Answer:
1) Dimensions of shear rate is ![[T^{-1}]](https://tex.z-dn.net/?f=%5BT%5E%7B-1%7D%5D)
.
2)Dimensions of shear stress are ![[ML^{-1}T^{-2}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D)
Explanation:
Since the dimensions of velocity 'v' are ![[LT^{-1}]](https://tex.z-dn.net/?f=%5BLT%5E%7B-1%7D%5D)
and the dimensions of distance 'y' are ![[L]](https://tex.z-dn.net/?f=%5BL%5D)
, thus the dimensions of 
become
![\frac{[LT^{-1}]}{[L]}=[T^{-1}]](https://tex.z-dn.net/?f=%5Cfrac%7B%5BLT%5E%7B-1%7D%5D%7D%7B%5BL%5D%7D%3D%5BT%5E%7B-1%7D%5D)
and hence the units become 
.
Now we know that the dimensions of coefficient of dynamic viscosity 
are ![[ML^{-1}T^{-1}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D)
thus the dimensions of shear stress can be obtained from the given formula as
![[\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}]](https://tex.z-dn.net/?f=%5B%5Ctau%20%5D%3D%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D%5Ctimes%20%5BT%5E%7B-1%7D%5D%5C%5C%5C%5C%5B%5Ctau%20%5D%3D%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D)
Now we know that dimensions of momentum are ![[MLT^{-1}]](https://tex.z-dn.net/?f=%5BMLT%5E%7B-1%7D%5D)
The dimensions of 
are ![[L^{2}T]](https://tex.z-dn.net/?f=%5BL%5E%7B2%7DT%5D)
Thus the dimensions of ![\frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]](https://tex.z-dn.net/?f=%5Cfrac%7BMoumentum%7D%7BArea%5Ctimes%20time%7D%3D%5Cfrac%7BMLT%5E%7B-1%7D%7D%7BL%5E%7B2%7DT%7D%3D%5BMLT%5E%7B-2%7D%5D)
Which is same as that of shear stress. Hence proved.
Answer:
B (exponential growth )
Explanation: wish u the best bby <33
Answer:
Explanation:
We can solve Von Karman momentum integral equation as seen below using following in the attached file
