1/8=.125, so 12 and a half % are blue
Answer:
5 x 4 + 2 - 3 / 1
Step-by-step explanation:
This took me so much tries to do but got it. Hope it helps! :)
Step-by-step explanation:
Consider an engineering material of initial length Lo, Area (A), Modulus of elasticity (E) and applied a force P due to which change in the length of the material is δ2 from it’s original length (Lo)
Initial length of the material is Lo. Hence, at time t = 0 when no force applied on the material the length of the material will not change (i.e., at time t=0, δ1 = 0)
Modulus of elasticity of the material:
![E=\frac{P \cdot L_{o}}{A\left[\delta_{2}-\delta_{1}\right]}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BP%20%5Ccdot%20L_%7Bo%7D%7D%7BA%5Cleft%5B%5Cdelta_%7B2%7D-%5Cdelta_%7B1%7D%5Cright%5D%7D)
Area of the material:
![E=\frac{P \cdot L_{o}}{A\left[\delta_{2}-\delta_{1}\right]}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BP%20%5Ccdot%20L_%7Bo%7D%7D%7BA%5Cleft%5B%5Cdelta_%7B2%7D-%5Cdelta_%7B1%7D%5Cright%5D%7D)
![A=\frac{P \cdot L_{o}}{E\left[\delta_{2}-\delta_{1}\right]}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BP%20%5Ccdot%20L_%7Bo%7D%7D%7BE%5Cleft%5B%5Cdelta_%7B2%7D-%5Cdelta_%7B1%7D%5Cright%5D%7D)
Length of the material:
![E=\frac{P \cdot L_{0}}{A\left[\delta_{2}-\delta_{1}\right]}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BP%20%5Ccdot%20L_%7B0%7D%7D%7BA%5Cleft%5B%5Cdelta_%7B2%7D-%5Cdelta_%7B1%7D%5Cright%5D%7D)
![L_{0}=\frac{E \cdot A\left[\delta_{2}-\delta_{1}\right]}{P}](https://tex.z-dn.net/?f=L_%7B0%7D%3D%5Cfrac%7BE%20%5Ccdot%20A%5Cleft%5B%5Cdelta_%7B2%7D-%5Cdelta_%7B1%7D%5Cright%5D%7D%7BP%7D)