Question

if you answer I will mark u brainliest ​

Answer 1

Answer:According to me the answer will be 6.25 m/sec or 22.5 Km/hour

Explanation:As the ship moves toward east with velocity 30 km/h or 8.33m/sec .

and the wind blows to the equator from southeast to ship with an angle 60° with a velocity 15 km/ h or 4.167 m /sec.

the velocity component in the direction of the ship will be 4.167Cos60°

=2.083

Now relative velocity = 8.33 - 2.083 ≈ 8.25 m/sec

Answer 2

Explanation:

Horizontal component of wind velocity = v - v₀ cos ∅

Vertical component of wind velocity = v₀ sin∅

Wind speed relative to ship :

= √(v - (-v₀ cos∅))² + (v₀sin∅)²

= √v² + v₀² + 2vv₀cos∅

Given,

v₀ = 30 km/hr, v = 15 km/hr.

Relative speed:

= √900 + 225 + 2 * 30 + 15 * cos60°

= √1575

 = 39.7 km/hr

Now,

Angle that relative velocity makes:

= tan⁻¹(15 * sin 60°)/(30 + 15 cos 60°)

= tan⁻¹(15√3/2)/(30 + 1/2)

= tan⁻¹(15√3/2)/(61/2)

= √3/5

= 19.1°

Therefore,

Wind velocity = 39.7 km/hr and

angle = 19.1°.

Hope it helps!