According to freud what part of the mind is concerned with morals and ethics

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

Mademuasel [1]

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

7 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! A IS NOT THE CORRECT ANSWER

patriot [66]

Answer: A

<u>Explanation:</u>

NOTES:

d = 650 meters

t = 10 seconds

**********************************

v = d/t

= 650 meters/10 seconds

= 65 meters/second

6 0

3 years ago

Read 2 more answers

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$