Answer:
DL = 1.5*10^-4[m]
Explanation:
First we will determine the initial values of the problem, in this way we have:
F = 60000[N]
L = 4 [m]
A = 0.008 [m^2]
DL = distance of the beam compressed along its length [m]
With the following equation we can find DL
![\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m]](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%20%3D%20Y%2A%5Cfrac%7BDL%7D%7BL%7D%20%5C%5Cwhere%3A%5C%5CY%20%3D%20young%27s%20modulus%20%3D%202%2A10%5E%7B11%7D%20%5BPa%5D%5C%5CDL%3D%5Cfrac%7BF%2AL%7D%7BY%2AA%7D%20%5C%5CDL%3D%5Cfrac%7B60000%2A4%7D%7B2%2A10%5E%7B11%7D%20%2A0.008%7D%20%5C%5CDL%3D%201.5%2A10%5E%7B-4%7D%20%5Bm%5D)
Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.
![A=\frac{\pi}{4} *D^{2} \\D = \sqrt{\frac{A*4}{\pi} } \\D = \sqrt{\frac{0.008*4}{\\pi } \\\\D = 0.1[m]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2AD%5E%7B2%7D%20%5C%5CD%20%3D%20%5Csqrt%7B%5Cfrac%7BA%2A4%7D%7B%5Cpi%7D%20%7D%20%5C%5CD%20%3D%20%20%5Csqrt%7B%5Cfrac%7B0.008%2A4%7D%7B%5C%5Cpi%20%7D%20%5C%5C%5C%5CD%20%3D%200.1%5Bm%5D)
Answer:
weathering breaks down the rocks while erosion moves them away from its original growth
The potential energy= mass times gravity times height. However, we are missing height. Gravity is a constant that is 9.8 on Earth. We then solve for height by dividing 350 by 10 and then 9.8 to get about 3.5.
TLDR: 3.5
<h2>Hello!</h2>
The answer is: B. Kinetic energy
<h2>
Why?</h2>
Since the ball is falling, speed increases because the gravity acceleration is acting. When speed increases, the kinetic energy increases too, so the ball is gaining kinetic energy.
The gravity acceleration is equal to
, it means that when falling, the ball will increase it's speed 9.81m every second.
We can calculate the kinetic energy by using the following formula:

Where:

Have a nice day!
<h2 />
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:

where q is the charge of the proton,

, with

being the elementary charge, and

and

are the initial and final voltage.
Substituting, we get (in electronvolts):

and in Joule: