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mr Goodwill [35]
3 years ago
14

Earth’s orbital speed is approximately _____.

Physics
2 answers:
Aleks04 [339]3 years ago
5 0
The answer that best fits the blank provided above is 30 KILOMETERS PER SECOND. The orbital speed of the Earth as recorded is 30 kilometers per second or <span>107000 km/h or</span> 18½ miles per second. The orbital period is 365 days which makes one year. Hope this helps.
liraira [26]3 years ago
4 0

THE CORRECT ANSWER IS 107,000 KILOMETER PER HOUR

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When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f
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Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer (r), measured in kilometers, can be represented by a right triangle:

r = \sqrt{x^{2}+y^{2}} (1)

Where:

x - Horizontal distance between the rocket and the observer, measured in kilometers.

y - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (\theta), measured in sexagesimal degrees, is defined by the following trigonometric relation:

\tan \theta = \frac{y}{x} (2)

If we know that x = 5\,km, then the expression is:

\tan \theta = \frac{y}{5}

And the rate of change of this angle is determined by derivatives:

\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y

\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}

\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}

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\dot \theta - Rate of change of the angle of elevation, measured in sexagesimal degrees.

\dot y - Vertical speed of the rocket, measured in kilometers per hour.

If we know that y = 4\,km and \dot y = 400\,\frac{km}{h}, then the rate of change of the angle of elevation is:

\dot \theta = 48.780\,\frac{\circ}{s}

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

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Answer:

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Explanation:

I hope this helped

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