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If a true heading of 135° results in a ground track of 130° and a true airspeed of 135 knots results in a groundspeed of 140 kno

vladimir2022 [97]

Answer:

245.1° and 13 knots

Explanation:

The given parameters are;

The true heading = 135°

The resultant ground track = 130°

The true airspeed = 135 knots

The ground speed = 140 knots

Given that the true airspeed the ground speed and the wind direction and magnitude form a triangle, we have;

From cosine rule, we have;

a² = b² + c² - 2×b×c×cos(A)

Where

a = The magnitude of the wind speed in knot

b = The true airspeed = 135 knots

c = The ground speed = 140 knots

A = The angle in between the true heading and the resultant ground track heading = 5°

Which gives;

a² = 135² + 140² - 2×135×140×cos(5 degrees) = 168.84 knots²

a = √168.84 = 12.9934 ≈ 13 knots

We have;

135 × sin(135 degrees) - 140× sin(130 degrees) = -11.7868

135 × cos(135 degrees) - 140× cos(130 degrees) = -5.469

Tan(θ) = -11.8/-5.5 = 2.155

θ = tan⁻¹(2.155) = 65.108°

Given that the wind is moving in opposite direction (slowing down the airplane, we add 180°, to get

Therefore, the angle direction = 180 + 65.108 = 245.1

Therefore, we have;

245.1° and 13 knots

3 0

3 years ago

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

MrMuchimi

Answer: $37^{\circ}$ North of west

Explanation:

Given

40,000-ton luxury line traveling 20 knots towards west and

60,000 ton freighter traveling towards North with 10 knots

suppose v is the common velocity after collision

conserving momentum in west direction

$20\times 40,000=(20,000+60,000)v\cos \theta$

suppose the final velocity makes \theta angle with x axis

$v\cos \theta =10----1$

Conserving Momentum in North direction

$10\times 60,000=80,000\times v\sin\theta$

$v\sin \theta =\frac{15}{2}----2$

divide 1 and 2

$\tan \theta =\frac{3}{4}$

$\theta =37^{\circ}$

so search in the area $37^{\circ}$ North of west to find the ship

3 0

3 years ago

Which way does a river flow??

spin [16.1K]

Downwards - from uphill towards the lowlands and eventually into the sea.

8 0

3 years ago

1. A cyclist accelerates from 0 m/s to 9 m/s in 3 seconds. What is his<br>acceleration?

Lapatulllka [165]

The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².

Answer:

Explanation:

Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.

Then acceleration = change in velocity/Time.

$Acceleration = \frac{Change in velocity}{Time taken}$