A force of 10N is making an angle of 30<br>with the horizontal. Its vertical component<br>will be;

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

Anton [14]

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.

8 0

3 years ago

Page 423 in the 8th grade science fusion textbook what at the answers

balu736 [363]

Go to slader.com and type in the full name of the text book and page number. It should have the andwers

3 0

2 years ago

Why pressure above the surface is greater than in the air

nexus9112 [7]

Answer:

At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.

Explanation:

7 0

3 years ago

A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.

Kobotan [32]

Explanation:

The given data is as follows.

F = $5.75 \times 10^{-16} N$

q = $1.6 \times 10^{-19} C$

v = 385 m/s

$sin (63.9^{o})$ = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

B = $\frac{F}{qv sin (\theta)}$

= $\frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}$

= $0.01065 \times 10^{3}$ T

= 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0

3 years ago

Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41

alexdok [17]

The Professor's centripetal acceleration is 0.044 m/s²

Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.

It is given by:

a = v²/r

where v is the velocity and r is the radius.

Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 = 0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

Find out more at: brainly.com/question/6082363

3 0

2 years ago

A force of 10N is making an angle of 30with the horizontal. Its vertical componentwill be;​

A force of 10N is making an angle of 30with the horizontal. Its vertical componentwill be;