Answer:
The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
Explanation:
Given that,
Thickness of A = 8.0 m
Conductivity = 25.0 m/d
Thickness of B = 2.0 m
Conductivity = 142 m/d
Thickness of C = 34 m
Conductivity = 40 m/d
We need to calculate the horizontal conductivity
Using formula of horizontal conductivity
![K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}](https://tex.z-dn.net/?f=K_%7BH%7D%3D%5Cdfrac%7BH_%7BA%7DK_%7BA%7D%2BH_%7BA%7DK_%7BA%7D%2BH_%7BA%7DK_%7BA%7D%7D%7BH_%7BA%7D%2BH_%7BB%7D%2BH_%7BC%7D%7D)
Put the value into the formula
![K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}](https://tex.z-dn.net/?f=K_%7BH%7D%3D%5Cdfrac%7B8.0%5Ctimes25%2B2%2C0%5Ctimes142%2B34%5Ctimes40%7D%7B8.0%2B2.0%2B34%7D)
![K_{H}=41.9\ m/d](https://tex.z-dn.net/?f=K_%7BH%7D%3D41.9%5C%20m%2Fd)
We need to calculate the vertical conductivity
Using formula of vertical conductivity
![K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}](https://tex.z-dn.net/?f=K_%7BV%7D%3D%5Cdfrac%7BH_%7BA%7D%2BH_%7BB%7D%2BH_%7BC%7D%7D%7B%5Cdfrac%7BH_%7BA%7D%7D%7BK_%7BA%7D%7D%2B%5Cdfrac%7BH_%7BB%7D%7D%7BK_%7BB%7D%7D%2B%5Cdfrac%7BH_%7BC%7D%7D%7BK_%7BC%7D%7D%7D)
Put the value into the formula
![K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}](https://tex.z-dn.net/?f=K_%7BV%7D%3D%5Cdfrac%7B8.0%2B2.0%2B34%7D%7B%5Cdfrac%7B8.0%7D%7B25%7D%2B%5Cdfrac%7B2.0%7D%7B142%7D%2B%5Cdfrac%7B34%7D%7B40%7D%7D)
![K_{V}=37.2\ m/d](https://tex.z-dn.net/?f=K_%7BV%7D%3D37.2%5C%20m%2Fd)
Hence, The horizontal conductivity is 41.9 m/d.
The vertical conductivity is 37.2 m/d.
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
![$J=\frac{\pi}{32}d^4$](https://tex.z-dn.net/?f=%24J%3D%5Cfrac%7B%5Cpi%7D%7B32%7Dd%5E4%24)
![$J=\frac{\pi}{32}\times (46)^4$](https://tex.z-dn.net/?f=%24J%3D%5Cfrac%7B%5Cpi%7D%7B32%7D%5Ctimes%20%2846%29%5E4%24)
J = 207.6 ![mm^4](https://tex.z-dn.net/?f=mm%5E4)
So the shear stress at point A is :
![$\tau_A =\frac{Tc_A}{J}$](https://tex.z-dn.net/?f=%24%5Ctau_A%20%3D%5Cfrac%7BTc_A%7D%7BJ%7D%24)
![$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$](https://tex.z-dn.net/?f=%24%5Ctau_A%20%3D%5Cfrac%7B85%20%5Ctimes%2010%5E3%5Ctimes%2012%20%7D%7B207.6%7D%24)
![$\tau_A = 4913.29 \ MPa$](https://tex.z-dn.net/?f=%24%5Ctau_A%20%3D%204913.29%20%5C%20MPa%24)
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Answer:
When the brakes are applied the in the typical double transverse wishbone front suspension, it "drives" the car ground due to the setting of the link-type system pivot points on the lower wishbone are have parallel alignment to the road
Explanation:
In order to minimize the car's reaction to the application of the brakes, the front and rear pivot are arranged with the lower wishbone's rear pivot made to be higher than the front pivot as such the inclined wishbone torque results in an opposing vertical force to the transferred extra weight from the back due to breaking.
The correct answer
would be d
Iron and carbon
hope this helps