I believe Box B will have a greater gravitational pull because the gravitational pull of an object depends on its mass. The more mass an object has, the greater its gravitational pull will become.
For example, we can take planets. Naturally, they are round because once upon a time there was a larger piece of rock that attracted others. But the size of the rock won't matter, it's the weight that matters. If the rock weighed nothing, the other rocks would just rebound upon contact. But if the rock weighed a lot, then things wouldn't so easily rebound and might actually stick to it.
Answer:
true
Explanation:
The statement being made is completely true. This layer of rock is called a Sedimentary Rock level and is slowly formed over millions of years with minerals and organic remains from the bottom of the Oceans that may no longer be covered in water anymore. Since it is made up of all these minerals and remains, it is studied widely by Geologists and Archeologists to better understand the Earth's past.
It's very hard to see the self-portrait, so I can't identify him.
Answer:
IT doesnt push down because the lawnmower has wheels and the ground is hard
Explanation:
Branliest
Answer:
Explanation:
<u>LC Circuit</u>
It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:
= charge of the capacitor in any time 
= initial charge of the capacitor
=angular frequency of the circuit
= current through the circuit in any time
The charge in an LC circuit is given by
The current is the derivative of the charge
We are given
It means that
![q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]](https://tex.z-dn.net/?f=q%28t_1%29%20%3D%20q_0%20%5C%2C%20cos%20%28%5Comega%20t_1%20%29%3Dq_1%5C%20.......%5Beq%201%5D)
![i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]](https://tex.z-dn.net/?f=i%28t_1%29%20%3D%20-%20%5Comega%20q_0%20%5C%2C%20sin%28%5Comega%20t_1%29%3Di_1.........%5Beq%202%5D)
From eq 1:
From eq 2:
Squaring and adding the last two equations, and knowing that
Operating
Solving for
Now we know the value of 
, we repeat the procedure of eq 1 and eq 2, but now at the second time 
, and solve for 
Solving for
Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.
Finally
