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3 years ago

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Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

1.55 × 105 m. Suppose further that a second group of engineers programmed the orbiter to go to 1.55 × 105 ft. What was the difference in kilometers between the two altitudes? How low did the probe go?

1 answer:

MAXImum [283]3 years ago

8 0

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

$h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m$

or in kilometers,

$h_2 = 47 km$

the first altitude in kilometers is

$h_1 = 155 km$

so the difference between the two altitudes is

$\Delta h = 155 km - 47 km = 108 km$

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Explanation:

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Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

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Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

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With this, we can calculate the density of gasoline at $21.7\°C$ :

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$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

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