Answer
(MG-24)
I pretty sure this the answer your looking for
Answer:
C. relatively large size
Explanation:
Earth's internal heat flow to the surface is 80% due to mantle convection while the remaining heat mostly originates in the Earth's crust and about 1% due to volcanic activity, earthquakes, and mountain building.
The earth's internal heat- stresses between the lithosphere and the upper mantle are responsible
for geological processes. These stresses are the results of motions in the upper mantle stemming from convection that push the lithosphere in different directions.
Larger planets cool slower than small planets.
Answer:
The law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as system's mass cannot change, so quantity can neither be added nor be removed.
Explanation:
Answer:
30.9 m
Explanation:
x = 129.9 m y = 30.9 m First of all, let's calculate the horizontal and vertical velocities involved h = 50.0cos(30) = 43.30127 m/s v = 50.0sin(30) = 25 m/s The horizontal distance is simply the horizontal velocity multiplied by the time, so 43.30127 m/s * 3 s = 129.9 m So the horizontal distance traveled is 129.9 m, so x = 129.9 m The vertical distance needs to take into account gravity which provides an acceleration of -9.8 m/s^2, so we get d = 25 m/s * 3s - 0.5*9.8 m/s^2 * (3 s)^2 d = 75 m - 4.9 m/s^2 * 9 s^2 d = 75 m - 44.1 m d = 30.9 m So the vertical distance traveled is 30.9 m, so y = 30.9 m
To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.
PART A)
Here,
M = Mass of Earth
R = Distance from center to the satellite
Replacing with our values we have,
PART B) The period of satellite is given as,
PART C) The gravitational force on the satellite is given by,