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nataly862011 [7]

3 years ago

12

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.A. How high would this material go on earth if it were ejected with the same speed as on Io?

1 answer:

Anton [14]3 years ago

8 0

Answer:

91.64 km

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

Explanation:

According to Newton Law of gravitation:

$g=\frac{Gm}{r^2}$

Where:

G is gravitational constant= $6.67*10^{-11} m^3/kg.s^2$

For Moon lo g is:

$g_M=\frac{6.67*10^{-11}*8.93*10^{22}}{(1821*10^3)^2m^2} \\g_M=1.7962 m/s^2$

According to law of conservation of energy

Initial Energy=Final Energy

$K.E_i+mgh_i=K.E_f+mgh_f$

$\frac{1}{2}m(v_0)^2+mgh_o= \frac{1}{2}m(v_f)^2+mgh_f\\At\ maximum\ height\ v_f=0\\\frac{1}{2}m(v_0)^2+0=mgh_f\\v_0=\sqrt{2gh_f}$

For Jupiter's moon Io:

Velocity is given by:

$v_0_M=\sqrt{2g_Mh_f_M}$

For Earth Velocity is given by:

$v_0_E=\sqrt{2g_Eh_f_E}$

Now:

$v_o_M=v_o_E$

$\sqrt{2g_Mh_f_M}=\sqrt{2g_Eh_f_E}\\h_f_E=\frac{g_Mh_f_M}{g_E}$

$g_E=9.8 m/s^2$

$g_m=1.7962 m/s^2, As\ Calculated\ above$

$h_f_E=\frac{1.7962*500*10^3m}{9.8} \\h_f_E=91642.85 m\\h_f_E=91.64Km$

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

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8 months ago

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Answer:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb$

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

$\int\limits^a_b$ γhxdy

we know that γ=62.5 lb/ $ft^{3}$

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

$m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}$

when substituting the x and y-values given on the graph, we get that the slope is:

$m=\frac{0-6}{7-0}=-\frac{6}{7}$

once we got this slope, we can substitute it in the point-slope form of the equation:

$y_{2}-y_{1}=m(x_{2}-x_{1})$

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we can now solve this equation for x, so we get that:

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with this last equation, we can substitute everything into our integral, so it will now look like this:

$\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy$

Now that it's all written in terms of y we can now simplify it, so we get:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy$

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

$62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}$

By using the fundamental theorem of calculus we get:

$62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]$

When solving we get:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb$

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