /45.2 9.8
μ = 0.725
I'm assuming the question is what is the robin's speed relative to to the ground...
Create an equation that describes its relative motion.
rVg = rVa + aVg
Substitute values.
rVg = 12 m/s [N] + 6.8 m/s [E]
Use vector addition.
| rVg | = √ | rVa |² + | aVg |²
| rVg | = √ 144 m²/s² + 46.24 m²/s²
| rVg | = √ 19<u>0</u>.24 m²/s²
| rVg | = 1<u>3</u>.78 m/s
Find direction.
tanФ = aVg / rVa
tanФ = 6.8 m/s / 12 m/s
Ф = 29°
Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]
