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Yuliya22 [10]
3 years ago
11

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

l as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×1022 kg and a radius of 1821 km . For this calculation, ignore any variation in gravity over the 500-km range of the debris.How high would this material go on earth if it were ejected with the same speed as on Io?(Express your answer to three significant figures and include the appropriate units.)
Physics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

91.5 km

Explanation:

Hi!

If we are to ignore the variation in gravity we can use the formula for teh potential energy near the surface of a planet:

mgh

If the energy of the material ejected from the volcano on Io's surface is the same on earth's surface we have:

mg_{Io}h_{Io} =mg_{e}h_{e}

where subindexes Io and e means Jupiter's moon Io, and Earth, respectively

solving for h_e

h_{e} = h_{Io} \frac{g_{e}}{g_{Io}}

The acceleration due to the gravity of a planet can be calculated as:

g = G \frac{m}{R^2}

Where R and m are the radius  and mass of the planet

Therefore:

h_{e} = h_{Io} (\frac{m_{Io}}{m_e})(\frac{R_e}{R_{Io}})^2

m_e = 5,972 × 10^24 kg

R_e = 6 371 km

Replacing all given values:

h_e = 500 km *(0.183) = 91.515 990 km

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Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}

Where x_i is the value of each measurement, n is the number of elements in the set, and \mu is the average or media of the values, defined as

\displaystyle \mu=\frac{\sum x_i}{n}

Let's analyze each set of data:

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Computing the stardard deviation:

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The average is

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Computing the stardard deviation:

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Computing the stardard deviation:

\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}

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The average is

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Computing the stardard deviation:

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\sigma=0.8

We can see the data set marked as B has the largest standard deviation

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