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2 years ago

10

The apparent height of a building 10.5 km away is 0.02 radians. What is the approximate height of the building to the nearest me

ter

1 answer:

Ksenya-84 [330]2 years ago

8 0

Answer:

Approximate height of the building is 23213 meters.

Explanation:

Let the height of the building be represented by h.

0.02 radians = 0.02 × $\frac{180^{o} }{\pi }$

= 0.02 x (180/ $\frac{22}{7}$ )

0.02 radians = 1.146°

10.5 km = 10500 m

Applying the trigonometric function, we have;

Tan θ = $\frac{opposite}{adjacent}$

So that,

Tan 1.146° = $\frac{h}{10500}$

⇒ h = Tan 1.146° x 10500

= 2.21074 x 10500

= 23212.77

h = 23213 m

The approximate height of the building is 23213 m.

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$\vec p = m\cdot \vec v$ (1)

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$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

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$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

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$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

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$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

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$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

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$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

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