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

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: