Question

Самолёт летит из города А в город Б со скоростью 328км/ч относительно воздуха .На трассе полета со скоростью 72км/ч дует ветер,направление которого перпендикулярно отрезку соединяющему эти города определите модуль скорости самолета относительно земли

Answer 1

|v|=√(328²+72²)≈335.8  км/час
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Answer 2

Calculation of the Ground Speed of the Airplane

To determine the ground speed of the airplane, we need to consider the relative velocity of the airplane with respect to the air and the velocity of the wind.

Given: - The airplane is flying from city A to city B with a speed of 328 km/h relative to the air. - There is a wind blowing perpendicular to the line connecting the two cities. - The wind is blowing at a speed of 72 km/h.

To find the ground speed of the airplane, we need to calculate the vector sum of the airplane's velocity and the wind velocity.

Let's denote the ground speed of the airplane as Vg, the velocity of the airplane relative to the air as Va, and the velocity of the wind as Vw.

The magnitude of the ground speed of the airplane can be calculated using the Pythagorean theorem:

Vg = sqrt(Va^2 + Vw^2)

In this case, the velocity of the airplane relative to the air is 328 km/h, and the velocity of the wind is 72 km/h.

Substituting these values into the equation, we get:

Vg = sqrt((328 km/h)^2 + (72 km/h)^2)

Calculating this expression, we find that the ground speed of the airplane is approximately 337.49 km/h.

Therefore, the module (magnitude) of the velocity of the airplane relative to the ground is approximately 337.49 km/h.

Please note that the direction of the wind is perpendicular to the line connecting the two cities, but the question does not provide information about the direction of the wind. If you need further assistance or have any additional information, please let me know!