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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)
The answer is axial precession. Axial precession refers to
the very slow motion of the Earth’s axis, which almost requires twenty-six
thousand (26,000) years to complete a full rotation. This Axial Precession is
caused by the effects of gravitational pull from the Sun and the Moon towards
the Earth.
FE is iron CO is cobalt CU is copper K is potassium NI is nickle MN is magnemese