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AleksandrR
2 months ago
5

Amelia flies her airplane through calm skies at a velocity v1. The direction of v1 is 15 degrees north of east, and the speed is

180 km/hr.
Eventually, however, she enters a windy part of the atmosphere and finds that her plane now moves at a velocity v2. The direction of v2 is due east, and the speed is 150 km/hr.

What is the speed of the wind?

In what direction is the wind blowing?
(between 0 and 360 degrees)

Mathematics
1 answer:
lawyer [12.5K]2 months ago
5 0

Answer:

Step-by-step explanation:

Initially, the plane is traveling at a velocity of 180\ km/hr

toward the 15^{\circ} north of east

Then, the wind began to blow, causing the plane to move eastward at a speed of 150\ km/hr

Assuming 1v_o is the wind's speed.

Thus,

\vec{v_2}=\vec{v_1}-\vec{v_o}

150\hat{i}=180[\cos 15\hat{i}+\sin 15\hat{j}]-\vec{v_o}

\vec{v_o}=\hat{i}[180\cos 15-150]+\hat{j}[180\sin 15]

\vec{v_o}=\hat{i}[173.866-150]+46.58\hat{j}

\vec{v_o}=23.86\hat{i}+46.58\hat{j}

The magnitude of the wind is

\mid v_o\mid=\sqrt{23.86^2+46.58^2}

\mid v_o\mid=\sqrt{2738.996}

\mid v_o\mid=52.33\ km/hr

The direction is \tan \theta=\frac{46.58}{23.86}

\theta =62.87^{\circ} north of east

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Answer: (97.98, 112.020)

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Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

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Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

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To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

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<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

</pfor>
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We have

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\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1

The equation for the tangent plane at the point \left(x_0,y_0,z_0\right)

\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}=1  (Given)

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