Lindsay should fly the plane in the direction [W 12.5° S] to get Hamilton.
Using Sine rule to solve this question
Sine rule => SinA/a = SinB/b = SinC/c = constant
The magnitude of wind is 50 with an angle of 60 degrees.
The magnitude of plane is 200 and the angle at which it should fly is unknown and should be θ.
One side is 50 km/hr at an angle of 60 degrees.
sin 60°/200 = sin θ / 50
50 × sin 60° = 200 × sin θ
√3/2 = 4 × sin θ
√3/8 = sin θ
sin θ = 0.2165
θ = sin⁻¹(0.2165)
θ = 12.5°
So Lindsay have to fly the plane in the direction of [W 12.5° S].
Learn more about Sine Rule here:
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Answer:
115 m/s, 414 km/hr
Explanation:
There are two forces acting on a skydiver: gravity and air resistance (drag). At terminal velocity, the two forces are equal and opposite.
∑F = ma
D − mg = 0
D = mg
Drag force is defined as:
D = ½ ρ v² C A
where ρ is the fluid density,
v is the velocity,
C is the drag coefficient,
and A is the cross sectional surface area.
Substituting and solving for v:
½ ρ v² C A = mg
v² = 2mg / (ρCA)
v = √(2mg / (ρCA))
We're given values for m and A, and we know the value of g. We need to look up ρ and C.
Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.
For a skydiver falling headfirst, C ≈ 0.7.
Substituting all values:
v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))
v = 115 m/s
v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)
v = 414 km/hr
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For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).
fringe = (delta t) / (λ/c)
We can find (delta t) with the equation:
delta t = [v^2(L1+L2)]/c^3
Derivation of this formula can be found in your physics text book. From here we find (delta t):
600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13
2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes
This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.
