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1 day ago

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An airplane cruises at 850 km/h relative to the air. It is flying from Denver, Colorado, due west to Reno, Nevada, a distance of

1200 km, and will then return. There is a steady 90km/h wind blowing to the east.

1 answer:

Yuliya22 [2.4K]1 day ago

8 0

Answer:

the difference in flight duration= 0.3023 hours

Explanation:

The question details are lacking, but I located it within your textbook.

Aircraft speed = 850 km/h

Wind speed against the aircraft = 90 km/h

As a result, the effective speed flying west = 850 - 90 = 760 km/hr

Distance traveled = 1200 km

time required = distance/speed = 1200/760 = 1.5789 hours

Returning, wind speed complements aircraft speed, thus

net speed for return journey = 850 + 90 = 940 km/hr

time taken for return = 1200/940 = 1.2765 hours

Consequently, the difference in flight duration= 1.5789 - 1.2765 = 0.3023 hours

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resulting in an overall improved speed

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The total journey lasts 13 hours

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Consequently, the total flight time is 13 - 2 = 11 hours

Now we apply the formula to calculate the time for traveling to Halifax

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$T = t_1 + t_2$

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d = 2021.6 km

Explanation:

This distance problem can be solved using vector analysis; it's best to find each plane's position components before applying the Pythagorean theorem to calculate the separation between them.

For Airplane 1:

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

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x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m = 7607 m

For Plane 2:

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 sin 25 = 8.452 103 m = 8452 m

To determine the distance between the planes using the Pythagorean theorem:

d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Now, we perform the calculations:

d² = (18126-16314)² + (1100-800)² + (8452-7607)²

d² = 3,283 106 + 9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √(408.7 10⁴)

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14 days ago

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Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

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Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

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$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

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In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

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