Ajet moving at 500.0 km/h due east is in a region where the wind is moving at 120.0 km/hr in a direction 30.0 degrees north of east. what is the speed of the aircraft relative to the ground?

v₃ ≈ 607 km/h,  ∅ = 5.67° north of east

Explanation:

Given

v₁ = 500.0 km/h

v₂ = 120.0 km/h

α = 30º

v₃ = ?  (speed of the aircraft relative to the ground)

We can apply the Law of Cosines as follows

v₃² = v₁² + v₂² - 2*v₁*v₂*Cos (180º-α)

⇒  v₃² = 500² + 120² - 2*500*120*Cos (180º-30º)

⇒  v₃² = 250,000 + 14,400 - 103,923.0485

⇒  v₃² = 368,323.0485

⇒  v₃ = √(368,323.0485) = 606.896 Km/h

⇒  v₃ ≈ 607 Km/h

then we can get the direction as follows

v₂ / Sin ∅ = v₃ / Sin 30º

⇒  Sin ∅ = Sin 30º*(v₂/v₃) = 0.5*(120/607) = 0.0988

⇒  ∅ = Sin⁻¹(0.0988) = 5.67º

Finally we have

v₃ ≈ 607 km/h,  ∅ = 5.67° north of east

i think the answer is a and b

i wish you luck

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach