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antiseptic1488 [7]
3 years ago
15

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 8.93×1022kg 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?
Physics
1 answer:
alisha [4.7K]3 years ago
6 0

Answer:

91.54 km

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Io = 1821000 m

m = Mass of the Io =  8.93 × 10²² kg

H_i = 500000 m

g = Acceleration due to gravity on Earth = 9.81 m/s²

Acceleration due to gravity on Io

g_i=\frac{GM}{r^2}\\\Rightarrow g_i=\frac{6.67\times 10^{-11}\times 8.93\times 10^{22}}{1821000^2}\\\Rightarrow g_i=1.796\ m/s^2

H_e=\frac{v^2}{2g}

H_i=\frac{v^2}{2g_l}

Dividing the above equations we get

\frac{H_e}{H_i}=\frac{\frac{v^2}{2g}}{\frac{v^2}{2g_i}}\\\Rightarrow \frac{H_e}{H_i}=\frac{g_i}{g}\\\Rightarrow \frac{H_e}{H_i}=\frac{1.796}{9.81}\\\Rightarrow H_e=H_i\times \frac{1.796}{9.81}\\\Rightarrow H_e=500000\times \frac{1.796}{9.81}\\\Rightarrow H_e=91539.245\ m=91.54\ km

The material would go 91.54 km on earth if it were ejected with the same speed as on Io.

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goldenfox [79]

Answer:

  • No
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The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

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Here, the net force needs to be ...

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Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

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  211.1 = 80.5v²/11.5

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The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

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3 years ago
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Answer:

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Explanation:

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B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B  \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 +  sin 45  \right )\\\\B = 3.4\times 10^{-5} T

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Answer:

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Answer:

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