Ask question

Ask question

All categories

2 years ago

5

An airplane pilot sets a compass course due west and maintains a speed of 220 km/h. after flying for a half hour she finds herse

lf over a town 120 km west and 20 km south of her starting point. (a) find the wind velocity (magnitude and direction). (b) if the wind velocity is 40 km/h due south, in what direction should the pilot set her course to travel due west? use the same airspeed of 220 km/h.

2 answers:

aivan3 [116]2 years ago

6 0

Answer:

$v_w = 44.7 km/h$

$\theta = 63.4 degree$

Explanation:

As we know that the velocity of the plane is

$v_1 = 220 km/h$ towards west

after half an hour the displacement of the plane in west direction is given as

$x = 120 km$

towards south the displacement is given as

$y = 20 km$

now we know that net velocity towards west

$v_p + v_w = \frac{120}{0.5}$

$220 + v_w = 240$

$v_w = 20 km/h$

Similarly towards south we have

$v_w = \frac{20}{0.5}$

$v_w = 40 km/h$

So the net speed of the wind is given as

$v_w = 20 \hat i + 40\hat j$

$v_w= \sqrt{20^2 + 40^2}$

$v_w = 44.7 km/h$

direction of wind is given as

$tan\theta = \frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{40}{20}$

$\theta = 63.4 degree$

timurjin [86]2 years ago

5 0

The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.

so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.

airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.

the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h

the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.

if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.

so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.

the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.

You might be interested in

How does a mirage occur in a deser?explain with a labelled diagram.

Ronch [10]

Answer:

Mirages happen when the ground is very hot and the air is cool.

Explanation:

They happen when light passes through two layers of air with different temperatures. The desert sun heats the sand, which in turn heats the air just above it. The hot air bends light rays and reflects the sky.

When you see it from a distance, the different air masses colliding with each other act as a mirror.

8 0

2 years ago

Read 2 more answers

77julia77 [94]

Answer:

What is the best description for the volume of air volume of air provided in a high quality rescue breath?

Explanation:

A) only Enough air to create a visible rise of the chest.

B) Until you can no longer force air in.

C) plenty of air make sure it is adequate to sustain life

D) clear and obvious rise of the chest, sustained over a few seconds

5 0

2 years ago

Assessing how well one variable predicts another variable is to blank as detecting cause-effect relationships between different

Reptile [31]

Are you asking about independent and dependent variables?

7 0

3 years ago

Read 2 more answers

At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is per

geniusboy [140]

Answer:

$\tau = 7.63 Nm$

Explanation:

As we know that moment of force is given as

$\tau = \vec r \times \vec F$

now we have

$\vec r = 1.2 m$

$\vec F = 14 N$

now from above formula we have

$\tau = r F sin\theta$

here we know that

$\theta = 27 degree$

so we have

$\tau = (1.2)(14) sin27$

$\tau = 7.63 Nm$

3 0

3 years ago

A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are

Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, $\Sigma F_y$ = 0

The vertical component of the tension, $T_y$ , in each half of the rope is given as follows;

$T_y$ = T × sin(θ)

∴ $\Sigma F_y$ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0

2 years ago