The answer is C. Filling a table with the changed and measured data values
Answer:
The axial force is ![P = 15.93 k N](https://tex.z-dn.net/?f=P%20%3D%20%2015.93%20k%20N)
Explanation:
From the question we are told that
The diameter of the shaft steel is ![d = 50mm](https://tex.z-dn.net/?f=d%20%3D%20%2050mm)
The length of the cylindrical bushing ![L =100mm](https://tex.z-dn.net/?f=L%20%3D100mm)
The outer diameter of the cylindrical bushing is ![D = 70 \ mm](https://tex.z-dn.net/?f=D%20%3D%20%2070%20%5C%20mm)
The diametral interference is ![\delta _d = 0.005 mm](https://tex.z-dn.net/?f=%5Cdelta%20_d%20%3D%200.005%20mm)
The coefficient of friction is ![\mu = 0.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.2)
The Young modulus of steel is ![207 *10^{3} MPa (N/mm^2)](https://tex.z-dn.net/?f=207%20%2A10%5E%7B3%7D%20MPa%20%28N%2Fmm%5E2%29)
The diametral interference is mathematically represented as
![\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}](https://tex.z-dn.net/?f=%5Cdelta_d%20%3D%20%5Cfrac%7B2%20%2Ad%20%2A%20P_B%20%2A%20D%5E2%7D%7BE%20%28D%5E2%20-d%5E2%29%7D)
Where
is the pressure (stress) on the two object held together
So making
the subject
![P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}](https://tex.z-dn.net/?f=P_B%20%3D%20%5Cfrac%7B%5Cdelta%20_d%20E%20%28D%5E2%20-%20d%5E2%29%7D%7B2%20%2A%20d%2A%20D%5E2%7D)
Substituting values
![P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }](https://tex.z-dn.net/?f=P_B%20%3D%20%5Cfrac%7B%280.005%29%20%28207%20%2A10%5E%7B3%7D%20%29%20%2A%20%2870%5E2%20-%2050%5E2%29%29%7D%7B2%20%2A%20%2850%29%20%2870%29%20%5E2%20%7D)
![P_B = 5.069 MPa](https://tex.z-dn.net/?f=P_B%20%3D%205.069%20MPa)
Now he axial force required is
![P = \mu * P_B * A](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cmu%20%2A%20P_B%20%2A%20A)
Where A is the area which is mathematically evaluated as
![\pi d l](https://tex.z-dn.net/?f=%5Cpi%20d%20l)
So ![P = \mu P_B \pi d l](https://tex.z-dn.net/?f=P%20%20%3D%20%20%5Cmu%20P_B%20%5Cpi%20d%20l)
Substituting values
![P = 0.2 * 5.069 * 3.142 * 50 * 100](https://tex.z-dn.net/?f=P%20%3D%20%200.2%20%2A%205.069%20%2A%203.142%20%2A%2050%20%2A%20100)
![P = 15.93 k N](https://tex.z-dn.net/?f=P%20%3D%20%2015.93%20k%20N)
Answer:
Plane mirror is similar to a concave mirror
the car travels 34 mi in one hour.
then, in 6 hours car travels
34 x 6 mi
= 204 mi
Distance between the two cars is increasing at the rate of 85 mph.
A passenger in Car-1 says that he is at rest in his own frame of reference,
and Car-2 is moving away from him at 85 mph, toward the west.