I really don’t know maybe energy from the poles just attaches and the soil comes and energy lifts the matter of soil and the magnet follows and lifts and goes up to the soil and the poles attach the magnet to the soil and the energy lifts the magnet and the soil follows and it is attracted to the magnet
The answer to this question is false.
Answer: The 6 kg rock sitting on a 3.2 m cliff.
Explanation:
The potential energy of an object of mass M that is at a height H above the ground us:
U = M*H*g
where g is the gravitational acceleration:
g = 9.8m/s^2
Then:
"An 8 kg rock sitting on a 2.2 m cliff"
M = 8kg
H = 2.2m
U = 8kg*2.2m*9.8 m/s^2 = 172.48 J
"a 6 kg rock sitting on a 3.2 m cliff"
M = 6kg
H = 3.2m
U = 6kg*3.2m*9.8m/s^2 = 188.16 J
You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.
Answer:
Magnesium and Bromine
Explanation:
I just took the test, and Magnesium has 7 electrons and Bromine has 2 valance electrons making the transfer a lot easier. In the first choice, Krypton already has 8 valance electrons therefore it cannot transfer or accept any more which rules it out as a possible answer. Calcium has 2 valance electrons and Potassium has 1 meaning it couldn't make a full shell of 8 and cannot make a ionic bond. Iodine has 7 electrons as well as Chlorine which wouldn't be the answer because it would have more than 8 valance electrons.
Answer:
The kinetic energy is 
Explanation:
From the question we are told that
The radius of the orbit is 
The gravitational force is 
The kinetic energy of the satellite is mathematically represented as

where v is the speed of the satellite which is mathematically represented as

=> 
substituting this into the equation

Now the gravitational force of the planet is mathematically represented as

Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
![KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A%5B%5Cfrac%7BGMm%7D%7Br%5E2%7D%5D%20%2A%20r)
=> 
substituting values

