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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:

 F_Balance = 46.6 N    ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

          ∑ F = 0

          Fe –W + F_Balance = 0

         F_Balance = - Fe + W

           

The electric force is given by Coulomb's law

          Fe = k q₁ q₂ / r₂

The weight is

          W = mg

Let's replace

           F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

          q₁ = + 8 μC = +8 10⁻⁶ C

          q₂ = - 3 μC = - 3 10⁻⁶ C

          r = 0.3 m = 0.3 m

Let's calculate

         F_Balance = 5 9.8 - 8.99 10⁹  8 10⁻⁶ 3 10⁻⁶ / (0.3)²

         F_Balance = 49 - 2,397

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This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

          m' = F_Balance / g

          m' = 46.6 /9.8

          m' = 4,755 kg

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