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:
B.KEEN EYESIGHT thats for predators !
Explanation:
Herbivores have multiple stomachs to process tough foods, coats to blend in so they don't get eaten, and tough teeth for chewing on the plants. THEY DO NOT HAVE PREY, THEY ARE THE PREY.
Answer:
The answer is "effective stress at point B is 7382 ksi
"
Explanation:
Calculating the value of Compressive Axial Stress:
![\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%20y%20%20%3D%5Cfrac%7BF%7D%7BA%7D%20%3D%20%5Cfrac%7B4%20F%7D%7B%28%20p%20d%20%5E2%20%29%7D%20%3D%20%5Cfrac%7B%284%20x%20%28%20-%2040000%20%5C%20lbf%29%29%7D%7B%5B%20p%20%5Ctimes%20%281%20%5C%20in%29%5E2%20%5D%7D%20%3D%20-%2050.9%20%5C%20ksi%20%5C%5C)
Calculating Shear Transverse:



![\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%27%20%3D%5B%20s%20y%5E2%20%2B3%28%20t%20%5Ctimes%20y%5E2%20%2B%20t%20yz%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C)
![= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi](https://tex.z-dn.net/?f=%3D%20%5B%20%28-50.9%29%5E2%20%2B3%28%2863.7%29%5E2%20%2B%280.17%29%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B%203%284057.69%29%2B0.0289%5D%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B12%2C173.07%2B0.0289%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D14763.9089%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%20%3D%207381.95445%20%5C%20ksi%5C%5C%5C%5C%20%3D%207382%20%5C%20ksi)
Answer:
The x component of the x axis is:

The y component of the y axis is:
The magnitude is given by:
The direction is given by: 
See the graphical magnitude and direction below.
Explanation:
The x component of the x axis is:

The y component of the y axis is:
The magnitude is given by:
The direction is given by: 