The magnitude force a charge of the 16nC particle of 0.40
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:
Data from a cross-sectional study or survey might need to incorporate weights or design effects in the analysis.The analysis plan should specify which variables are most
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